An inequality on length of two curves [closed]

Question

I am looking for a proof, reference, comment of an inequality as follows:

If $f(x)$ and $g(x)$ be two continuous derivative funcions in interval $[a, b]$. Such that:

1. $f(a)=g(a)$ and $f(b)=g(b)$

2. $($$f''(x) > 0$ or $f''(x) < 0$$)$ and $($$g''(x) > 0$ or $g''(x) < 0$$)$

3. $$|f(x)-\frac{f(b)-f(a)}{b-a}(x-a)-f(a)|\ge|g(x)-\frac{g(b)-g(a)}{b-a}(x-a)-g(a)|$$

Then

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

Answer 1

Your way of asking could have been simplified by things like changing condition $3$ (as one can) to $f(x) \leq g(x)$ and setting $(a,f(a))=(0,0).$ I'll mention that after providing the comments you request.

My comment is that you are asking, in a complicated way, about a fundamental geometric inequality of curve lengths. That inequality is the foundation for arc length integrals, which you use for asking the question. You ask about the second of the two inequalities below.

• Of all curves from $P$ to $Q$ the shortest is the straight line segment $PQ$.
• If $PRQ$ and $PSQ$ are convex curves both on the same side of segment $PQ$ and $PRQ$ is included between $PQ$ and $PSQ$ (allowing them to coincide at times if desired) then the (arc) length of $PRQ$ is less than that of $PSQ$.

The first of those implies that in your illustration $\int_{a}^{b}{ \sqrt{1+f'(x)^2}}dx \ge \int_{a}^{b}{ \sqrt{1+k^2}}dx$ where $k=\frac{f(b)-f(a)}{b-a}.$

Is that functional analysis or geometry?

I'll have to explain some of that. Your definition of convex is (essentially) that the two curves are twice differentiable with positive second derivative. But the inequality holds for a more general (and basic) definition of convex :

• A curve $\mathcal{C}$ is convex if for any two points $x,y \in \mathcal{C},$ all the points between $x$ and $y$ on $\mathcal{C}$ lie on the same side of the line $xy.$ Here we would allow some of the points to be on the line but not some on one side and some on the other.

So in the second picture segment $AC$ is shorter than convex curve $AFC$ which in turn is shorter than convex curve $ABC.$ Similarly $CE \lt CGE \lt CDE.$

If I am not mistaken, those two inequalities and the definition are pretty close to what Archimedes said in a book On Measurement of a Circle where the inequalities are axioms.

He used this to give a method of computing $\pi$ to desired accuracy using polygons approximating a circle, as on the left, and letting the number of sides grow. Related is our definition for arc length integrals.

DIGRESSION: Here is the simplest case of the main inequality: If $R$ is an internal point of triangle $PSQ$ then $PR+RQ \lt PS+SQ$

It seems it must follow from the triangle inequality, and it does. The proof is short, but not that easy to find. I know I have seen it discussed someplace but don't find it at the moment. Maybe someone has a link?

To be fair, you never mentioned arc length, you just gave two integrals. I don't know for sure but it seems possible to me that for that kind of perverse way of asking the question one could do an end run as follows:

First use geometry to prove the geometric result for polygonal paths, probably using the digression above. Then consider the two integrals in question. An integral is the limit of Riemann sums. For the kind of integral in question these Riemann sums are the length of polygonal paths. Since the inequality holds for the sums, it holds for the limits.

I'm not totally sure. I do view it a little like proving the first inequality (a line segment is the shortest path) via arc length integrals.

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They way you state the problem makes it needlessly complicated. Even if you are going to state the problem the way you did, you could simplify.

• wlog you can assume in $2$ that $f''(x),g''(x) \geq 0$
• then your condition $3$ can be simplified (wlog) to $f(x) \leq g(x)$

Let $L(x)=f(a)+\frac{f(b)-f(a)}{b-a}(x-a)$ be the line connecting the points. Then your expression $|f(x)-\frac{f(b)-f(a)}{b-a}(x-a)-f(a)|=|L(x)-f(x)|$ , a vertical distance.

Why not state it that way?

You can assume that $(a,f(a))=(a,g(a))=(0,0)$ , $b=1$ and $f(1)=g(1)=k.$ Now your line $L(x)$ is $y=kx$ and your complicated condition $3$

$$|f(x)-\frac{f(b)-f(a)}{b-a}(x-a)-f(a)|\ge|g(x)-\frac{g(b)-g(a)}{b-a}(x-a)-g(a)|$$

which was already simplified to

$$|L(x)-f(x)| \geq |L(x)-g(x)|$$

becomes

$$|kx-f(x)|\geq |kx-g(x)|.$$

You can also assume that $f''$ and $g''$ are positive so that the curves are under the line (i.e. $f(x),g(x) \leq kx$ ) and then the condition is simply that for $a \leq x \leq b,$

$$f(x) \leq g(x)$$

Why not state it that way?

Why can we assume that $f''$ is positive?

Let $L(x)$ be the line. In case $f''(x) \lt 0$ flip $f$ under that line by using instead $f^*(x)=2L(x)-f(x)=2kx-f(x)$ Check that $(f^*)''=-f'' \geq 0$ and $|kx-f(x)| =|kx-f^*(x)|.$

Answer 2

Assume that $f''(x),\ g''(x) >0$.

If $D_f =\{ (x,y)| f(x)\leq y\leq k(x) \},\ D_g=\{(x,y)|g(x)\leq y\leq k(x)\}$ where $k(x)=\frac{f(b)-f(a)}{b-a}(x-a)+f(a)$, then $D_g\subset D_f\ \ast$.

And $D_f,\ D_g$ are convex so that from $\ast$, ${\rm length}\ \partial D_f\geq {\rm length}\ \partial D_g$.