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integral inequality of length of curve


An integral inequalityHow to show the inequality is strict?Prove an integral inequality: $ left(int|f|^2dxright)^2le 4left(int|xf(x)|^2dxright)left(int|f'|^2dxright) $A tricky integral inequalityIntegral inequality problem (cauchy-schwarz)Need help on proving triangle inequality for norm.A particular proof of the mean value inequality for vector valued functionsAn integral inequality concerning compositionsIntegral inequality with a strange conditionAn inequality involving integrals and square root













3












$begingroup$


Let $f:mathbb{R}to mathbb{R}$ be a continuously differentiable function. Prove that for any $a.bin mathbb{R}$
$$left (int_a^bsqrt{1+(f'(x))^2},dxright)^2ge (a-b)^2+(f(b)-f(a))^2$$.





i think mean value theorem kills it but can't do it ...even try cauchy-schwarz inequality but nothing conclution










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    the smallest distance between the two points $(a, f(a))$ and $(b, f(b))$ is the straight line distance which is your RHS (the square root of that of course but same applies to LHS ; conclude...
    $endgroup$
    – Conrad
    3 hours ago












  • $begingroup$
    @Conrad But this is exactly what is to be proved, since the LHS is the definition of arc length.
    $endgroup$
    – Matematleta
    2 hours ago










  • $begingroup$
    This is classic stuff - can do it locally using Taylor approximation so make the curve piecewise linear and use elementary geometry or as done in various answers with various inequalities
    $endgroup$
    – Conrad
    44 secs ago
















3












$begingroup$


Let $f:mathbb{R}to mathbb{R}$ be a continuously differentiable function. Prove that for any $a.bin mathbb{R}$
$$left (int_a^bsqrt{1+(f'(x))^2},dxright)^2ge (a-b)^2+(f(b)-f(a))^2$$.





i think mean value theorem kills it but can't do it ...even try cauchy-schwarz inequality but nothing conclution










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    the smallest distance between the two points $(a, f(a))$ and $(b, f(b))$ is the straight line distance which is your RHS (the square root of that of course but same applies to LHS ; conclude...
    $endgroup$
    – Conrad
    3 hours ago












  • $begingroup$
    @Conrad But this is exactly what is to be proved, since the LHS is the definition of arc length.
    $endgroup$
    – Matematleta
    2 hours ago










  • $begingroup$
    This is classic stuff - can do it locally using Taylor approximation so make the curve piecewise linear and use elementary geometry or as done in various answers with various inequalities
    $endgroup$
    – Conrad
    44 secs ago














3












3








3


1



$begingroup$


Let $f:mathbb{R}to mathbb{R}$ be a continuously differentiable function. Prove that for any $a.bin mathbb{R}$
$$left (int_a^bsqrt{1+(f'(x))^2},dxright)^2ge (a-b)^2+(f(b)-f(a))^2$$.





i think mean value theorem kills it but can't do it ...even try cauchy-schwarz inequality but nothing conclution










share|cite|improve this question











$endgroup$




Let $f:mathbb{R}to mathbb{R}$ be a continuously differentiable function. Prove that for any $a.bin mathbb{R}$
$$left (int_a^bsqrt{1+(f'(x))^2},dxright)^2ge (a-b)^2+(f(b)-f(a))^2$$.





i think mean value theorem kills it but can't do it ...even try cauchy-schwarz inequality but nothing conclution







real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









uniquesolution

8,8161823




8,8161823










asked 4 hours ago









RAM_3RRAM_3R

593214




593214








  • 3




    $begingroup$
    the smallest distance between the two points $(a, f(a))$ and $(b, f(b))$ is the straight line distance which is your RHS (the square root of that of course but same applies to LHS ; conclude...
    $endgroup$
    – Conrad
    3 hours ago












  • $begingroup$
    @Conrad But this is exactly what is to be proved, since the LHS is the definition of arc length.
    $endgroup$
    – Matematleta
    2 hours ago










  • $begingroup$
    This is classic stuff - can do it locally using Taylor approximation so make the curve piecewise linear and use elementary geometry or as done in various answers with various inequalities
    $endgroup$
    – Conrad
    44 secs ago














  • 3




    $begingroup$
    the smallest distance between the two points $(a, f(a))$ and $(b, f(b))$ is the straight line distance which is your RHS (the square root of that of course but same applies to LHS ; conclude...
    $endgroup$
    – Conrad
    3 hours ago












