p-seminorms on smooth functions are equivalent 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?
 A: 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.
