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On $\mathbb{R}$, we define the Schwartz class functions as infinitely differentiable functions such that

$$ \lim\limits_{|x|\to \infty} | x^{m}f^{(n)}(x) | = 0 $$

for all $m, n \in \mathbb{N}$ and $f^{(n)}$ denotes the $n^{th}$ derivative of $f$.

However, I was asked to define Schwartz class on integers, and I came up with the following:

$$ \mathcal{S}(\mathbb{Z}) = \{ f : \mathbb{Z} \to \mathbb{C} \;\; | \;\; \lim\limits_{|n| \to \infty} n^{k}f(n) = 0 \;\; \forall\,\, k \in \mathbb{N} \}$$

The first examples that come to my mind are the restrictions of Schwartz functions on $\mathbb{R}$ to $\mathbb{Z}$. For example, $f(n) = e^{-n^{2}}$ being the restriction of the Gaussian function.

I was wondering if the converse is true. That is, given a function in $\mathcal{S}(\mathbb{Z})$, can it be extended to a function in $\mathcal{S}(\mathbb{R})$?

I was thinking that we could smoothly interpolate the function in between the integers, where it is already defined. The question remains about the rapid decay of this function. However, since original function goes to zero rather faster, I expect this to go to zero fast enough as well, since the slopes cannot change erratically in this case.

So, is the statement true? And if so, what is the justification?

Vishal Gupta
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  • At the first glance, I think the statement is not true. For a function in S(R), we know the information of derivative near the infinity. However, in you discrete case, I think we at least need the information of higher order of difference to make your idea works. – yaoxiao Aug 22 '13 at 12:58
  • @Vishal, you forgot $=0$ in the first equation. – njguliyev Aug 22 '13 at 13:00
  • @yaoxiao Could you provide some rationale? – Vishal Gupta Aug 22 '13 at 13:01
  • in you discrete case, I think we at least need the information of higher order of difference to make your idea works – yaoxiao Aug 22 '13 at 13:04
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    @yaoxiao: I respectfully disagree with you, I think that the analogy between $\mathcal{S}(\mathbb{R})$ and the space $\mathcal{S}(\mathbb{Z})$ defined here is complete. Indeed, we have that $$\mathcal{S}(\mathbb{Z})=\mathcal{F} (C^\infty (\mathbb{T})), $$ where $\mathbb{T}$ is the torus and $\mathcal{F}$ is the Fourier transform on it (namely, the operator which passes from a $2\pi$-periodic function to its Fourier coefficients). – Giuseppe Negro Aug 22 '13 at 13:40
  • @GiuseppeNegro Could you please add more details and write an answer? – Vishal Gupta Aug 22 '13 at 13:42
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    @Vishal: I don't think that it is worth an answer since it is only a (vague) observation. Let me clarify a little what I thought. We are asked to produce an "analogue of the Schwartz class on the integers" (which is already a rather vague request). The Schwartz class on the line is defined so that it extends the class of smooth functions with compact support and is left invariant by the Fourier transform. (I believe that it is also the smallest subspace of $L^1(\mathbb{R})$ with this property, but I have not proved this). [...] – Giuseppe Negro Aug 22 '13 at 18:58
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    [...] On the integer group $\mathbb{Z}$ things are different because the group is not self-dual (meaning that the Fourier transform you define on $L^2(\mathbb{Z})$, namely the map $$(c_n){n\in \mathbb{Z}}\mapsto \sum{n\in \mathbb{Z}} c_n e^{i n x}$$ takes values in $L^2(\mathbb{T})$ rather than in $L^2(\mathbb{Z})$ itself) and there is no concept of "smoothness". But on the dual group, the torus $\mathbb{T}$, there is such a concept and we have the space $C^\infty(\mathbb{T})$. Since $\mathbb{T}$ is compact, any function in $C^\infty(\mathbb{T})$ has compact support. [...] – Giuseppe Negro Aug 22 '13 at 22:28
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    [...] So we can guess that the "analogue of the Schwartz class on the torus" is the space $C^\infty(\mathbb{T})$ itself. If the analogy were correct, then the Fourier transform should map $C^\infty(\mathbb{T})$ onto the space $\mathcal{S}(\mathbb{Z})$ bijectively. And indeed this is true. That's what I meant above when I said "the analogy between $\mathcal{S}(\mathbb{Z})$ and $\mathcal{S}(\mathbb{R})$ is complete". – Giuseppe Negro Aug 22 '13 at 22:33

1 Answers1

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What GiuseppeNegro says is true. Indeed, let $f_0$ be in $\mathcal{S}(\mathbb Z)$ as defined above, and extend it to a function $f:\mathbb R\to\mathbb R$ as follows: between $n$ and $n+1$, let $f$ be equal to $f_0(n)$ on $\left[n,n+\frac{1}{3}\right]$, and equal to $f_0(n+1)$ on $\left[n+\frac{2}{3},n+1\right]$. In $\left[n+\frac{1}{3},n+\frac{2}{3}\right]$, extend $f$ linearly. Then, for all $x\in\mathbb R$, you have that $|f(x)|\leq\max\{|f([x])|,|f([x]+1)|\}$, where $[x]$ is the floor function of $x$.

Now, let $\phi$ be an infinitely differentiable function, with support in $\left[-\frac{1}{3},\frac{1}{3}\right]$ and integral equal to $1$, and define $g$ to be the convolution of $f$ and $\phi$: $g=f\ast\phi$. You can check that, if $k\in\mathbb N$, then the $k$-th derivative of $g$ at $x$ is $g^{(k)}(x)=\left(f\ast\phi^{(k)}\right)(x)$. Also, if $n\in\mathbb Z$, then $$g(n)=(f\ast\phi)(n)=\int_{\mathbb R}\phi(y)f(n-y)\,dy=\int_{-\frac{1}{3}}^{\frac{1}{3}}\phi(y)f(n-y)\,dy=\int_{-\frac{1}{3}}^{\frac{1}{3}}\phi(y)f(n)\,dy=f(n)=f_0(n),$$ so $g$ is equal to $f_0$ on $\mathbb Z$.

Let $k,l\in\mathbb N$. $|\phi^{(k)}|$ is bounded above by some $C_k>0$, so you have that $$\left|x^lg^{(k)}(x)\right|\leq|x|^l\int_{-\frac{1}{3}}^{\frac{1}{3}}\left|\phi^{(k)}(y)f(x-y)\right|\,dy\leq|x|^lC_k\int_{x-\frac{1}{3}}^{x+\frac{1}{3}}|f(y)|,$$ which is bounded by $$\frac{2}{3}C_k|x|^l\sup\left\{|f(y)|:x-\frac{1}{3}\leq y\leq x+\frac{1}{3}\right\}.$$ Then, using the property $|f(x)|\leq\max\{|f([x])|,|f([x]+1)|\}$ and the fact that $f_0$ is in $\mathcal{S}(\mathbb Z)$, you have that $$\lim_{|x|\to\infty}\left|x^lg^{(k)}(x)\right|=0,$$ so $g$ is in $\mathcal{S}(\mathbb R)$ and extends $f_0$.

detnvvp
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