  • $begingroup$
    @Conrad But this is exactly what is to be proved, since the LHS is the definition of arc length.
    $endgroup$
    – Matematleta
    2 hours ago










  • $begingroup$
    This is classic stuff - can do it locally using Taylor approximation so make the curve piecewise linear and use elementary geometry or as done in various answers with various inequalities
    $endgroup$
    – Conrad
    44 secs ago








3




3




$begingroup$
the smallest distance between the two points $(a, f(a))$ and $(b, f(b))$ is the straight line distance which is your RHS (the square root of that of course but same applies to LHS ; conclude...
$endgroup$
– Conrad
3 hours ago






$begingroup$
the smallest distance between the two points $(a, f(a))$ and $(b, f(b))$ is the straight line distance which is your RHS (the square root of that of course but same applies to LHS ; conclude...
$endgroup$
– Conrad
3 hours ago














$begingroup$
@Conrad But this is exactly what is to be proved, since the LHS is the definition of arc length.
$endgroup$
– Matematleta
2 hours ago




$begingroup$
@Conrad But this is exactly what is to be proved, since the LHS is the definition of arc length.
$endgroup$
– Matematleta
2 hours ago












$begingroup$
This is classic stuff - can do it locally using Taylor approximation so make the curve piecewise linear and use elementary geometry or as done in various answers with various inequalities
$endgroup$
– Conrad
44 secs ago




$begingroup$
This is classic stuff - can do it locally using Taylor approximation so make the curve piecewise linear and use elementary geometry or as done in various answers with various inequalities
$endgroup$
– Conrad
44 secs ago










4 Answers
4






active

oldest

votes


















2












$begingroup$

Notice that the function $y mapsto sqrt{1+y^2}$ is strictly convex. So by the Jensen's inequality,



$$ frac{1}{b-a} int_{a}^{b} sqrt{1 + f'(x)^2} , mathrm{d}x geq sqrt{1 + left(frac{1}{b-a}int_{a}^{b} f'(x) , mathrm{d}xright)^2} = sqrt{1 + left(frac{f(b) - f(a)}{b-a} right)^2}. $$



Multiplying both sides by $b-a$ and squaring proves the desired inequality. Moreover, by the strict convexity, the equality holds if and only if $f'$ is constant over $[a, b]$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    This really nice!
    $endgroup$
    – Nastar
    42 mins ago



















1












$begingroup$

An easy way to do this is to note that since distance is invariant under rotations, without loss of generality, we may assume that $f(a)=f(b).$ And now, since $sqrt{1-f'(x)}ge 0$ on $[a,b]$, the function in $C^1([a,b])$ that minimizes the integral coincides with the function $f$ that minimizes the integrand, and clearly, this happens when $f'(x)=0$ for all $xin [a,b].$ That is, when $f$ is constant on $[a,b].$ Then, $f(x)=f(a)$ and the result follows.



If you want to do this without the wlog assumption, then argue as follows:



Let $epsilon>0, fin C^1([a,b])$ and choose a partition $P={a,x_1,cdots,x_{n-2},b}$.



The length of the polygonal path obtained by joining the points



$(x_i,f(x_i))$ is $sum_i sqrt{(Delta x_i)^2+(Delta y_i)^2}$ and this is clearly $ge (b-a)^2+(f(b)-f(a))^2$. (You can make this precise by using an induction argument on $n$.)



And this is true for $textit{any}$ partition $P$.



But the above sum is also $sum_isqrt{1+frac{Delta y_i}{Delta x_i}}Delta x_i $ and now, upon applying the MVT, we see that what we have is a Riemann sum for $sqrt{1+f'(x)}$.



To finish, choose $P$ such that $left |int^b_asqrt{1+f'(x)}dx- sum_isqrt{1+f'(c_i)}Delta x_i right |<epsilon $. (The $c_i$ are the numbers $x_i<c_i<x_{i-1}$ obtained from the MVT). Then,



$(b-a)^2+(f(b)-f(a))^2le sum_isqrt{1+f'(c)}Delta x_i<int^b_asqrt{1+f'(x)}+epsilon.$



Since $epsilon$ is arbitrary, the result follows.



For a slick way to do this, use a variational argument: assuming a minimum $f$ exists, consider $f+tphi$ where $t$ is a real parameter and $phi$ is arbitrary $C^1([a,b])$.



Subsitute it into the integral:



$l(t)=int_a^b sqrt{1+(f'+tphi')^2}dx$.



