Does there exists an approximate identity in Fréchet algebra $\mathcal{S}(\mathbb R)$?

Question

We put $\|f\|_{(N, \alpha)}:= \sup_{x\in \mathbb R} (1+|x|)^{\alpha} | D^{\beta}f(x)|; $ and he Schwartz space,

$S(\mathbb R): = \{f\in C^{\infty}(\mathbb R): \|f\|_{(N, \alpha)}< \infty , \forall \alpha, N \in \mathbb N \cup \{0\} \}.$

It is well-known that (1) $\mathcal{S}(\mathbb R)$ is a Fréchet space with the topology defined by the norms $\|\cdot\|_{(N, \alpha)}$ (2) $\mathcal{S}(\mathbb R) \ast \mathcal{S} (\mathbb R) \subset \mathcal{S}(\mathbb R).$ [For the proof you may see Folland , Real Analysis chapter 8]

And we may say $\mathcal{S}(\mathbb R)$ is a Fréchet algebra with respect to convolution.

My naive questions are:

(1) What is a notion of approximate identity in Fréchet algebra $\mathcal{S}(\mathbb R)$? How to define it? (I am familiar with the notion of approximate identity in $(L^1, \ast)$; and in fact in $L^{1},$ the approximate identity is uniformly bounded; for instance take we may take, $\phi_{t}(x)=t^{-1}\phi(t^{-1 }x), t>0$ for $\phi \in L^{1}$)

(2) Does there exists an approximate identity in $\mathcal{S}(\mathbb R)$? Is it uniformly bounded?

Answer 1

There's certainly an approximate identity in $\mathcal S$.

For $\phi\in\mathcal S$ and $t>0$ define $\phi_t(x)=t^{-1}\phi(x/t)$. Then if $\int\phi=1$ it follows that $\phi_t*f\to f$ in $\mathcal S$ for every $f\in\mathcal S$.

Say $\psi=\hat\phi$. It's easiest to verify that you have an approximate identity if you choose $\phi$ so that $\psi=1$ in a neighborhood of the origin. Note that $\hat\phi_t=\psi^t$, where $\psi^t(x)=\psi(tx)$. Since the Fourier transform is an isomorphism you only need to check that $$(1-\psi^t)\,\hat f\to0$$in $\mathcal S$, which is not too hard.

"Not too hard"... Ok, let's assume that $0\le\psi\le1$ and $\psi(x)=1$ for all $|x|\le 1$. Then $1-\psi^t$ vanishes on $[-1/t,1/t]$. So $$\sup_x|x|^N|(1-\psi^t(x))f(x)|\le\sup_{|x|>1/t}|x|^N|f(x)|\le t\sup_x|x|^{N+1}|f(x)|\to0.$$That takes care of $||.||_{N,0}$. I'll do $\alpha=1$; the rest is the same but the notation gets complicated. First,$$D((1-\psi^t)f)(x)=-\frac1t\psi'\left(\frac xt\right)f(x)+(1-\psi^t(x))f'(x)=\chi_t(x),$$say. Now $$\sup_x|x|^N|\chi_t(x)|=\sup_{|x|>1/t}|x|^N|\chi_t(x)|\le t^2\sup_x|x|^{N+2}|\chi_t(x)|\le ct.$$

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If by bounded you mean bounded as a subset of $\mathcal S$ then no it's not bounded.

Answer 2

Concerning bounded approximate identities: $\mathcal S$ is a Montel spaces, that is bounded sets are relatively compact ($\mathcal S$ is even nuclear). If it had a bounded approximate identity compactness would give a limit which then would be an identity element which certainly does not exist.