I have recently encountered this problem in my studies of Sobolev spaces and generalized functions (distributions), on which I can say I might have some intuition but cannot stumble across a final solution, but first some notations:

$ \mathcal{S} $ is the Schwartz class of functions and $ \mathcal{S}' $ is the class of distributions ("generalized functions") on $ \mathcal{S} $

$ \partial ^ {\alpha} $ is the multi-index distribution derivative on $ L^2 $

$ \Lambda _ s f = { [(1+|\xi|^2)^{\frac{s}{2}} \widetilde{f} ]^{\vee} } $ is a continuous linear operator on the Schwartz distributions.

The Sobolev space $ H_s = \{ f \in S' | \Lambda_s f \in L^2 \} $

The Sobolev Embedding Theorem: suppose that $ s > k + \frac{1}{2}n $

a. If $ f \in H_s $ then $ \widehat{\partial ^ \alpha f} \in L^1 $ and $ ||\widehat{\partial ^ \alpha f}||_1 \leq C ||f||_{(s)} $ for multi-index $ |\alpha| < k $ where C depends only on $ k-s $

b. $ H_s \subset C_0 ^ k $ and the inclusion map is continuous.

A reference exercise I managed to do which the problem states as relevant:

If $ f \in L^1(R^{n+m}) $ we define $ P(f) = \int{f(x,y)dy} $ where we have $ x \in R^n $ and $ y \in R^m $ then $ Pf \in L^1(R^n) $ and also we have $ ||Pf||_1 \leq ||f||_1 $ and $ \widehat{ Pf(\xi) } = \widehat{f}(\xi,0) $

And now the actual exercise/problem I am faced with, from Folland's real analysis second edition page $309$ exercise #$35$:

The Sobolev theorem states that if $ s > \frac{1}{2}n $, then it makes sense to evaluate functions in $ H_s $ at a point. For $ 0 \leq s \leq \frac{1}{2}n $ functions in $ H_s $ are defined almost everywhere but if $ s > \frac{1}{2}k $ with $ k < n $ it makes sense to restrict functions in $ H_s $ to subspaces of codimension k. More precisely let us write $ R^n = R^{n-k} \times R^k $, $x=(y,z)$ and dual coordinates $ \xi = (\eta,\zeta) $ and define $ R : \mathcal{S}^n \to \mathcal{S}^{n-k} $ by $ Rf(y) = f(y,0) $

a. We are to show $ \widehat{(Rf)}(\eta) = \int{\widehat{f}(\eta,\zeta)}d\zeta $ via the mentioned exercise stated earlier.

b. We are to show that if $ s > \frac{1}{2}k $ then the following inequality holds:

$ |\widehat{Rf}(\eta)|^{2} \leq C_s (1+|\eta|^2)^{\frac{k}{2}-s} \int{|\widehat{f}(\eta,\zeta)}|^2 (1+|\eta|^2 + |\zeta|^2)^s d\zeta $

c. We are to show that $R$ extends to a bounded map from $ H_s(R^n) $ to $ H_{s-\frac{k}{2}} (R^{n-k}) $ provided that $ s > \frac{k}{2} $

Now I can say I have some intuition for part a via the aforementioned exercise but I still cannot seem to actually find the solutions (rigorous ones) to this problem, so to sum up I really need the help on this seemingly difficult question. I would appreciate any kind of help.


It would help if you actually copied down the problem statement correctly. The inequality in (b) should be $$\left|\widehat{Rf}(\eta)\right|^{2}\leq C_{s}(1+\left|\eta\right|^{2})^{\frac{k}{2}-s}\int_{\mathbb{R}^{k}}\left|\widehat{f}(\eta,\zeta)\right|^{2}(1+\left|\eta\right|^{2}+\left|\zeta\right|^{2})^{s}d\zeta$$

