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Let $f:[0,\infty)\rightarrow [0,\infty) $ be a concave function and $ p\in [0,\dfrac{1}{2}]$
a) Show that f is integrable on each interval $[a,b] \subset [0,\infty).$
b) Show that $ F:[0,\infty)\rightarrow \Bbb{R}$, $F(x)=\int_{0}^{x^p} f(t)dt$ is concave.

[Attempt] I proved point a) using a well known inequality: If $x<y<z$ and $f$ is concave then $$ \frac{f(x)-f(y)}{x-y} \ge \frac{f(y)-f(z)}{y-z}$$
and by taking five points $\ a<b<c<d<e \ $ it can be proven that the function is continuous therefore integrable. But at point b) we know that the function $x^p$ is concave and because $f(x)\ge 0, F$ is increasing so $$ F(tx_1+(1-t)x_2)= \int_{0}^{(tx_1+(1-t)x_2)^{p}} f(t) dt \ge \int_{0}^{t(x_1)^{p} +(1-t)(x_2)^{p}} f(t) dt $$ but from here i don't know how to separate them. In order to solve it I think the following inequalities may have to be used: if $f$ is concave and $f \ge 0 $ then $f(tx)\ge tf(x)$ for any $x\ge 0$ and $ t\in [0,1]$ and $f(a+b)\le f(a)+f(b)$ for any $a,b \ge 0$

Maish Petronas
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1 Answers1

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You can compute the derivative of $F$ by looking at the limit of the following quantity when $h$ tends to $0$:

$$ \frac{F(x+h) - F(x)}{h} = \frac{1}{h} \int_{x^p}^{(x+h)^p} f(t) dt. $$

An immediate lower-bound (by non-decreaseness of $f$) is

$$ \frac{(x+h)^p - x^p}{h} f(x^p), $$ which tends to $p x^{p-1} f(x^p)$ as $h$ tends to $0$.

We will obtain an upper bound with a bit more work.

$$ \frac{F(x+h) - F(x)}{h} = \frac{(x+h)^p - x^p}{h} \int_{x^p}^{(x+h)^p} f(t) \frac{dt}{(x+h)^p - x^p}, $$ and Jensen's inequality gives $$ \frac{F(x+h) - F(x)}{h} \leq \frac{(x+h)^p - x^p}{h} f \left(\int_{x^p}^{(x+h)^p} t \frac{dt}{(x+h)^p - x^p} \right). $$ By continuity of $f$, $f \left(\int_{x^p}^{(x+h)^p} t \frac{dt}{(x+h)^p - x^p} \right)$ tends to $f(x^p)$ as $h$ tends to $0$. Furthermore, the first term tends to $p x^{p-1}$ in the same limit regime.

In other words, we have shown that $F$ is derivable and that for any $x$, $F'(x) = p x^{p-1} f(x^p)$.

EDIT : I just realized that all of that was really overkill for proving the derivability of $F$. The derivability and the expression are direct by composition.

We will now prove that $F'$ is non-increasing, yielding concavity.

Since $f$ is concave and positive, you have stated that for any $x$ and $t \in [0, 1]$, $f(tx) \geq t f(x)$. Hence, if $x_2 \geq x_1$, $$ x_1^{p-1} f(x_1^p) \geq x_1^{p-1} f(x_2^p) (x_1/x_2)^p ,$$ and since $x_2 \geq x_1$ and $p \in [0, 1/2]$, $x_2^{p-1} \leq x_1^{p-1} (x_1/x_2)^p$. Hence, $$ F'(x_2) \leq F'(x_1),$$ proving concavity of F.

Beleth
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