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Let $K \subset R^n$ be compact, and let $C_0^{\infty}(K)$ be the space of smooth functions on with support in $K$.

For $p \in [1,\infty), \alpha$ multiindex, let $|f|_{\alpha,p} = (\int_K |D^{\alpha}f|^p)^{\frac{1}{p}}$, and $|f|_{\alpha,\infty} = sup_K |D^{\alpha}f|$ - as usual.

For $p \in [1,\infty]$, let $\tau_p$ be the locally convex topology generated on $C_0^{\infty}(K)$ by the seminorms $(|f|_{\alpha,p})_{\alpha}$.

Claim: All the $\tau_p$ are the same.

One proof of this might utilize the Sobolev theory (Kondrashow embedding theorem) which is too heavy as tool for my taste.

Unfortunately, I don't know a more elementary proof. Furthermore I guess a beautiful proof of this might work similarly for other spaces like the Schwartz-space or the space of smooth functions on the whole of $R^n$. Do you have any good idea?

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so does this mean showing that the seminorms that you have are within constant factors of each other? –  user1709 Nov 17 '10 at 21:27
    
Well, almost, because to go from low $p$ to high $p$ you need to sacrifice derivatives. –  Willie Wong Nov 17 '10 at 23:39
    
I'm also not convinced that the claim can be generalized to whole of $\mathbb{R}^n$; The function $(\sqrt{1 + x^2})^{-1/2}$ is a function in $W^{\infty,3}$ but not in $L^1$ of $\mathbb{R}$. –  Willie Wong Nov 17 '10 at 23:47
    
Also, you don't need the full strength of the Kondrachov embedding, you just need Sobolev embedding, whose proof (the GNS version) I happen to think is quite beautiful. –  Willie Wong Nov 17 '10 at 23:58
    
Where did you read this claim? My PDE knowledge does not yet go much further than the book by Evans. –  Jonas Teuwen Nov 18 '10 at 14:33

1 Answer 1

up vote 1 down vote accepted

Here's a remark added to original remarks below.

As mentioned below, Hölder's inequality reduces this to showing that $\tau_1$ is finer than $\tau_\infty$. This means showing that the $\tau_1$ to $\tau_\infty$ identity map is continuous, while we already know that its inverse is continuous. Hence if you prove that $\tau_1$ and $\tau_\infty$ are complete, this follows from the open mapping theorem.


I'm not sure this is worth posting, because I can only finish the easy case $n=1$, but here's a proof for that case from someone ignorant of Sobolev embedding.

If $1\leq p\leq q\leq\infty$, then because $K$ has finite measure, Hölder's inequality yields $|f|_{\alpha,p}\leq C|f|_{\alpha,q}$ for a constant $C$ depending only on $p$, $q$, and the measure of $K$. Thus $\tau_q$ is finer than $\tau_p$, and the work is in showing that $\tau_1$ is finer than $\tau_\infty$. Here, as Willie Wong pointed out, it is no longer possible to do so by directly comparing seminorms with the same multi-index.

Suppose that $(f_k)$ is a sequence of functions such that $|f_k|_{\alpha,1}\to 0$ for all $\alpha$ as $k\to\infty$. This implies that the same holds for the sequence $(D^{\alpha_0}f_k)$ for each fixed multi-index $\alpha_0$, so it is enough to check that $|f_k|_{0,\infty}\to0$, that is, $(f_k)$ converges uniformly to $0$. Here's a way to see this when $n=1$. Let $y$ be a point where all of the functions vanish, say $y=\inf K$ for definiteness. For each $x\in K$, $|f_k(x)|=|f_k(x)-f_k(y)|=|\int_y^x f_k'(t)dt|\leq \int_K|f_k'(t)|dt=|f_k|_{1,1}$, which goes to $0$ independent of $x$. Hence $(f_k)$ converges uniformly to $0$.

Actually, with slightly more work the proof that $\tau_1$ is finer than $\tau_\infty$ for $n=1$ can be extended to the case of smooth functions with arbitrary support. With the same assumptions on $(f_k)$, for each $x\in\mathbb{R}$, $|f_k(x)-f_k(0)|=|\int_0^x f_k'(t)dt|\leq \int|f_k'(t)|dt=|f_k|_{1,1}$, which goes to $0$ independent of $x$. This means that there is a sequence $(\varepsilon_k)$ of positive numbers converging to $0$ such that each $f_k$ is within $\varepsilon_k$ of the constant function with value $f_k(0)$. Since $|f_k|_{0,1}$ goes to $0$, the corresponding constants must converge to $0$, and this implies that $(f_k)$ converges uniformly to $0$. As Willie Wong also pointed out, you don't get $\tau_q$ finer than $\tau_p$ for $q\gt p$ in this case.

To extend this argument to $n\gt 1$, I would need bounds on $|f(x)-f(y)|$ in terms of volume integrals of sums of (a finite number of) partial derivatives of $f$, and this is where my ignorance keeps me from going further.

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Your approach is essentially the $n=1$ proof of the Sobolev inequality. For $n > 1$, you either appeal to Sobolev and Morrey's inequalities, or you reproduce them by hand. You can also use a modified version of the Sobolev inequality: you can control $|f_k|_{1,1}$ on the line connecting $f(x)$ and $f(y)$ by $|f_k|_{n+1,1}$ on the volume using the high-codimension trace inequality. The usual proof of which, however, goes through Morrey's inequality again. –  Willie Wong Nov 18 '10 at 12:09
    
@Willie Wong: Thank you. Part of what made me decide to post is the hope that someone would inform me on what I'm missing. Paging through Evans led me to suspect that Sobolev and Morrey inequalities would be useful, but I hadn't thought through the details of putting it together. –  Jonas Meyer Nov 18 '10 at 19:38

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