A question involving sharpening the bound on Sobolev type inequality with Sobolev spaces in terms of distributions of Schwartz functions

I have met this problem recently in my real analysis class involving sharpening the bound on a Sobolev type inequality, from Folland's real analysis, but first I should mention the notations used here:

$\partial ^ {\alpha}$ 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 rest ought to be standard notations.

(A sharper Sobolev theorem) For $0 < \alpha < 1$ let

$\Lambda _ {\alpha} (R^n) = \{ f \in BC(R^n) | \sup_{x \neq y} \frac{f(x)-f(y)}{|x-y|^{\alpha}} < \infty \}$

a. We are to show that if $s = \frac{1}{2}n+\alpha$ then $|| \tau_x \delta - \tau_y \delta ||_{(-s)} < C_{\alpha} |x-y|^{\alpha}$. Instructions: We know that the Fourier transform of $\tau_x \delta$ is $e^{-2 \pi i \xi x}$. We are to write the integral defining ${|| \tau_x \delta - \tau_y \delta ||_{(-s)}}^2$ as the sum of the integrals over the regions $|\xi| \leq R$ and $|\xi| > R$ where $R = |x-y|^{-1}$ and we are to use the mean value theorem to estimate the Fourier transform of $\tau_x \delta - \tau_y \delta$ on the first region.

b. We are to show that if $s = \frac{1}{2}n+\alpha$ then $H_s \subset \Lambda_{\alpha}(R^n )$

c. We are to show that if $s = \frac{1}{2}n+a + k$ where k is some natural and $0 < a < 1$ then we have $H_s \subset \{ f \in C_0 ^ k | \partial ^ \alpha f \in \Lambda _{\alpha}(R^n) \wedge |\alpha| < k \}$

The rest I believe are all standard notations. My problem here is this was mentioned as a challenge problem and it was recommended to use part a to start off and deduce parts b and c and of course as luck would have it I cannot even interpret the instructions given in part a which are supposed to be a hint in order to do it so basically I have no clue on the related three parts (I cannot even approach any of them nor can I even assume one and do the others) so I am sorry but I do need the help on these, and truly appreciate the help given by anyone. Thanks to all of you.

