Prove that $\max(||f||_{\infty},||f'||_{\infty})\leq \frac12 \|f'\|_{L^2([0,1])}$

Question

i want to prove this: $$\max\left(||f||_{\infty},||f'||_{\infty}\right)\leq \frac12 \|f'\|_{L^2([0,1])}$$ where $f\in C^1([0,1])$ and $f(0)=f(1)=0$.

What I did is: (using Cauchy Schwartz inequality) $$ |f(t)|=\left|\int_0^t f'(t) \,dt\right|\le \int_0^t |f'(t)| \,dt\leq \left(\int_0^1 |f'(t)|^2 \,dt\right)^{\frac12}=\|f'\|_{L^2([0,1])}$$

I don't know how to continue. Any help please?

Answer 1

You were on the right track, but you didn't use $f(1) = 0$.

For $t \in [0,1]$ we have \begin{align} |f(t)| &= \frac12 \left|\int_0^t f'(t)\,dt\right| + \frac12\left|\int_t^1 f'(t)\,dt\right| \\ &\leq \frac12 \int_0^t |f'(t)|\,dt + \frac12\int_t^1 |f'(t)|\,dt\\ &= \frac12 \int_0^1 |f'(t)|\,dt \\ &\le \frac12\left(\int_0^1 |f'(t)|^2 \,dt\right)^{\frac12}\\ &=\frac12\|f'\|_{L^2([0,1])} \end{align} again using Cauchy-Schwartz. Therefore $\|f\|_\infty \le \frac12\|f'\|_{L^2([0,1])}$.

The inequality $\|f'\|_\infty \le \frac12\|f'\|_{L^2([0,1])}$ isn't true.

As suggested by @Martin R, consider $f(x) = x(1-x)$. We have $\|f'\|_\infty = 1$ but $\frac12\|f'\|_{L^2([0,1])} = \frac1{2\sqrt{3}}$.

Answer 2

I don't think that the inequality holds. For example, let $f(t)=\sin(2\pi t)$. Then $$ f\in C^1[0,1], f(0)=f(1)=0. $$ Clearly $$ \max_{t\in[0,1]}|f'(t)|=2\pi,\|f'\|_{L^2[0,1]}=\sqrt2\pi. $$ But $$ \max_{t\in[0,1]}|f'(t)|>\frac12\|f'\|_{L^2[0,1]}. $$