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In my obsevation:

If $f(x)$ and $g(x)$ be two continuous funcions, and they have derivative and second derivative are also continuous in interval $[a, b]$. If $f(x)$ is more curvature than $g(x)$ in interval $[a, b]$ and $f(a)=g(a)$, $f(b)=g(b)$ then length of $f(x)$ from $a$ to $b$ seem longer than length of $g(x)$ from $a$ to $b$ ?

Example: Length of any curve (curve is not a line) is longer than a line in every interval.

This observation can fomulate that:

$$\frac{|f''(x)|}{(1+f'(x)^2)^{\frac{3}{2}}} \ge \frac{|g''(x)|}{(1+g'(x)^2)^{\frac{3}{2}}}$$ for all $x \in [a, b]$ and $f(a)=g(a)$, $f(b)=g(b)$

then $$\int_{a}^{b}{ \sqrt{1+f'(x)^2}}dx \ge \int_{a}^{b}{ \sqrt{1+g'(x)^2}}dx$$

My question: I think this is true and well-known, can you share me a reference. If this result is not known can you give a proof or a comment? Thank You very much.

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    $\begingroup$ For simplicity, assume that $f'' \ge 0$ and $g'' \ge 0$ to get rid of the modulus, integrate on $[a,b]$ and see what you get. Next try the general case. This should have been asked on [Mathematics], though. $\endgroup$
    – Alex M.
    Commented Jul 12, 2018 at 19:12

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The answer is no. Indeed, if this conjecture were true, then the conditions $f(a)=g(a)$, $f(b)=g(b)$, and \begin{equation} k_f(x):=\frac{|f''(x)|}{(1+f'(x)^2)^{3/2}}=k_g(x) \end{equation} for all $x \in [a, b]$ would imply that \begin{equation} \ell_f:=\int_a^b\sqrt{1+f'(x)^2}\,dx=\ell_g; \end{equation} that is, we would have that the length of the graph of a smooth enough function $f$ over $[a,b]$ would be determined by the curvature of the graph and the values of $f$ at $a,b$.

Let us show that this is not so. Indeed, let \begin{equation} h(x):=\frac12 \sqrt{\frac{(1+4 x^2)^{3/2}-3 x-8 x^3}{x+3 x^3}}. \end{equation} Everywhere here $x>0$, unless otherwise specified. It is not hard to see that the expression under the square root is $>0$, so that the definition of $h$ is correct. Let now \begin{equation} f(x):=\int_a^x h(u)\,du \end{equation} and \begin{equation} g(x):=x^2-a^2, \end{equation} where $a:=1/100$. Then \begin{equation} \frac{f''(x)}{(1+f'(x)^2)^{3/2}}=\frac{-2}{(1+4x^2)^{3/2}}=-\frac{g''(x)}{(1+g'(x)^2)^{3/2}}. \end{equation} So, the function $f$ is concave, the function $g$ is convex, $f(a)=g(a)$, and \begin{equation} k_f=k_g. \end{equation} Moreover, $f'(a)>g'(a)$. So, there is a unique real $b>a$ such that $f(b)=g(b)$; in fact, $b=0.710526\ldots\;$. However, \begin{equation} \ell_f=0.920463\ldots\ne0.894858\ldots=\ell_g, \end{equation} which does disprove the conjecture.

This counterexample is illustrated here:

enter image description here

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    $\begingroup$ Your answer is very nice. $\endgroup$
    – Đào Thanh Oai
    Commented Jul 12, 2018 at 23:23
  • $\begingroup$ @IosifPinelis: "we would have that the length of the graph of a smooth enough function $f$ over $[a,b]$ would be determined by the curvature of the graph and the values of $f$ at $a,b$." But don't we have this? It seems to me that this is an immediate consequence of the fundamental theorem of curves: given the curvature and torsion (in this case $0$), there exists a unique curve possessing them, modulo isometries (theorem 2.1.7). In particular, its length will be unambiguously determined by the curvature function alone. $\endgroup$
    – Alex M.
    Commented Jul 13, 2018 at 12:42
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    $\begingroup$ @AlexM. I think the point of the example is that we have equality of the two curvatures at points with the same $x$-coordinate, not at points the same distance along the curves. $\endgroup$
    – Andreas Blass
    Commented Jul 13, 2018 at 13:03
  • $\begingroup$ @AlexM. : Andreas Blass stated the point of the example exactly right. $\endgroup$
    – Iosif Pinelis
    Commented Jul 13, 2018 at 16:33

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