Equivalence of norms on Schwarz space

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 $$x^\alpha f$$.

• "Consider the following norms on the Schwarz space": Schwarz space is not a normed space. – Surb Apr 28 at 18:51
• In one dimension: $|f(x)|=|f(0)+\int_0^x f'(t)dt| \le |f(0)|+\|f'\|_1$ implies $\|f\|_\infty\le |f(0)|+\|f'\|_1$. – 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.
• 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. – Rhys Steele Apr 29 at 15:57