Since $f$ minimizes this integral, the derivative of $l$ at $t=0$ must be equal to zero. Then,



$0=l'(0)= int_a^b dfrac{f'phi'}{sqrt{1+(f')^2}}dx$.



After an integration by parts, we get



$dfrac{f'}{sqrt{1+(f')^2}} = c$ for some constant $cin mathbb R,$ from which it follows that $f'=c$. And this means, of course, that the graph of $f$ is a straight line connecting $(a,f(a))$ and $(b,f(b)).$ The desired inequality follows.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Note that for every complex valued integrable function $phi :[a,b]to Bbb C$, it holds that
    $$
    left|int_a^b phi(x) dxright|le int_a^b|phi(x)| dx.
    $$
    Let $phi(x)=1+if'(x)$. Then we can see that
    $$begin{align*}
    left|int_a^b phi(x) dxright|&=left|(b-a)+i(f(b)-f(a))right|\&=sqrt{(b-a)^2+(f(b)-f(a))^2}
    end{align*}$$
    and
    $$
    int_a^b|phi(x)| dx=int_a^b sqrt{1+(f'(x))^2} dx.
    $$
    Now, the desired inequality follows.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      (+1) Amazing, this should the accepted answer! Anyway, is there any reason to work with $mathbb{C}$ rather than $mathbb{R}^2$ with $phi(x) = gamma'(x)$ and $gamma(x) = (x, f(x))$?
      $endgroup$
      – Sangchul Lee
      11 mins ago












    • $begingroup$
      Umm, sorry, I see no specific reason, since both are essentially the same version of the triangle inequality in integral form. But I just prefered $Bbb C$-version because it can be easily derived from the real triangle inequality; if $f$ is real-valued, integrable, $int_a^b fle int_a^b |f|$. Thank you!
      $endgroup$
      – Song
      1 min ago





















    0












    $begingroup$

    Expanding upon what @Conrad said, the shortest distance between two points is the distance of the line between, which is what your RHS is measuring (it is actually the square of the distance from $(a, f(a))$ to $(b, f(b))$.



    Now if we assume $left (int_a^bsqrt{1+(f'(x))^2},dxright)^2 < (a-b)^2+(f(b)-f(a))^2$, then we have contradicted the fact that that the shortest distance between $(a, f(a))$ and $(b, f(b))$ is $sqrt{(a-b)^2+(f(b)-f(a))^2}$. Therefore, it must be the case that $left (int_a^bsqrt{1+(f'(x))^2},dxright)^2 geq (a-b)^2+(f(b)-f(a))^2$






    share|cite|improve this answer








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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Notice that the function $y mapsto sqrt{1+y^2}$ is strictly convex. So by the Jensen's inequality,



      $$ frac{1}{b-a} int_{a}^{b} sqrt{1 + f'(x)^2} , mathrm{d}x geq sqrt{1 + left(frac{1}{b-a}int_{a}^{b} f'(x) , mathrm{d}xright)^2} = sqrt{1 + left(frac{f(b) - f(a)}{b-a} right)^2}. $$



      Multiplying both sides by $b-a$ and squaring proves the desired inequality. Moreover, by the strict convexity, the equality holds if and only if $f'$ is constant over $[a, b]$.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        This really nice!
        $endgroup$
        – Nastar
        42 mins ago
















      2












      $begingroup$

      Notice that the function $y mapsto sqrt{1+y^2}$ is strictly convex. So by the Jensen's inequality,



      $$ frac{1}{b-a} int_{a}^{b} sqrt{1 + f'(x)^2} , mathrm{d}x geq sqrt{1 + left(frac{1}{b-a}int_{a}^{b} f'(x) , mathrm{d}xright)^2} = sqrt{1 + left(frac{f(b) - f(a)}{b-a} right)^2}. $$



      Multiplying both sides by $b-a$ and squaring proves the desired inequality. Moreover, by the strict convexity, the equality holds if and only if $f'$ is constant over $[a, b]$.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        This really nice!
        $endgroup$
        – Nastar
        42 mins ago














      2












      2








      2





      $begingroup$

      Notice that the function $y mapsto sqrt{1+y^2}$ is strictly convex. So by the Jensen's inequality,



      $$ frac{1}{b-a} int_{a}^{b} sqrt{1 + f'(x)^2} , mathrm{d}x geq sqrt{1 + left(frac{1}{b-a}int_{a}^{b} f'(x) , mathrm{d}xright)^2} = sqrt{1 + left(frac{f(b) - f(a)}{b-a} right)^2}. $$



