How to prove this bound of $L^\infty$ norm. A differentiable function $ f:\mathbb R\to \mathbb R$ satisfies such conditions,
$ $\begin{cases} \lim_{x\to\infty} f(x)=\lim_{x\to-\infty} f(x)=0, &\\ 
\int_{-\infty}^{\infty}|{f(x)}|^{2}dx<\infty, \ \    \int_{-\infty}^{\infty}|{f'(x)}|^{2}dx<\infty. &
\end{cases}$$
Prove
$$\max_{x\in \mathbb R}|f(x)|\leq \left(\int_{-\infty}^{\infty}|{f(x)}|^{2}dx<\infty\right)^{1\over4}\left(\int_{-\infty}^{\infty}|{f'(x)}|^{2}dx<\infty\right)^{1\over4}.$$

I'll assume $f'$ is continuous. Let $a\in \mathbb {R}.$ Then for $h>0,$ the FTC gives
$$f(a+h)^2/2-f(a)^2/2 = \int_a^{a+h} f\,f'$$
and
$$f(a-h)^2/2-f(a)^2/2 = -\int_{a-h}^{a} f\,f'. $$
Add these together, then take absolute values to get
$$|f(a+h)^2/2-f(a-h)^2/2 -f(a)^2| \le \int_{a-h}^{a+h} |f|\,|f'| \le \int_{\mathbb R} |f|\,|f'|$$
Cauchy-Schwartz then shows
$$|f(a+h)^2/2-f(a-h)^2/2 -f(a)^2| \le (\int_{\mathbb {R}} |f|^2)^{1/2}(\int_{\mathbb {R}} |f'|^2)^{1/2}.$$
Let $h\to \infty$ and use the assumption $f(x)\to 0$ as $|x| \to 0$ to see
$$|f(a)|^2 \le (\int_{\mathbb {R}} |f|^2)^{1/2}(\int_{\mathbb {R}} |f'|^2)^{1/2}.$$
Taking square roots gives
$$\tag 1 |f(a)|  \le (\int_{\mathbb {R}} |f|^2)^{1/4}(\int_{\mathbb {R}} |f'|^2)^{1/4}.$$
Since $a$ was an arbitrary point, $\max_{\mathbb {R}}|f|$ is less than or equal to the right side of $(1)$ as desired. (The assumption $f'(x) \to 0$ was not used.)
