# Estimate involving $L^{\infty}$ norm of function in $H^2(\mathbb{R}) \cap H^{-1/2}(\mathbb{R})$

I have a question concerning a Gagliardo-Nirenberg type inequality involving the $$L^{\infty}$$-norm on the unbounded domain $$\mathbb{R}$$:

To be more precise, I have encountered the following estimate

$$\| f \|_{L^{\infty}} \leq C \| f \|_{L^2}^{1/2} \| f_x \|_{L^2}^{1/2}$$

for some constant $$C$$ depending only on $$f$$, where $$f$$ is an element of $$H^2(\mathbb{R}) \cap \dot{H}^{-1/2}(\mathbb{R})$$, the latter being the homogeneous fractional Sobolev space in terms of Fourier transforms and $$f_x$$ denoting the (first) derivative .

For a bounded domain $$I \subset \mathbb{R}$$, the analogous statement follows immediately from the Gagliardo-Nirenberg inequality which may be found in Brezi's Functional Analysis book (Chapter 8, Comment 1 (iii), Equation (42)) and the Poincaré inequality.

I, however, am interested in the inequality for the unbounded domain $$\mathbb{R}$$.

Is anyone aware of general interpolation inequalities taking care of this case? Help is much appreciated!

Thank you!

A way to do it is with the Fourier transform $$\hat{f}(y) = \int_{\mathbb{R}} e^{-2i\pi\,xy}f(x)\,\mathrm{d} x$$. Using Fourier inversion theorem, the definition of the Fourier transform, multiplying and dividing by $$\sqrt{1+|2\pi x|^2}$$ and then using the Cauchy-Schwarz inequality yields $$\|f\|_{L^∞} = \|\hat{\hat{f}}\|_{L^∞} ≤ \int_{\mathbb{R}}|\hat{f}| ≤ \left(\int_{\mathbb{R}} \frac{\mathrm{d}x}{1+|2\pi x|^2}\right)^\frac{1}{2} \left(\int_{\mathbb{R}} (1+|2\pi x|^2)\,|\hat{f}(x)|^2\,\mathrm{d}x\right)^\frac{1}{2}.$$ Notice that the first integral is not difficult to compute $$\int_{\mathbb{R}} \frac{\mathrm{d}x}{1+|2\pi x|^2} = \frac{1}{2\pi}\int_{\mathbb{R}} \frac{\mathrm{d}y}{1+|y|^2} = \left[\frac{\arctan(x)}{2\pi}\right]_{x=-\infty}^∞ = \frac{1}{2}.$$ By the properties of the Fourier transform, the second integral is nothing but $$\left(\int_{\mathbb{R}} (1+|2\pi x|^2)\,|\hat{f}(x)|^2\,\mathrm{d}x\right)^\frac{1}{2} = \left(\left\|f\right\|_{L^2}^2 + \left\|\nabla f\right\|_{L^2}^2\right)^{1/2}.$$ Your inequality now just follows by scaling. Just apply the above inequality to $$f_r(x) := f(x/r^2)$$ and noticing that $$\|f_r\|_{L^\infty} = \|f\|_{L^\infty}$$, $$\|f_r\|_{L^2} = r\,\|f\|_{L^2}$$ and $$\|\nabla f_r\|_{L^2} = \frac{1}{r}\,\|\nabla f\|_{L^2}$$, this leads to $$\|f\|_{L^\infty} \leq \frac{1}{\sqrt{2}} \left(r \,\left\|f\right\|_{L^2}^2 + \frac{1}{r} \,\left\|\nabla f\right\|_{L^2}^2\right)^{1/2}$$ Now optimize with respect to $$r$$, i.e. take $$r = \left\|\nabla f\right\|_{L^2} / \left\|f\right\|_{L^2}$$, to get $$\|f\|_{L^\infty} \leq \left\|f\right\|_{L^2}^{1/2} \,\left\|\nabla f\right\|_{L^2}^{1/2},$$ i.e. your inequality holds with the constant $$C = 1$$.
Remark: The constant $$1$$ is sharp. Indeed, taking $$f$$ such that $$\widehat{f}(y) = (1+|2\pi y|^2)^{-1}$$, then $$\|f\|_{L^\infty} = \int \widehat{f} = 1/2$$ and (using for instance the Beta function) one obtains $$\left\|f\right\|_{L^2} = \left\|\nabla f\right\|_{L^2} = 1/2$$.
• Thank you for your help! I will definitely keep in mind the trick of representing $f$ as $\hat{\hat{f}}$ to inject integrals into my equations/inequalities! That was the key idea (besides Plancherel and integration by parts) that, for whatever reasons, escaped my mind. Commented Sep 5, 2022 at 11:24