      Multiplying both sides by $b-a$ and squaring proves the desired inequality. Moreover, by the strict convexity, the equality holds if and only if $f'$ is constant over $[a, b]$.






      share|cite|improve this answer









      $endgroup$



      Notice that the function $y mapsto sqrt{1+y^2}$ is strictly convex. So by the Jensen's inequality,



      $$ frac{1}{b-a} int_{a}^{b} sqrt{1 + f'(x)^2} , mathrm{d}x geq sqrt{1 + left(frac{1}{b-a}int_{a}^{b} f'(x) , mathrm{d}xright)^2} = sqrt{1 + left(frac{f(b) - f(a)}{b-a} right)^2}. $$



      Multiplying both sides by $b-a$ and squaring proves the desired inequality. Moreover, by the strict convexity, the equality holds if and only if $f'$ is constant over $[a, b]$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 52 mins ago









      Sangchul LeeSangchul Lee

      95.1k12170277




      95.1k12170277








      • 1




        $begingroup$
        This really nice!
        $endgroup$
        – Nastar
        42 mins ago














      • 1




        $begingroup$
        This really nice!
        $endgroup$
        – Nastar
        42 mins ago








      1




      1




      $begingroup$
      This really nice!
      $endgroup$
      – Nastar
      42 mins ago




      $begingroup$
      This really nice!
      $endgroup$
      – Nastar
      42 mins ago











      1












      $begingroup$

      An easy way to do this is to note that since distance is invariant under rotations, without loss of generality, we may assume that $f(a)=f(b).$ And now, since $sqrt{1-f'(x)}ge 0$ on $[a,b]$, the function in $C^1([a,b])$ that minimizes the integral coincides with the function $f$ that minimizes the integrand, and clearly, this happens when $f'(x)=0$ for all $xin [a,b].$ That is, when $f$ is constant on $[a,b].$ Then, $f(x)=f(a)$ and the result follows.



      If you want to do this without the wlog assumption, then argue as follows:



      Let $epsilon>0, fin C^1([a,b])$ and choose a partition $P={a,x_1,cdots,x_{n-2},b}$.



      The length of the polygonal path obtained by joining the points



      $(x_i,f(x_i))$ is $sum_i sqrt{(Delta x_i)^2+(Delta y_i)^2}$ and this is clearly $ge (b-a)^2+(f(b)-f(a))^2$. (You can make this precise by using an induction argument on $n$.)



      And this is true for $textit{any}$ partition $P$.



      But the above sum is also $sum_isqrt{1+frac{Delta y_i}{Delta x_i}}Delta x_i $ and now, upon applying the MVT, we see that what we have is a Riemann sum for $sqrt{1+f'(x)}$.



      To finish, choose $P$ such that $left |int^b_asqrt{1+f'(x)}dx- sum_isqrt{1+f'(c_i)}Delta x_i right |<epsilon $. (The $c_i$ are the numbers $x_i<c_i<x_{i-1}$ obtained from the MVT). Then,



      $(b-a)^2+(f(b)-f(a))^2le sum_isqrt{1+f'(c)}Delta x_i<int^b_asqrt{1+f'(x)}+epsilon.$



      Since $epsilon$ is arbitrary, the result follows.



      For a slick way to do this, use a variational argument: assuming a minimum $f$ exists, consider $f+tphi$ where $t$ is a real parameter and $phi$ is arbitrary $C^1([a,b])$.



      Subsitute it into the integral:



      $l(t)=int_a^b sqrt{1+(f'+tphi')^2}dx$.



      Since $f$ minimizes this integral, the derivative of $l$ at $t=0$ must be equal to zero. Then,



      $0=l'(0)= int_a^b dfrac{f'phi'}{sqrt{1+(f')^2}}dx$.



      After an integration by parts, we get



      $dfrac{f'}{sqrt{1+(f')^2}} = c$ for some constant $cin mathbb R,$ from which it follows that $f'=c$. And this means, of course, that the graph of $f$ is a straight line connecting $(a,f(a))$ and $(b,f(b)).$ The desired inequality follows.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        An easy way to do this is to note that since distance is invariant under rotations, without loss of generality, we may assume that $f(a)=f(b).$ And now, since $sqrt{1-f'(x)}ge 0$ on $[a,b]$, the function in $C^1([a,b])$ that minimizes the integral coincides with the function $f$ that minimizes the integrand, and clearly, this happens when $f'(x)=0$ for all $xin [a,b].$ That is, when $f$ is constant on $[a,b].$ Then, $f(x)=f(a)$ and the result follows.



