# Specific Integral Question: $\int_0^1(f'(t))^2dt \geq 3(\int_0^1f(t)dt)^2$

I'm in the midst of studying for my real analysis final tomorrow and I came across this question in a practice final and am stumped on how to even start it.

Let $f: [0,1] \rightarrow \mathbb{R}$ be a differentiable function such that $f'$ is continuous and $f(1)=0$. Prove that the following inequality holds: $\int_{0}^{1}(f'(t))^2dt \geq 3(\int_{0}^{1}f(t)dt)^2$.

Any guidance on how to start would be of help..

• I'm not sure but I would have a look over the Cauchy Schwarz inequality
– Stu
Aug 9 '17 at 23:34

First apply Integration by Parts:

$$\int_0^1f(x)\;dx=-\int_0^1xf'(x)\;dx$$

Here we have used the fact that $xf(x)$ vanishes at both $0$ and $1$.

Now invoke Cauchy-Schwarz. We have $$\left(\int_0^1f(x)\;dx\right)^2=\left(\int_0^1xf'(x)\;dx\right)^2≤\int_0^1\left(f'(x)\right)^2\;dx\;\times\;\int_0^1x^2\;dx$$

And your inequality follows at once.

• And equality holds if and only if $f'$ is a scalar multiple of $x$, which combined with $f(1) = 0$ implies $f(x) = \lambda (1 - x^2)$. Aug 10 '17 at 0:09
• @lulu: Very nice! (+1) Aug 10 '17 at 6:48