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Please, can someone help me with this; I'm seriously stuck. Is it true that since the Sobolev space $H^{\alpha}(\mathbb{R})$ for $\alpha > 0$ is a dense subset of $\mathbf{L}^2(\mathbb{R})$, then one can find a constant $C$ such that for every function $f \in H^{\alpha}(\mathbb{R})$, we have that $$\Vert f \Vert_{\mathbf{L}^{2}(\mathbb{R})} \leqslant C \Vert f \Vert_{H^{\alpha}(\mathbb{R})}?$$

Secondly, if $\Omega$ and $\Omega^{\prime}$ are two open bounded sets in $\mathbb{R}$ such that $\overline{\Omega} \subset \Omega^{\prime}$, can I find such inequality as follows: there exists a positive real constant $C$ such that $$\Vert f \Vert_{H^{\alpha}(\Omega)} \leqslant C \Vert f \Vert_{\mathbf{L}^{2}(\Omega^{\prime})}?$$

Please, I'll also need some reliable sources to read these up if they are true.

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  • $\begingroup$ Please, correct spelling of Sobolev's name in the title. $\endgroup$ Aug 11, 2013 at 2:24

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You did not specify your choice of a norm on $H^\alpha$. The most convenient here is the norm involving the Fourier transform $\widehat f$, namely $$\|f\|_{H^\alpha} = \left(\int_{\mathbb R}(1+\xi^2)^{\alpha/2}|\widehat f(\xi)|^2\,d\xi \right)^{1/2} \tag1$$ Comparing (1) to Parseval's identity $$\|f\|_{L^2} = \left(\int_{\mathbb R}|\widehat f(\xi)|^2\,d\xi \right)^{1/2} \tag2$$ you can see that $\|f\|_{L^2}\le \|f\|_{H^\alpha}$.

You could probably also argue from the containment $H^\alpha \subset L^2$, but I think this is circular: the reason we know that $H^\alpha$ is contained in $L^2$ is the inequality $\Vert f \Vert_{\mathbf{L}^{2}(\mathbb{R})} \leqslant C \Vert f \Vert_{H^{\alpha}(\mathbb{R})}$.

The reverse inequality $\Vert f \Vert_{H^{\alpha}(\Omega)} \leqslant C \Vert f \Vert_{\mathbf{L}^{2}(\Omega^{\prime})}$ fails for general $f$. For one thing, a general element of $L^2(\Omega')$ does not restrict to a function in $H^\alpha(\Omega)$. But even if it does, the $H^\alpha$ norm is not controlled by the $L^2$ norm. Let $f(x)=\max(0,1-|x|)$ and consider the sequence $f_n=nf(n^2x)$ defined on on $(-1,1)$. The rescaling is chosen so that $\|f_n\|_{L^2} = \|f\|_{L^2}$ for all $n$. On the Fourier side, $\widehat{f_n}(\xi)=n^{-1}\widehat{f}(n^{-2}\xi)$. Thus, $$\|f_n\|_{H^\alpha}^2 = \int_{\mathbb R}(1+\xi^2)^{\alpha/2} n^{-2} |\widehat f(n^{-2}\xi)|^2\,d\xi = \int_{\mathbb R}(1+n^4\zeta^2)^{\alpha/2} |\widehat f(\zeta)|^2\,d\zeta \tag3$$ after substitution $\zeta=n^{-2}\xi$. As $n\to\infty$, the right hand side of (3) grows indefinitely.

A typical situation in which you get a reverse inequality on a smaller domain is when $f$ is a solution of some differential equation. You can look up interior regularity for elliptic PDE.

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