        If you want to do this without the wlog assumption, then argue as follows:



        Let $epsilon>0, fin C^1([a,b])$ and choose a partition $P={a,x_1,cdots,x_{n-2},b}$.



        The length of the polygonal path obtained by joining the points



        $(x_i,f(x_i))$ is $sum_i sqrt{(Delta x_i)^2+(Delta y_i)^2}$ and this is clearly $ge (b-a)^2+(f(b)-f(a))^2$. (You can make this precise by using an induction argument on $n$.)



        And this is true for $textit{any}$ partition $P$.



        But the above sum is also $sum_isqrt{1+frac{Delta y_i}{Delta x_i}}Delta x_i $ and now, upon applying the MVT, we see that what we have is a Riemann sum for $sqrt{1+f'(x)}$.



        To finish, choose $P$ such that $left |int^b_asqrt{1+f'(x)}dx- sum_isqrt{1+f'(c_i)}Delta x_i right |<epsilon $. (The $c_i$ are the numbers $x_i<c_i<x_{i-1}$ obtained from the MVT). Then,



        $(b-a)^2+(f(b)-f(a))^2le sum_isqrt{1+f'(c)}Delta x_i<int^b_asqrt{1+f'(x)}+epsilon.$



        Since $epsilon$ is arbitrary, the result follows.



        For a slick way to do this, use a variational argument: assuming a minimum $f$ exists, consider $f+tphi$ where $t$ is a real parameter and $phi$ is arbitrary $C^1([a,b])$.



        Subsitute it into the integral:



        $l(t)=int_a^b sqrt{1+(f'+tphi')^2}dx$.



        Since $f$ minimizes this integral, the derivative of $l$ at $t=0$ must be equal to zero. Then,



        $0=l'(0)= int_a^b dfrac{f'phi'}{sqrt{1+(f')^2}}dx$.



        After an integration by parts, we get



        $dfrac{f'}{sqrt{1+(f')^2}} = c$ for some constant $cin mathbb R,$ from which it follows that $f'=c$. And this means, of course, that the graph of $f$ is a straight line connecting $(a,f(a))$ and $(b,f(b)).$ The desired inequality follows.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          An easy way to do this is to note that since distance is invariant under rotations, without loss of generality, we may assume that $f(a)=f(b).$ And now, since $sqrt{1-f'(x)}ge 0$ on $[a,b]$, the function in $C^1([a,b])$ that minimizes the integral coincides with the function $f$ that minimizes the integrand, and clearly, this happens when $f'(x)=0$ for all $xin [a,b].$ That is, when $f$ is constant on $[a,b].$ Then, $f(x)=f(a)$ and the result follows.



          If you want to do this without the wlog assumption, then argue as follows:



          Let $epsilon>0, fin C^1([a,b])$ and choose a partition $P={a,x_1,cdots,x_{n-2},b}$.



          The length of the polygonal path obtained by joining the points



          $(x_i,f(x_i))$ is $sum_i sqrt{(Delta x_i)^2+(Delta y_i)^2}$ and this is clearly $ge (b-a)^2+(f(b)-f(a))^2$. (You can make this precise by using an induction argument on $n$.)



          And this is true for $textit{any}$ partition $P$.



          But the above sum is also $sum_isqrt{1+frac{Delta y_i}{Delta x_i}}Delta x_i $ and now, upon applying the MVT, we see that what we have is a Riemann sum for $sqrt{1+f'(x)}$.



          To finish, choose $P$ such that $left |int^b_asqrt{1+f'(x)}dx- sum_isqrt{1+f'(c_i)}Delta x_i right |<epsilon $. (The $c_i$ are the numbers $x_i<c_i<x_{i-1}$ obtained from the MVT). Then,



          $(b-a)^2+(f(b)-f(a))^2le sum_isqrt{1+f'(c)}Delta x_i<int^b_asqrt{1+f'(x)}+epsilon.$



          Since $epsilon$ is arbitrary, the result follows.



          For a slick way to do this, use a variational argument: assuming a minimum $f$ exists, consider $f+tphi$ where $t$ is a real parameter and $phi$ is arbitrary $C^1([a,b])$.



          Subsitute it into the integral:



          $l(t)=int_a^b sqrt{1+(f'+tphi')^2}dx$.



          Since $f$ minimizes this integral, the derivative of $l$ at $t=0$ must be equal to zero. Then,



          $0=l'(0)= int_a^b dfrac{f'phi'}{sqrt{1+(f')^2}}dx$.



