Showing an inequality in the Banach space $C^1([a,b])$?

Question

Given $f \in C^1([a,b]),$ how can it be shown that $$\|f\|_\infty^2 \le \dfrac{\|f\|_2^2}{b-a} + 2 \|f\|_2 \|f'\|_2?$$

I suppose I could use Cauchy-Schwarz, but how to handle the derivative? And I don't think Holder's applies...

This is a problem from an old Preliminary exam.

Answer 1

One idea is to let $x\in\arg \min |f^2(x)|$, fix $y\in [a,b]$, and apply the FTC to $f^2$: $$ f(y)^2-f(x)^2=2\int_x^yf(t)f'(t)dt, $$ so that by Holder, $$ \sup_{y\in[a,b]}f(y)^2\le f(x)^2+2\|f\|_2\|f'\|_2. $$ To deal with the $f(x)^2$ term, $$ f(x)^2=\frac{1}{b-a}\int_a^bf(x)^2dt\le \frac{1}{b-a}\int_a^bf(t)^2dt $$ since $f(x)^2\le f(t)^2$ for all $t$.