While reading about Function Spaces here ; from the chain of inclusions I had some questions that: whether $\mathcal{S}(\mathbb R^{n})$ (the Schwartz class of functions) is included in $C_{0} (\mathbb R^{n})$ (Space of all continuous functions vanishing at infinity) AND if affirmative, is the inclusion is dense?? i.e. Can we approximate a function in $C_{0} (\mathbb R^{n})$ by a sequence of functions in $\mathcal{S}(\mathbb R^{n})$ uniformly???

I found an article here .

So my question is: Can we approximate a function in $C_{0} (\mathbb R^{n})$ by a sequence of functions in $\mathcal{S}(\mathbb R^{n})$ uniformly???


Yes, $\mathcal{S}(\mathbb{R}^n)$ is dense in $C_0(\mathbb{R}^n)$.

Let $f$ be in $C_0(\mathbb{R}^n)$. Recall that this implies that $f$ is uniformly continuous.

Your claim is easy using mollifiers. Let me show you how this works.

Pick a smooth positive function $\psi$ with support contained in the open unit ball $B_1(0)$ such that $\int_{\mathbb{R}^n} \psi = 1$. Notice that $\psi$ is clearly a Schwartz function. For $\delta>0$ we define $\psi_\delta(x)=\frac1{\delta^n} \psi(\frac x{\delta})$.

Then $\psi_{\delta}$ is supported inside $B_\delta(0)$ and $\int_{\mathbb{R}^n} \psi_\delta=1$ (this is why we multiplied by $\delta^{-n}$). So as $\delta$ tends to zero, $\psi_\delta$ becomes more concentrated around the origin. Define

$$f_\delta := f*\psi_\delta.$$

Then $f_\delta\in\mathcal{S}(\mathbb{R}^n)$, because $\psi_\delta$ is Schwartz. We claim that $f_\delta\to f$ in $C_0(\mathbb{R}^n)$ as $\delta\to 0$.

Let $\varepsilon>0$. Since $f$ is uniformly continuous, there exists $\delta_0>0$ such that $|f(x)-f(y)|<\varepsilon$ for all $|x-y|<\delta_0$. Thus, if $\delta<\delta_0$ we have for $x\in\mathbb{R}^n$ arbitrary:

$$|f*\psi_\delta(x)-f(x)| \le \int_{\mathbb{R}^n} |\psi_\delta(y)(f(x-y)-f(x))| dy< \varepsilon\int_{\mathbb{R}^n} \psi_\delta(y) dy=\varepsilon.$$

In the first inequality we used the triangle inequality and used that $\int_{\mathbb{R}^n} \psi_\delta = 1$. In the second inequality we used that $\psi_\delta$ is supported in $B_\delta(0)$.

This proves the claim.

  • $\begingroup$ So, a function in $C_{0}(\mathbb R^{n})$ can be approximated by a sequence of functions in $\mathcal S(\mathbb R^{n})$ uniformly; So,$\mathcal S(\mathbb R^{n})$ is dense in $C_{0}(\mathbb R^{n})$. Isn't it?? $\endgroup$ – user92360 Nov 23 '15 at 9:51
  • $\begingroup$ Yes, that is correct. $\endgroup$ – J.R. Nov 23 '15 at 9:55
  • $\begingroup$ So, with due respect may I suggest you to edit the first line of your answer..?? $\endgroup$ – user92360 Nov 23 '15 at 10:03
  • $\begingroup$ Oh, now I see, horrible typo. Sorry. $\endgroup$ – J.R. Nov 23 '15 at 10:05

There is something I don't understand in J.R.'s answer. I don't see why $f_\delta$ must be Schwartz. Take, in one dimension, $f(x)=1/(1+x^2)$. It's easy to see that for $\delta$ small one has $f_\delta(x) \ge f(x)/2$, hence $f_\delta$ is not Schwartz. Alternatively, the Fourier transform of $f_\delta$ is $$ \widehat{f_\delta} = \widehat f \; \widehat{\psi_\delta} = c e^{-|x|} \; \widehat{\psi_\delta} $$ which is not Schwartz, since it is not $C^\infty$. Thus $f_\delta$ cannot be Schwartz. Probably one must truncate $f$ before applying the mollifier.

  • $\begingroup$ Rather comment on his answer than answering yourself. $\endgroup$ – user370967 Apr 3 '17 at 8:37
  • $\begingroup$ I know, I tried that, but my reputation apparently is not good enough for posting a comment to somebody's answer. $\endgroup$ – brian0 Apr 3 '17 at 9:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.