          After an integration by parts, we get



          $dfrac{f'}{sqrt{1+(f')^2}} = c$ for some constant $cin mathbb R,$ from which it follows that $f'=c$. And this means, of course, that the graph of $f$ is a straight line connecting $(a,f(a))$ and $(b,f(b)).$ The desired inequality follows.






          share|cite|improve this answer











          $endgroup$



          An easy way to do this is to note that since distance is invariant under rotations, without loss of generality, we may assume that $f(a)=f(b).$ And now, since $sqrt{1-f'(x)}ge 0$ on $[a,b]$, the function in $C^1([a,b])$ that minimizes the integral coincides with the function $f$ that minimizes the integrand, and clearly, this happens when $f'(x)=0$ for all $xin [a,b].$ That is, when $f$ is constant on $[a,b].$ Then, $f(x)=f(a)$ and the result follows.



          If you want to do this without the wlog assumption, then argue as follows:



          Let $epsilon>0, fin C^1([a,b])$ and choose a partition $P={a,x_1,cdots,x_{n-2},b}$.



          The length of the polygonal path obtained by joining the points



          $(x_i,f(x_i))$ is $sum_i sqrt{(Delta x_i)^2+(Delta y_i)^2}$ and this is clearly $ge (b-a)^2+(f(b)-f(a))^2$. (You can make this precise by using an induction argument on $n$.)



          And this is true for $textit{any}$ partition $P$.



          But the above sum is also $sum_isqrt{1+frac{Delta y_i}{Delta x_i}}Delta x_i $ and now, upon applying the MVT, we see that what we have is a Riemann sum for $sqrt{1+f'(x)}$.



          To finish, choose $P$ such that $left |int^b_asqrt{1+f'(x)}dx- sum_isqrt{1+f'(c_i)}Delta x_i right |<epsilon $. (The $c_i$ are the numbers $x_i<c_i<x_{i-1}$ obtained from the MVT). Then,



          $(b-a)^2+(f(b)-f(a))^2le sum_isqrt{1+f'(c)}Delta x_i<int^b_asqrt{1+f'(x)}+epsilon.$



          Since $epsilon$ is arbitrary, the result follows.



          For a slick way to do this, use a variational argument: assuming a minimum $f$ exists, consider $f+tphi$ where $t$ is a real parameter and $phi$ is arbitrary $C^1([a,b])$.



          Subsitute it into the integral:



          $l(t)=int_a^b sqrt{1+(f'+tphi')^2}dx$.



          Since $f$ minimizes this integral, the derivative of $l$ at $t=0$ must be equal to zero. Then,



          $0=l'(0)= int_a^b dfrac{f'phi'}{sqrt{1+(f')^2}}dx$.



          After an integration by parts, we get



          $dfrac{f'}{sqrt{1+(f')^2}} = c$ for some constant $cin mathbb R,$ from which it follows that $f'=c$. And this means, of course, that the graph of $f$ is a straight line connecting $(a,f(a))$ and $(b,f(b)).$ The desired inequality follows.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 59 mins ago

























          answered 2 hours ago









          MatematletaMatematleta

          11.5k2920




          11.5k2920























              1












              $begingroup$

              Note that for every complex valued integrable function $phi :[a,b]to Bbb C$, it holds that
              $$
              left|int_a^b phi(x) dxright|le int_a^b|phi(x)| dx.
              $$
              Let $phi(x)=1+if'(x)$. Then we can see that
              $$begin{align*}
              left|int_a^b phi(x) dxright|&=left|(b-a)+i(f(b)-f(a))right|\&=sqrt{(b-a)^2+(f(b)-f(a))^2}
              end{align*}$$
              and
              $$
              int_a^b|phi(x)| dx=int_a^b sqrt{1+(f'(x))^2} dx.
              $$
              Now, the desired inequality follows.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                (+1) Amazing, this should the accepted answer! Anyway, is there any reason to work with $mathbb{C}$ rather than $mathbb{R}^2$ with $phi(x) = gamma'(x)$ and $gamma(x) = (x, f(x))$?
                $endgroup$
                – Sangchul Lee
                11 mins ago












              • $begingroup$
                Umm, sorry, I see no specific reason, since both are essentially the same version of the triangle inequality in integral form. But I just prefered $Bbb C$-version because it can be easily derived from the real triangle inequality; if $f$ is real-valued, integrable, $int_a^b fle int_a^b |f|$. Thank you!
                $endgroup$
                – Song
                1 min ago


