Let $f\in\mathcal{S}(\mathbb{R}^{n})$. For (a), observe that by $n$-dimensional Fourier inversion and Fubini's theorem, $$f(y,0)=\int_{\mathbb{R}^{n}}\widehat{f}(\xi)e^{2\pi i\xi\cdot(y,0)}d\xi=\int_{\mathbb{R}^{n-k}}e^{2\pi i\eta\cdot y}d\eta\int_{\mathbb{R}^{k}}\widehat{f}(\eta,\zeta)d\zeta, \quad\forall y\in\mathbb{R}^{n-k}$$ Set $g(\eta):=\int_{\mathbb{R}^{k}}\widehat{f}(\eta,\zeta)d\zeta$. By your previous exercise, $g\in L^{1}(\mathbb{R}^{n-k})$ and observe that the RHS above may be written as $${g}^{\vee}(y),\quad\forall y\in\mathbb{R}^{n-k}$$ where we take the $(n-k)$-dimensional inverse Fourier transform. Since $\widehat{g}^{\vee}(y)=f(y,0)$ is in $L^{1}$ (actually Schwartz), by $(n-k)$-dimensional Fourier inversion, $$\int_{\mathbb{R}^{k}}\widehat{f}(\eta,\zeta)d\zeta=g(\eta)=\int_{\mathbb{R}^{n-k}}g^{\vee}(y)e^{-2\pi iy\cdot\eta}dy=\int_{\mathbb{R}^{n-k}}(Rf)(y)e^{-2\pi iy\cdot\eta}y=\widehat{Rf}(\eta),\quad\forall\eta\in\mathbb{R}^{n-k}$$

Note that an easy way to see (a) is also by the density of tensor products of functions $\mathcal{S}(\mathbb{R}^{n-k})$ and $\mathcal{S}(\mathbb{R}^{k})$ in $\mathcal{S}(\mathbb{R}^{n})$.

For (b), observe first that \begin{align*} \left|\int_{\mathbb{R}^{k}}\widehat{f}(\eta,\zeta)d\zeta\right|^{2}&\leq\left(\int_{\mathbb{R}^{k}}\left|\widehat{f}(\eta,\zeta)\right|d\zeta\right)^{2}\\ &=\left(\int_{\mathbb{R}^{k}}\left|\widehat{f}(\eta,\zeta)\right|(1+\left|\eta\right|^{2}+\left|\zeta\right|^{2})^{\frac{s}{2}}(1+\left|\eta\right|^{2}+\left|\zeta\right|^{2})^{-\frac{s}{2}}d\zeta\right)^{2}\\ &\leq\left(\int_{\mathbb{R}^{k}}\left|\widehat{f}(\eta,\zeta)\right|^{2}(1+\left|\eta\right|^{2}+\left|\zeta\right|^{2})^{s}d\zeta\right)\left(\int_{\mathbb{R}^{k}}(1+\left|\eta\right|^{2}+\left|\zeta\right|^{2})^{-s}d\zeta\right), \end{align*} where we use Cauchy-Schwarz to obtain the ultimate inequality. Since $2s>k$, the second factor in the last inequality is finite. By dilation invariance, we see that \begin{align*} \int_{\mathbb{R}^{k}}(1+\left|\eta\right|^{2}+\left|\zeta\right|^{2})^{-s}d\zeta&=(1+\left|\eta\right|^{2})^{-s}\int_{\mathbb{R}^{k}}\left(1+\frac{\left|\zeta\right|^{2}}{1+\left|\eta\right|^{2}}\right)^{-s}d\zeta\\ &=(1+\left|\eta\right|^{2})^{\frac{k}{2}-s}\int_{\mathbb{R}^{k}}(1+\left|\zeta\right|^{2})^{-s}d\zeta \end{align*} Using part (a), we obtain the desired conclusion.

For (c), multiply $\left|\widehat{Rf}(\eta)\right|^{2}$ by the factor $(1+\left|\eta\right|^{2})^{s-\frac{k}{2}}$, and use Fubini's theorem to obtain \begin{align*} \int_{\mathbb{R}^{k}}(1+\left|\eta\right|^{2})^{s-\frac{k}{2}}\left|\widehat{Rf}(\eta)\right|^{2}d\eta&\leq\int_{\mathbb{R}^{k}}\int_{\mathbb{R}^{n-k}}\left|\widehat{f}(\eta,\zeta)\right|^{2}(1+\left|\eta\right|^{2}+\left|\zeta\right|^{2})^{s}d\zeta\\ &=\left\|\Lambda_{s}f\right\|_{L^{2}}^{2}<\infty \end{align*} By density of Schwartz functions in $H^{s}(\mathbb{R}^{n})$, we obtain the desired conclusion.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.