Convexity and shortest curve

Question

I found a PDF in which there were the two following questions:

1. For any function $g \in C^1$, let $F(g) = \int_{0}^{1} \sqrt{1 + (g’(t))^2}dt$. Show that if f, g are $C^1$ and $\lambda \in ]0, 1[$, then $F(\lambda f + (1 - \lambda)g) \leq \lambda F(f) + (1-\lambda) F(g)$.

2. If $a, b$ are given in $\mathbb{R}$, show that there is a function f with $f(0)=a, f(1)=b$ such that for each function g with $g(0)=a, g(1)=b$, we have $F(f) \leq F(g)$.

I’m good with question 1) (I used the convexity of $x \mapsto \sqrt{1+x^2}$).

I get the meaning of question 2 (basically, the goal is to show that amongst all $C^1$ functions, then it’s the linear function satisfying the conditions that is the shortest). However, i can’t really see how to use question 1 to answer question 2 ...

Thank you.

Answer 1

The function $\phi(x)= \sqrt{1+x^2}$ is convex and the map $f \to f'$ is linear.

Since $\phi$ is convex we have $\phi(x+h) \ge \phi(x) + \phi'(x)h$.

Hence $F(f+h)=\int_0^1\sqrt{1+((f+h)'(t))^2 } dt \ge F(f)+\int_0^1 {f'(t) \over \sqrt{1+(f'(t))^2}}h'(t) dt $.

Let $f(t) = a+t(b-a)$, note that $f'(t) = b-a$ and so $F(f+h) \ge F(f) + {{b-a} \over \sqrt{1+(b-a)^2}} (h(1)-h(0))$.

If $h=g-f$ then we see that $h(0)=h(1)$ and so $F(g) \ge F(f)$.

Answer 2

The following argument (provided I didn't mess it up) is a standard line of attack in this situations. The keyword is calculus of variations.

There are some analysis details here you should worry about. For instance, can we really just integrate under the integral sign?

There are two steps:

1. Convexity of $F$ implies that if $f$ is a local min of $F$, then it is a global min of $F$. This is one of the the key properties of convexity, especially when you are minimizing.

What exactly does 'local min' mean in this context? To answer this, let's consider some other $g$, and define $h_t = (1 - t)f + t g$, which is the line between $f$ and $g$ in $C^1[a,b]$. So we have $h_0 = f, h_1 = g$. Also, define and $\phi(t) = F(h_t)$, and note that part1) shows the function $\phi$ is convex; in particular, if $F(g) < F(f)$, then $\phi'(0) < 0$. So, to show that if $f$ is linear, then for all choices of $g$, $\phi'(0) \geq 0$.

One can think of $\phi'(0)$ as the directional derivative of $F$, at $f$, in the direction $g$. Showing that it is $\geq 0$ for all directions is the usual calculus notion of a local min.

2. We have to use somewhere the definition of $F$, so now would be a good time to start writing things out. Explicitly we have $f(x) = a + (b - a)x$. So, $h'_t(x) = (1 - t)(b - a) + t g'(x) = t(a - b + g'(x)) + ( b - a) $.

$$\phi(t) = F( h_t) = \int_0^1 \sqrt{ 1 + (h'_t(x))} dx = \int_0^1 \sqrt{1 + ( t(a - b + g'(x)) + ( b - a) )^2} dx $$

Let's try to compute $\phi'(0)$ by differentiating under the integral. Setting $m = (a - b + g'(x)), n = ( b -a)$, I had wolfram do the algebra for me.

$$\frac{d}{dt} \phi(t) = \int_0^1 \frac{m (mt + n)}{ \sqrt{(mt + n)^2 + 1}} dx $$

So,

$$(\frac{d}{dt}\phi)(0) = \int_0^1 \frac{mn}{ \sqrt{(mt + n)^2 + 1}} dx = \int_0^1 \frac{(a - b + g'(x)) g'(x)}{\sqrt{(mt + n)^2 + 1}} dx $$

In the special case that $a = b$, the integral here is $\geq 0$ since the integrand, being a square, is always non-negative. There should be a simple trick to reduce to this case, but I'm not seeing it right now. Maybe if you do the algebra by hand you'll see it? (You can translate and scale to make $a = 0, b = 1$. I thought about applying a sheer transformation, but showing that that preserves minima of $F$ seems yucky.)