              1












              $begingroup$

              Note that for every complex valued integrable function $phi :[a,b]to Bbb C$, it holds that
              $$
              left|int_a^b phi(x) dxright|le int_a^b|phi(x)| dx.
              $$
              Let $phi(x)=1+if'(x)$. Then we can see that
              $$begin{align*}
              left|int_a^b phi(x) dxright|&=left|(b-a)+i(f(b)-f(a))right|\&=sqrt{(b-a)^2+(f(b)-f(a))^2}
              end{align*}$$
              and
              $$
              int_a^b|phi(x)| dx=int_a^b sqrt{1+(f'(x))^2} dx.
              $$
              Now, the desired inequality follows.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                (+1) Amazing, this should the accepted answer! Anyway, is there any reason to work with $mathbb{C}$ rather than $mathbb{R}^2$ with $phi(x) = gamma'(x)$ and $gamma(x) = (x, f(x))$?
                $endgroup$
                – Sangchul Lee
                11 mins ago












              • $begingroup$
                Umm, sorry, I see no specific reason, since both are essentially the same version of the triangle inequality in integral form. But I just prefered $Bbb C$-version because it can be easily derived from the real triangle inequality; if $f$ is real-valued, integrable, $int_a^b fle int_a^b |f|$. Thank you!
                $endgroup$
                – Song
                1 min ago
















              1












              1








              1





              $begingroup$

              Note that for every complex valued integrable function $phi :[a,b]to Bbb C$, it holds that
              $$
              left|int_a^b phi(x) dxright|le int_a^b|phi(x)| dx.
              $$
              Let $phi(x)=1+if'(x)$. Then we can see that
              $$begin{align*}
              left|int_a^b phi(x) dxright|&=left|(b-a)+i(f(b)-f(a))right|\&=sqrt{(b-a)^2+(f(b)-f(a))^2}
              end{align*}$$
              and
              $$
              int_a^b|phi(x)| dx=int_a^b sqrt{1+(f'(x))^2} dx.
              $$
              Now, the desired inequality follows.






              share|cite|improve this answer









              $endgroup$



              Note that for every complex valued integrable function $phi :[a,b]to Bbb C$, it holds that
              $$
              left|int_a^b phi(x) dxright|le int_a^b|phi(x)| dx.
              $$
              Let $phi(x)=1+if'(x)$. Then we can see that
              $$begin{align*}
              left|int_a^b phi(x) dxright|&=left|(b-a)+i(f(b)-f(a))right|\&=sqrt{(b-a)^2+(f(b)-f(a))^2}
              end{align*}$$
              and
              $$
              int_a^b|phi(x)| dx=int_a^b sqrt{1+(f'(x))^2} dx.
              $$
              Now, the desired inequality follows.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 14 mins ago









              SongSong

              16.1k1739




              16.1k1739












              • $begingroup$
                (+1) Amazing, this should the accepted answer! Anyway, is there any reason to work with $mathbb{C}$ rather than $mathbb{R}^2$ with $phi(x) = gamma'(x)$ and $gamma(x) = (x, f(x))$?
                $endgroup$
                – Sangchul Lee
                11 mins ago












              • $begingroup$
                Umm, sorry, I see no specific reason, since both are essentially the same version of the triangle inequality in integral form. But I just prefered $Bbb C$-version because it can be easily derived from the real triangle inequality; if $f$ is real-valued, integrable, $int_a^b fle int_a^b |f|$. Thank you!
                $endgroup$
                – Song
                1 min ago




















              • $begingroup$
                (+1) Amazing, this should the accepted answer! Anyway, is there any reason to work with $mathbb{C}$ rather than $mathbb{R}^2$ with $phi(x) = gamma'(x)$ and $gamma(x) = (x, f(x))$?
                $endgroup$
                – Sangchul Lee
                11 mins ago












              • $begingroup$
                Umm, sorry, I see no specific reason, since both are essentially the same version of the triangle inequality in integral form. But I just prefered $Bbb C$-version because it can be easily derived from the real triangle inequality; if $f$ is real-valued, integrable, $int_a^b fle int_a^b |f|$. Thank you!
                $endgroup$
                – Song
                1 min ago


















              $begingroup$
              (+1) Amazing, this should the accepted answer! Anyway, is there any reason to work with $mathbb{C}$ rather than $mathbb{R}^2$ with $phi(x) = gamma'(x)$ and $gamma(x) = (x, f(x))$?
              $endgroup$
              – Sangchul Lee
              11 mins ago






