Consider the following inequality: $$\tag{1} \|f\|_{L^\infty(\mathbb R)} \le C_{\rm{best}} \|f\|_{H^1(\mathbb R)}, $$ where $\|f\|_{H^1(\mathbb R)}^2=\|f\|_{L^2(\mathbb R)}^2 + \|f'\|_{L^2(\mathbb R)}^2$ and $$C_{\rm{best}} = \sup\left\{ \frac{ \|f\|_{L^\infty} }{\| f \|_{H^1}}\ :\ f\in C^\infty_c(\mathbb R)\right\}.$$ Using the Cauchy-Schwarz inequality in Fourier space one sees that $C_{\rm{best}}\le\frac{1}{\sqrt{2}}$: $$\begin{split} |f(x)|=\left|\int_{\mathbb R} \hat{f}(\xi) e^{i x \xi}\frac{d\xi}{2\pi}\right|&\le \sqrt{\int_{\mathbb R} |\hat{f}(\xi)|^2(1+\xi^2)\, \frac{d\xi}{2\pi}}\sqrt{\int_{\mathbb R} (1+\xi^2)^{-1}\,\frac{d\xi}{2\pi}}\\ &=\|f\|_{H^1}\frac{1}{\sqrt 2}. \end{split}$$
Note that
$$\frac{1}{\sqrt{2}}= \frac{ \|e^{-|x|}\|_{L^\infty} }{\| e^{-|x|} \|_{H^1}}.$$Primary question. Is it true that $C_{\rm{best}}=\frac{1}{\sqrt{2}}$?
Secondary question. If the answer to the previous question is affirmative, is it true that equality is attained in (1) if and only if $f(x)=A e^{-|x+x_0|}$ for some $A, x_0\in \mathbb R$?
The Fourier approach presented above is not suited to those problems, because it uses the rough estimate $|e^{i x \xi}|\le 1$. An affirmative answer to those questions would be interesting for me, since I have never seen an example of an extremal function that is not smooth.