As a warm-up, we first prove that a tempered distribution $f\in H^{s}(\mathbb{R}^{n})$ coincides with a $C_{0}(\mathbb{R}^{n})$ function (i.e. continuous and vanishing at $\infty$). Since the Fourier transform is an isomorphism on $\mathcal{S}'(\mathbb{R}^{n})$, it suffices to show that $\widehat{f}\in L^{1}(\mathbb{R}^{n})$. Observe that by Holder's inequality, \begin{align*} \int_{\mathbb{R}^{n}}\left|\widehat{f}(\xi)\right|d\xi&=\int_{\mathbb{R}^{n}}(1+\left|\xi\right|^{2})^{-s/2}(1+\left|\xi\right|^{2})^{s/2}\left|\widehat{f}(\xi)\right|d\xi\\ &\leq\left(\int_{\mathbb{R}^{n}}(1+\left|\xi\right|^{2})^{-s}d\xi\right)^{1/2}\left\|\Lambda_{s}f\right\|_{L^{2}}\\ &\lesssim_{n,\alpha}\left\|\Lambda_{s}f\right\|_{L^{2}} \end{align*} where we use that $2s=n+2\alpha>n$ and therefore the first factor is finite. In particular, we see that $\left\|f\right\|_{\infty}\lesssim_{n,\alpha}\left\|\Lambda_{s}f\right\|_{L^{2}}$.
We now show that $f$ is $\alpha$-Holder continuous. Since for $x,y$ with $\left|x-y\right|>1$, we have the trivial estimate $$\dfrac{\left|f(x)-f(y)\right|}{\left|x-y\right|^{\alpha}}\leq 2\left\|f\right\|_{L^{\infty}}\lesssim_{n,\alpha}\left\|\Lambda_{s}f\right\|_{L^{2}},$$ it suffices to consider $\left|x-y\right|\leq 1$. By basic properties of the Fourier transform and Fourier inversion, we may write \begin{align*} f(x)-f(y)&=\int_{\mathbb{R}^{n}}\widehat{f}(\xi)e^{2\pi ix\cdot\xi}d\xi-\int_{\mathbb{R}^{n}}\widehat{f}(\xi)e^{2\pi iy\cdot\xi}d\xi=\int_{\mathbb{R}^{n}}\widehat{f}(\xi)\left[e^{2\pi ix\cdot\xi}-e^{2\pi iy\cdot\xi}\right]d\xi \end{align*} So by Holder, \begin{align*} \left|f(x)-f(y)\right|&\leq\int_{\mathbb{R}^{n}}(1+\left|\xi\right|^{2})^{s/2}\left|\widehat{f}(\xi)\right|(1+\left|\xi\right|^{2})^{-s/2}\left|e^{2\pi i x\cdot\xi}-e^{2\pi iy\cdot\xi}\right|d\xi\\ &\leq\left\|\Lambda_{s}f\right\|_{L^{2}}\left(\int_{\mathbb{R}^{n}}(1+\left|\xi\right|^{2})^{-s}\left|e^{2\pi ix\cdot \xi}-e^{2\pi iy\cdot\xi}\right|^{2}d\xi\right)^{1/2} \end{align*} To estimate the integral factor, we follow the suggestion in the problem statement and break up the integral into the piece over the region $\left\{\left|\xi\right|\leq\left|x-y\right|^{-1}\right\}$ and the piece of the complement. By the mean value theorem, \begin{align*} \int_{\left|\xi\right|\leq\left|x-y\right|^{-1}}(1+\left|\xi\right|^{2})^{-s}\left|e^{2\pi i x\cdot\xi}-e^{2\pi i y\cdot\xi}\right|^{2}d\xi&\lesssim\int_{\left|\xi\right|\leq\left|x-y\right|^{-1}}(1+\left|\xi\right|^{2})^{-s}\left|\xi\right|^{2}\left|x-y\right|^{2} \end{align*} We seek an estimate for the integral factor in the last inequality. Making the change of variable $\eta=\left|x-y\right|\xi$, we see that \begin{align*} \left(\int_{\left|\xi\right|\leq\left|x-y\right|^{-1}}(1+\left|\xi\right|^{2})^{-s}\left|\xi\right|^{2}d\xi\right)&=\left|x-y\right|^{-n-2}\int_{\left|\eta\right|\leq 1}(1+\left|x-y\right|^{-2}\left|\eta\right|^{2})^{-s}\left|\eta\right|^{2}d\eta\\ &\lesssim\left|x-y\right|^{2s-n-2}\int_{\left|\eta\right|\leq 1}(1+\left|\eta\right|^{2})^{-s}\left|\eta\right|^{2}d\eta\\ &\lesssim_{n}\left|x-y\right|^{2\alpha-2} \end{align*} Taking $1/2$ roots gives us the desired estimate.
For the second integral, we make the change of variable $\eta=\left|x-y\right|\xi$ to obtain \begin{align*} \int_{\left|\xi\right|>\left|x-y\right|^{-1}}(1+\left|\xi\right|^{2})^{-s}\left|e^{2\pi i x\cdot\xi}-e^{2\pi iy\cdot\xi}\right|d\xi&\lesssim\int_{\left|\xi\right|>\left|x-y\right|^{-1}}(1+\left|\xi\right|^{2})^{-s}d\xi\\ &=\left|x-y\right|^{-n}\int_{\left|\eta\right|>1}(1+\left|x-y\right|^{-2}\left|\eta\right|^{2})^{-s}d\eta\\ &\lesssim_{n}\left|x-y\right|^{2\alpha}\\ \end{align*} Taking the $1/2$ root completes the argument.
For part (c), let $\alpha$ be a multi-index with $\left|\alpha\right|\leq k$. Recall that $\widehat{\partial^{\alpha}f}=(-2\pi i\xi)^{\alpha}\widehat{f}(\xi)$, where the partial derivative is in the distributional sense. Observe that \begin{align*} \int_{\mathbb{R}^{n}}(1+\left|\xi\right|^{2})^{(\frac{n}{2}+a)}\left|\widehat{\partial^{\alpha}f}(\xi)\right|^{2}d\xi&=\int_{\mathbb{R}^{n}}(1+\left|\xi\right|^{2})^{(\frac{n}{2}+a)}\left|(-2\pi i\xi)^{\alpha}\right|^{2}\left|\widehat{f}(\xi)\right|^{2}d\xi\\ &\lesssim_{n,\alpha,k}\int_{\mathbb{R}^{n}}(1+\left|\xi\right|^{2})^{\frac{n}{2}+a+\left|\alpha\right|}\left|\widehat{f}(\xi)\right|^{2}d\xi\\ &\leq\left\|\Lambda_{s}f\right\|_{L^{2}}, \end{align*} since $\left|\alpha\right|\leq k$. So by part (b), $\partial^{\alpha}f\in \Lambda_{a}(\mathbb{R}^{n})$. Recall from distribution theory that if the weak partial partial derivatives $\partial_{x_{i}}f\in C_{0}$ for $i=1,\ldots,n$, then $f$ is $C^{1}$ and $\partial_{x_{i}}f$ coincide with the strong (or classical) derivatives of $f$. Alternatively, one can show directly using the Fourier transform, dominated convergence, and induction to show that $\partial^{\alpha}f\in C^{1}$ for $\left|\alpha\right|<k$.
• @zbigniew2015: You're still using $\alpha$ twice. As the multi-index in $\partial^{\alpha}$ and as a scalar $\alpha\in(0,1)$. – Matt Rosenzweig Oct 15 '15 at 20:30
• Your are right, sorry I changed it, I changed it so it is a for the scalar and $\alpha$ for the multi-index. Could you please help with part c as I asked you, like mention how to approach that part? – kroner Oct 15 '15 at 20:43