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The Problem:

Let $H^s(\mathbb{R})$ be the Sobolev Space of order $s$ on the real line, with norm $$ ||u||_{H^s} = \bigg(\int_\mathbb{R}(1+|\xi|^2)^s|\hat{u}(\xi)|^2\,d\xi\bigg)^\frac{1}{2}, $$ where $\hat{u}$ denotes the Fourier transform of $u$, given by $\hat{u}(\xi) = \frac{1}{2\pi}\int_\mathbb{R} u(x)e^{-i\xi x}\,dx$.

Give $r<s<t$, all real, and $\epsilon > 0$ show that there exists a constant $C>0$ such that $$ ||u||_{H^s} \leq \epsilon ||u||_{H^t} +C ||u||_{H^r},$$

for all $u\in H^t(\mathbb{R})$.

Where I'm at:

I've really hit a wall with this analysis prelim problem. None of the standard Sobolev inequalities I've studied (such as Gagliardo-Nirenberg, Poincaré,etc.) seem relevant. I've tried several algebraic manipulations on the integrand as well that have led nowhere. The only thing I can think of is trying to write the integral over $\mathbb{R}$ as the sum of the integral over two regions, such as a ball of radius $\epsilon$ and it's complement. However that has also gone nowhere so far.

Any help would be appreciated, even if it's only the first few steps or just explains what the key manipulation should be.

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    $\begingroup$ The inequality means that the $H^r$ and $H^s$-topologies coincide on the unit ball of $H^t$. This would follow from the compactness of that ball in $H^s$ because there is no strictly coarser Hausdorff topology on a compact space. I think this argument is sometimes called Ehrling's lemma. $\endgroup$
    – Jochen
    Aug 29, 2019 at 6:59
  • $\begingroup$ Thanks! I'll look into the proof of this lemma, since we didn't cover it, and see if proving this special case is feasible. $\endgroup$ Aug 29, 2019 at 22:37

2 Answers 2

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I would suggest a good-old cut-off argument. We have that$\newcommand{\bbR}{\mathbb{R}}$ $$\begin{align*}\lVert u \rVert_{H^s}^2 =& \int_\bbR (1 + |\xi|^2)^s |\hat u(\xi)|^2\;d\xi \\=& \int_{|\xi| \leq R} (1 + |\xi|^2)^s |\hat u(\xi)|^2\;d\xi + \int_{|\xi| > R} (1 + |\xi|^2)^s |\hat u(\xi)|^2\;d\xi\\ \leq& (1+R^2)^{r-s}\int_{|\xi| \leq R} (1 + |\xi|^2)^r |\hat u(\xi)|^2\;d\xi + \frac{1}{(1+R^2)^{t-s}}\int_{\bbR} (1 + |\xi|^2)^t |\hat u(\xi)|^2\;d\xi\\ \leq & (1+R^2)^{r-s}\int_{|\xi| \leq R} (1 + |\xi|^2)^r |\hat u(\xi)|^2\;d\xi + \frac{1}{(1+R^2)^{t-s}}\int_{\bbR} (1 + |\xi|^2)^t |\hat u(\xi)|^2\;d\xi\\ \leq& (1+R^2)^{r-s} \lVert u \rVert_{H^r}^2 + (1+R^2)^{s-t}\lVert u \rVert_{H^t}^2 \end{align*}$$ Thus, $$\lVert u \rVert_{H^s} \leq (1+R^2)^\frac{r-s}{2} \lVert u \rVert_{H^r} + (1 + R^2)^\frac{s-t}{2} \lVert u \rVert_{H^t}$$ By choosing $R$ such that $(1 + R^2)^\frac{s-t}{2} \leq \epsilon$, we obtain the result.


Notice that this proof has a conceptual interpretation: for any function $u$, we can consider it's high frequency components and low frequency components. For the high frequency bits, the $H^t$ norm is much larger than the $H^s$ norm, so we can control $H^s$ with just a little $H^t$. On the other hand, the low frequency bits are easy to control, so any $H^r$ will do.

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  • $\begingroup$ Thank you, I hadn't tried keeping my cutoff radius undefined until the and then picking it conveniently. I'll keep this in mind in the future. $\endgroup$ Sep 5, 2019 at 23:42
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    $\begingroup$ @EdgarJaramilloRodriguez Glad to help! Giving yourself an extra parameter or two often helps make estimates easier (since it can e hard to find the right value for some constant ex nihilo). $\endgroup$
    – user88319
    Sep 6, 2019 at 13:36
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Let $\alpha\in (0,1)$ be such that $s=\alpha t+(1-\alpha)r$. Let $p=1/\alpha$ and $q=1/(1-\alpha)$. Use the inequality $$ ab\leq \frac{a^p}p+\frac{b^q}q $$ where $a=\varepsilon(1+\left\vert \xi\right\rvert ^2)^{\alpha t}$ and $b=(1+\left\vert \xi\right\rvert ^2)^{s-\alpha t}/\varepsilon$.

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