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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!

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1 Answer 1

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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 $\left\|f\right\|_{L^2} = \left\|\nabla f\right\|_{L^2} = 1/2$.

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  • $\begingroup$ 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. $\endgroup$
    – Lukic
    Commented Sep 5, 2022 at 11:24

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