Why does the metric on a Schwartz space generate the same topology as the family of seminorms

harmonic-analysisnormed-spacesschwartz-space

I am reading Rauch's "Partial Differential Equations", and he makes a jump I don't understand.

He defines the Schwarz space as the space of $C^\infty$ functions that decrease faster than any polynomial. He assigns it the topology generated by the seminorms
$$ \lVert f \rVert_{\alpha \beta} := \sup \lvert x^\alpha \partial^\beta f \rvert. $$

Then, he proceeds to define the metric given by
$$ \rho(f, g) := \sum 2^{-\lvert \alpha \rvert – \lvert \beta \rvert} \sigma(\lVert f – g \rVert_{\alpha \beta}),$$
where $\sigma$ is the function $t \mapsto 1 – \frac1{1+t}$, but (I presume?) something like $\min( \cdot, 1)$ would also work. It is easy to check that this is a metric, if we allow metrics to take infinite values. However, especially given some of the propositions that follow, I believe this is in fact always finite. For example, I wish to show that the topology generated by $\rho$ is the same as the one generated by the norms. I was able to check that any ball in any of the seminorms contains a $\rho$-ball, but the other implication (any $\rho$-ball contains an intersection of balls in the seminorm) eludes me. If I could show $\rho$ finite, however, the path is clear: given an $\varepsilon$-ball in the $\rho$ norm, simply split the series in a small tail (if we split it far enough, this tail can surely be made to be less than $\varepsilon/2$) and a finite number of terms, which we can make as small as desired by making the corresponding norms small.

If $\rho$ is always finite, how would I go about showing that? The only estimates I could think of were gross, but they seemed asymptotically more or less optimal:
$$ \rho \leq \sum_{a, b \geq 0} 2^{-a-b} n^a n^b = \sum_{j = 0}^\infty 2^{-j} n^j (j+1). $$

Where I bounded the number of multi-indices $\alpha$ with $\lvert \alpha \rvert = a$ by $n^a$. A counting argument would give me the exact number, but I don't think it makes much of a big difference for big $n$.

Anyway, the series above converges for $n = 1$, but not for bigger $n$. So either I'm missing a better estimate and $\rho$ does converge in general, or there is an argument I'm not seeing to show that $\rho$ generates the topology. Which is it?

Note: this question might seem similar to Frechet spaces: topology induced by metric is the same as the topology induced by the family of seminorms, but there is a big difference: the exponent is $- \lvert \alpha \rvert – \lvert \beta\rvert$ and not $-n$.

Best Answer

It looks as if the following should get you through (in dimension $n$, say):

$$ \sum_{\alpha,\beta} 2^{-|\alpha|-|\beta|} = \sum_{\alpha_1,...,\alpha_n, \beta_1,...,\beta_n \geq 0} 2^{-(\alpha_1+\cdots \alpha_n)-(\beta_1+\cdots+\beta_n)} = (\sum_{\alpha_1\geq 0}2^{-\alpha_1})^{2n} = 2^{2n} $$

Related Question