Challenge I: Concavity, solved by Millenial

[Greg Bernhardt]

Written by micromass:

I have recently posted a challenge in my signature. The challenge read as follows:

Let $f:[0,+\infty )\rightarrow [0,+\infty )$ be a twice differentiable function such that $f(0) = 0$. Prove that if $f^{\prime\prime}(x)\leq 0$ for all $x$, then $f(x+y)\leq f(x) + f(y)$ for all $x$ and $y$.

The first answer I got was from Millenial. He gave the following correct solution:

Let g(x) = f'(x). Then, the question tells us that g(x) is decreasing. We are only interested in cases where x and y are greater than or equal to 0, because that's where f is defined. Define
\int^{x}_{y} g(t)\,dt = A(x,y)
Then, A(x,0) = f(x) where equality follows because f(0) = 0.

The question asks us to prove A(x,0) + A(y,0) \geq A(x+y,0). Without loss of generality, we assume x\geq y. Then, the inequality is reduced to 2A(y,0) + A(x,y) \geq A(y,0) + A(x,y) + A(x+y,x), which can be simplified to A(y,0) \geq A(x+y,x).

Using the fact that A denotes integrals, we can put lower and upper bounds on A using maxima and minima of g in the given intervals. We find that A(y,0) \geq y\cdot g(y) and A(x+y, x) \leq y\cdot g(x) because g(x) is decreasing. This means that our question is reduced to proving y\cdot g(y) \geq y\cdot g(x). Since y\geq 0 and the case where y = 0 is trivial, we may divide by y to get g(y) \geq g(x), and since x \geq y, this is equivalent to saying g is decreasing; which we already know.

This solution is very elegant. But there are other solutions. For example, we can prove the following chain of implications:

• $f^{\prime\prime}(x)\leq 0$ for all $x\in \mathbb{R}$.
• $f^\prime$ is decreasing.
• $\frac{f(x)}{x}$ is decreasing.
• For each $t\in [0,1]$ and for all $x$ holds that $tf(x)\leq f(tx)$.
• For each $x$ and $y$ holds that $f(x+y)\leq f(x) + f(y)$.

The proof is as follows:

From $(1)$ to $(2)$ is just elementary calculus.
From $(2)$ to $(3)$ is as follows. We need to prove that the derivative of $\frac{f(x)}{x}$ is negative. By taking derivatives, we need to prove that $f^\prime(x) x\leq f(x)$ for each $x$. Apply the mean value theorem on the interval $[0,x]$. So we know that there is a constant $c\in [0,x]$ such hat
f(x)= f^{\prime}(c) x
Using that $f^\prime$ is decreasing, the inequality follows.
From $(3)$ to $(4)$ follows since $tx\leq x$ and thus $\frac{f(x)}{x}\leq \frac{f(tx)}{tx}$. Hence the inequality follows.
From $(4)$ to $(5)$ follows from
\begin{eqnarray*}<br /> f(x) + f(y) <br /> &amp; = &amp; f\left(\frac{x}{x+y}(x+y)\right) + f\left(\frac{y}{x+y} (x+y)\right)\\<br /> &amp; \leq &amp; \frac{x}{x+y}f(x+y) + \frac{y}{x+y} f(x+y) \\<br /> &amp; = &amp; f(x+y)<br /> \end{eqnarray*}

Yet another proof consists of proving that $f$ is a concave function. This means that for each $x$ and $y$ and $t\in [0,1]$ that
f(tx + (1-t)y) \geq tf(x) + (1-t)f(y)
Visually, it means that



So, the straight line between two points is always below the actual function.

Here is the proof that any function such that $f^{\prime\prime}<0$ is concave:

Assume that $f^{\prime\prime}\leq 0$. Take $x<y$ and $t\in (0,1)$. Let $z = tx + (1-t)y$ and thus $x< z < y$. By the mean value theorem, there are $u$ and $v$ such that $x<u<z<v<y$ and such that $f(z) - f(x) = f^\prime(u) (z-x)$ and $f(y) - f(z) = f^\prime(v) (y-z)$. Since $f^\prime$ is decreasing, we have $f^\prime(v)\leq f^\prime(u)$. Hence
\frac{f(y) - f(z)}{y-z}\leq \frac{f(z) - f(x)}{z-x}
Since $z = tx + (1-t)y$, we see that
\frac{f(y) - f(z)}{t(y-x)} \leq \frac{f(z) - f(x)}{(1-t)(y-x)}
And thus
(1-t)( f(y) - f(z) ) \leq t(f(z) - f(x))
The desired inequality follows.

Now, we can immediately prove again that $tf(x)\leq f(tx)$ for each $x$ and $t\in [0,1]$. Indeed, apply concavity with $y=0$ (and remember that $f(0) = 0$, then
tf(x) = tf(x) + (1-t)f(0) \leq f(tx + (1-t)y)\leq f(tx)
We can then, as before, prove $f(x+y)\leq f(x) + f(y)$.

Many congratulations to Millenial for solving the problem!

 

[fresh_42]

There are some theorems in calculus which are especially important: the mean value theorem (or Rolle's theorem), the theorem of Bolzano-Weierstrass, the implicit function theorem and of course the many variants of Stoke's theorem.