Does $\Vert f\Vert_{L^\infty}^2\leq \Vert f\Vert_{L^2}\Vert f'\Vert_{L^2}$ hold?

Question

I know that Sobolev's embedding implies that $H^1(\mathbb{R})\hookrightarrow C(\mathbb{R})\cap L^\infty(\mathbb{R})$, which in particular implies that $$ \Vert f\Vert_{L^\infty}\leq \Vert f\Vert_{H^1}. $$ Now, I was wondering about the following slightly more specific (?) inequality. More precisely, I was wondering if $f\in H^1(\mathbb{R})$, then the following inequality holds: $$ \Vert f\Vert_{L^\infty}^2\leq \Vert f\Vert_{L^2}\Vert f'\Vert_{L^2}. $$ The problem here is to be able to put a factor only with the $L^2$ norm of $f$. Does this inequality hold?

Answer 1

Let $f \in \mathcal{S} (\mathbb{R})$. Let $x^*$ be any point realizing the maximum of $|f|$. Then

$$2|f(x^*)| = \left|\int_{-\infty}^{x^*} f'(t) \ dt\right| + \left|\int_{x^*}^{+\infty} f'(t) \ dt\right| \leq \|f'\|_{\mathbb{L}^1}.$$

Hence, $\|f\|_{\mathbb{L}^\infty} \leq \frac{1}{2} \|f'\|_{\mathbb{L}^1}$. Applying this inequality to $f^2$ and using the Cauchy-Schwarz inequality,

$$\|f\|_{\mathbb{L}^\infty}^2 = \|f^2\|_{\mathbb{L}^\infty} \leq \frac{1}{2} \|(f^2)'\|_{\mathbb{L}^1} = \|ff'\|_{\mathbb{L}^1} \leq \|f\|_{\mathbb{L}^2} \|f'\|_{\mathbb{L}^2}.$$

Finally, $\mathcal{S} (\mathbb{R})$ is dense in $H^1 (\mathbb{R})$, and all of $f \mapsto \|f\|_{\mathbb{L}^\infty}$, $f \mapsto \|f\|_{\mathbb{L}^2}$ and $f \mapsto \|f'\|_{\mathbb{L}^2}$ are continuous in the $H^1$ norm (the later two by definition of the $H^1$ norm, the former by the Sobolev embedding). Hence the claimed inequality actually holds in $H^1 (\mathbb{R})$.