              $begingroup$
              (+1) Amazing, this should the accepted answer! Anyway, is there any reason to work with $mathbb{C}$ rather than $mathbb{R}^2$ with $phi(x) = gamma'(x)$ and $gamma(x) = (x, f(x))$?
              $endgroup$
              – Sangchul Lee
              11 mins ago














              $begingroup$
              Umm, sorry, I see no specific reason, since both are essentially the same version of the triangle inequality in integral form. But I just prefered $Bbb C$-version because it can be easily derived from the real triangle inequality; if $f$ is real-valued, integrable, $int_a^b fle int_a^b |f|$. Thank you!
              $endgroup$
              – Song
              1 min ago






              $begingroup$
              Umm, sorry, I see no specific reason, since both are essentially the same version of the triangle inequality in integral form. But I just prefered $Bbb C$-version because it can be easily derived from the real triangle inequality; if $f$ is real-valued, integrable, $int_a^b fle int_a^b |f|$. Thank you!
              $endgroup$
              – Song
              1 min ago













              0












              $begingroup$

              Expanding upon what @Conrad said, the shortest distance between two points is the distance of the line between, which is what your RHS is measuring (it is actually the square of the distance from $(a, f(a))$ to $(b, f(b))$.



              Now if we assume $left (int_a^bsqrt{1+(f'(x))^2},dxright)^2 < (a-b)^2+(f(b)-f(a))^2$, then we have contradicted the fact that that the shortest distance between $(a, f(a))$ and $(b, f(b))$ is $sqrt{(a-b)^2+(f(b)-f(a))^2}$. Therefore, it must be the case that $left (int_a^bsqrt{1+(f'(x))^2},dxright)^2 geq (a-b)^2+(f(b)-f(a))^2$






              share|cite|improve this answer








              New contributor




              se2018 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$


















                0












                $begingroup$

                Expanding upon what @Conrad said, the shortest distance between two points is the distance of the line between, which is what your RHS is measuring (it is actually the square of the distance from $(a, f(a))$ to $(b, f(b))$.



                Now if we assume $left (int_a^bsqrt{1+(f'(x))^2},dxright)^2 < (a-b)^2+(f(b)-f(a))^2$, then we have contradicted the fact that that the shortest distance between $(a, f(a))$ and $(b, f(b))$ is $sqrt{(a-b)^2+(f(b)-f(a))^2}$. Therefore, it must be the case that $left (int_a^bsqrt{1+(f'(x))^2},dxright)^2 geq (a-b)^2+(f(b)-f(a))^2$






                share|cite|improve this answer








                New contributor




                se2018 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Expanding upon what @Conrad said, the shortest distance between two points is the distance of the line between, which is what your RHS is measuring (it is actually the square of the distance from $(a, f(a))$ to $(b, f(b))$.



                  Now if we assume $left (int_a^bsqrt{1+(f'(x))^2},dxright)^2 < (a-b)^2+(f(b)-f(a))^2$, then we have contradicted the fact that that the shortest distance between $(a, f(a))$ and $(b, f(b))$ is $sqrt{(a-b)^2+(f(b)-f(a))^2}$. Therefore, it must be the case that $left (int_a^bsqrt{1+(f'(x))^2},dxright)^2 geq (a-b)^2+(f(b)-f(a))^2$






                  share|cite|improve this answer








                  New contributor




                  se2018 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  $endgroup$



                  Expanding upon what @Conrad said, the shortest distance between two points is the distance of the line between, which is what your RHS is measuring (it is actually the square of the distance from $(a, f(a))$ to $(b, f(b))$.



                  Now if we assume $left (int_a^bsqrt{1+(f'(x))^2},dxright)^2 < (a-b)^2+(f(b)-f(a))^2$, then we have contradicted the fact that that the shortest distance between $(a, f(a))$ and $(b, f(b))$ is $sqrt{(a-b)^2+(f(b)-f(a))^2}$. Therefore, it must be the case that $left (int_a^bsqrt{1+(f'(x))^2},dxright)^2 geq (a-b)^2+(f(b)-f(a))^2$







                  share|cite|improve this answer








                  New contributor




                  se2018 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  share|cite|improve this answer



                  share|cite|improve this answer






                  New contributor




                  se2018 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  answered 3 hours ago









                  se2018se2018

                  873




                  873




                  New contributor




                  se2018 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





                  New contributor





                  se2018 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  se2018 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






























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