# Prove $\int _0^\infty f^2 dx \leq \cdots$ for $f$ convex

Prove $$\int _0^\infty f^2(x) dx \leq \frac{2}{3}\cdot \max_{x \in \mathbb R^+} f(x) \cdot \int _0^\infty f(x) dx$$ for $f(x) \geq 0$ and convex.

I know via Holder's we can get without the constant $\frac{2}{3}$ but that should also be true for $f$ not convex. I tried some approaches using convexity but didn't get the desired result.

A hint or reference would also help. Also, it would be interesting to know when equality holds.

• Equality is when $f$ is piecewise linear, say when $f(x)=1-x, x\in [0,1]$, zero elsewhere... – Macavity Dec 30 '17 at 8:59
• Thanks Macavity I was aware of this and linear examples (apologize for not mentioning this). I was interested in the converse, probably that would mean seeing a proof of the inequality and figuring it out. – MUB Dec 30 '17 at 9:02

The r.h.s. is finite if and only if the convex function $f$ is monotone decreasing in $[0,+\infty)$ with $\lim_{x\to +\infty} f(x) = 0$.

Let us consider such a function. In order to simplify the proof, assume also that $f\in C^1([0,\infty))$ (this assumption can easily be removed).

Let us define the auxiliary function $\varphi\colon [0,+\infty)\to \mathbb{R}$, $$\varphi(x) := \frac{2}{3} f(x) \int_x^{+\infty} f(t)\, dt - \int_x^{+\infty} f(t)^2\, dt, \qquad x\geq 0.$$ We are going to prove that $\varphi(x) \geq 0$ for every $x\geq 0$, hence the required inequality follows from $\varphi(0) \geq 0$.

We have that $$\varphi'(x) = \frac{2}{3}f'(x) \int_x^{+\infty} f(t)\, dt -\frac{2}{3} f(x)^2 + f(x)^2 = \frac{2}{3}f'(x) \int_x^{+\infty} f(t)\, dt +\frac{1}{3} f(x)^2.$$ Since $f$ is convex, if $f'(x) \neq 0$ (hence $f'(x) < 0$) it holds $$(*) \qquad \int_x^{+\infty} f(t)\, dt \geq \int_x^{x - f(x)/f'(x)} \left(f(x) + (t-x)f'(x)\right)\, dt = - \frac{f(x)^2}{2 f'(x)}\,,$$ so that $\varphi'(x) \leq 0$. On the other hand, if $f'(x) = 0$, then necessarily $f(x) = 0$ so that $\varphi'(x) = 0$.

In conclusion, $\varphi'(x) \leq 0$ for every $x\geq 0$ and $\lim_{x\to +\infty} \varphi(x) = 0$, hence $\varphi(x)\geq 0$ for every $x\geq 0$.

The $C^1$ regularity requirement can be removed observing that $\varphi$ is Lipschitz continuous and the inequality $\varphi'(x) \leq 0$ holds at every point of differentiability.

The equality holds true if and only if $\varphi(0) = 0$, i.e., if and only if $\varphi'(x) = 0$ for every $x\geq 0$. From the discussion leading to (*), it follows that this can happen if and only if there exists $K\geq 0$ such that $f(x) = 0$ for every $x\geq K$ and $f$ is affine on the interval $[0,K]$.

• thank you for helping me out. just a couple of comments while i am reading the proof. f(x)=c is possible but its trivial in the case of inequality so we can leave out f'(x)=0. i am still not sure why f'(x) cannot be >0 ? – MUB Dec 30 '17 at 10:39
• Since $f$ is convex, the if $f'(x) > 0$ as some point then $f$ stays above an increasing affine function, so that $\int f = +\infty$. – Rigel Dec 30 '17 at 10:47
• This is really an elegant proof and I referred to it here. – Another possibility to avoid the regularity requirement is to work with a one-sided derivative (which exists everywhere since $f$ is convex). – Martin R Apr 2 at 12:10