Consider the following norms on the Schwarz space, for $1\leq q\leq \infty$ $$\lVert f \rVert_{\alpha,\beta, p}=\lVert x^{\alpha}\partial^\beta f\lVert_{L^p}$$

I want to show that the norms $\lVert \cdot \rVert_{\alpha,\beta, p}$ define the same topology as the standard one on Schwarz space where the topology is given by $$\lVert f \rVert_{\alpha,\beta}=\lVert (1+|x|)^{\alpha}\partial^\beta f\lVert_{u}$$

It suffices to show that either set of seminorms provide a local basis at 0 in the topology generated by the other.

For one direction we fixed $\alpha', \beta', \epsilon'$, and want to show there is some $\alpha, \beta, \epsilon$ such that $\{\Vert \cdot \rVert_{\alpha,\beta, p}<\epsilon\}\subset$ $\{\Vert \cdot \rVert_{\alpha',\beta'}<\epsilon'\}$.

For this we truncate the integral $\int|f(x)|dx$ by the unit disc and its complement, and then $\{\Vert \cdot \rVert_{0,0, p}<\epsilon\}$ is in $\{\Vert \cdot \rVert_{0,0}<\epsilon'\}\bigcap \{\Vert \cdot \rVert_{N,0}<\epsilon''\}$ for large enough $N$. Since $x^{\alpha}\partial^\beta f $ is in Schwarz space if $f$ is in Schwarz space, we have $\{\Vert \cdot \rVert_{\alpha,\beta, p}<\epsilon\}\subset$ $\{\Vert \cdot \rVert_{\alpha',\beta'}<\epsilon'\}$.

I don't know how to deal with the other direction. It seems I need to bound sup norm of a Schwarz function $f$ by $L^p$ norm of some $x^\alpha f$.

Thanks for your help.

  • $\begingroup$ "Consider the following norms on the Schwarz space": Schwarz space is not a normed space. $\endgroup$ – Surb Apr 28 at 18:51
  • $\begingroup$ In one dimension: $|f(x)|=|f(0)+\int_0^x f'(t)dt| \le |f(0)|+\|f'\|_1$ implies $\|f\|_\infty\le |f(0)|+\|f'\|_1$. $\endgroup$ – Jochen Apr 29 at 11:06

I suspect the direction you've done gets the inclusion the wrong way around. The estimate you easily get for $\|f\|_{L^p}^p$ is along the lines of $$\int |f|^p dx = \int_{B(1)} |f|^p dx + \int_{B(1)^c} |f|^p dx \leq \|f\|_{0,0} + \int_{B(1)^c} (1+|x|)^{-N} dx \|f\|_{N,0} = \|f\|_{0,0} + C \|f\|_{N,0}$$ where we chose $N$ (independent of $f$) large enough that $\int_{B(1)^c} (1+|x|)^{-N} dx < \infty$. This implies that $$\{\|\cdot\|_{0,0} < \frac{\varepsilon}{2}\} \cap \{\|\cdot\|_{N,0} < \frac{\varepsilon}{2C}\} \subseteq \{\|\cdot\|_{0,0,p} \leq \varepsilon\}.$$ It's then not too hard to replace the subscript $0,0,p$ with $\alpha,\beta,p$ for whatever $\alpha, \beta$ you want.

The other direction requires us to show that $\varepsilon'>0$ and $\alpha',\beta'$ we can find $\varepsilon$ and finitely many pairs $(\alpha,\beta)$ such that $$\bigcap_{\alpha,\beta} \{\|\cdot \|_{\alpha,\beta,p}<\varepsilon\} \subseteq \{\|\cdot\|_{\alpha',\beta'} < \varepsilon'\}.$$ This means we want to convert finitely many bounds on seminorms of the type $\|\cdot\|_{\alpha,\beta,p}$ into a bound on the seminorm $\|\cdot\|_{\alpha',\beta'}$. One way to do this is via Sobolev embeddings.

First let's demonstrate the idea in the case $\alpha' = \beta' = 0$. Here I will always take $\alpha = 0$ also. For $k > \frac{n}{p}$ (where $n$ is the dimension of your underlying Euclidean space), $W^{k,p}(\mathbb{R}^n)$ is continuously embedded in $C(\mathbb{R}^n)$ and so $\|\cdot\|_u \leq C \|\cdot\|_{W^{k,p}(\mathbb{R}^n)}$ for some constant $C$. Now $$\|\cdot\|_{W^{k,p}(\mathbb{R}^n)}^p = \sum_{|\beta| \leq k} \|\cdot\|_{0,\beta,p}^p$$ so let $N = |\{ \beta: |\beta| \leq k\}|$. Then the above shows that $$\bigcap_{|\beta| \leq k} \{ \|\cdot\|_{0,\beta,p} < \frac{\varepsilon'}{NC}\} \subseteq \{\|\cdot\|_{0,0} < \varepsilon'\}.$$ For general $\alpha',\beta'$, the difficult term to control is $\|x^{\alpha'}\partial^{\beta'}f\|_u$. We can do this using the above method as long as we have control on $\|\partial^{\gamma}(x^{\alpha'}\partial^{\beta'}f)\|_{L^p}$ for a suitable collection of multi-indices $\gamma$. This will follow from control on a finite family of the seminorms $\|\cdot\|_{\alpha,\beta,p}$ by the Leibniz rule.

  • $\begingroup$ Just a remark: if you knew from the start that both families of seminorms make $\mathcal{S}^\prime(\mathbb{R}^n)$ a Frechet space then it suffices to check one inclusion since the other would then follow immediately from the open mapping theorem. This trick often saves you a little bit of work. $\endgroup$ – Rhys Steele Apr 29 at 15:57
  • $\begingroup$ Thanks! I didn't notice that open mapping theorem applies here. $\endgroup$ – user136592 Apr 29 at 18:32

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