3
$\begingroup$

Let $X$ be the closed subspace of Schwartz space $\mathcal{S}(\mathbb{R}^N)$ defined by

\begin{equation*} X=\left\{f\in\mathcal{S}(\mathbb{R}^N):\quad \int f\; dx=0\right\}. \end{equation*}

My question: Is $C_0^\infty(\mathbb{R}^N)\cap X$ dense in $X$, w.r.t. to the topology of $\mathcal{S}(\mathbb{R}^N)$?


Note that the usual way to approximate Schwartz functions by $C_0^\infty$-functions does not yield this, at least not directly:

Take $\eta_k$ to be radial functions on $[0,k]$ which decrease smoothly to zero on $[k,k+1]$. Then, for any Schwartz function $f$, we have $\eta_k f\rightarrow f$ in the topology of $\mathcal{S}(\mathbb{R}^N)$ and of course $\eta_k f\in C_0^\infty(\mathbb{R}^N)$. However, using the Fourier transform,

\begin{equation*} \int \eta_k f \;dx= \widehat{\eta_k f}(0)=(\hat{\eta}_k * \hat{f})(0). \end{equation*}

Now, a weak version of the uncertainty principle tells us that $\hat{\eta}_k$ cannot have compact support. Thus, even if $\hat{f}$ vanishes in an entire neighborhood around zero (which is actually the case I'm interested in), the convolution with $\hat{\eta}_k$ will generally ``smear out'' $\hat{f}$ so that $(\hat{\eta}_k * \hat{f})(0)\neq 0$, which means $\eta_kf\notin X$.

$\endgroup$
  • $\begingroup$ @Christian Remling:It's not clear to me that the approximation will work near the boundary of $B_{2^n}$ in the very strong Schwartz topology. What about using dominated convergence to prove that it is closed? Pointwise convergence is trivial and from uniform convergence of $(1+x^2)^{N/2+1}|f_n-f|$ one gets a dominating function. $\endgroup$ – M Lemm Jun 30 '14 at 0:20
  • $\begingroup$ @M Lemm: You're right, I was confused. In my "example", we would have to make $\varphi_n\sim 2^{-n}$ on $2^n<x<2^{n+1}$, and this part is $\sim 1/x$, so does not go to zero in $\mathcal S$. (The DC argument is fine, of course.) $\endgroup$ – Christian Remling Jun 30 '14 at 0:29
6
$\begingroup$

Yes. Your approximations (or any other reasonable method) will have small integrals $\epsilon_k=\int \eta_k f = O(k^{-N})$ for all $N$, so you can fix this by adding $-\epsilon_k \varphi(x)$, $\varphi\in C_0^{\infty}$, $\int\varphi = 1$.

$\endgroup$
  • $\begingroup$ I see. Why do you need to shift $\varphi$, though? $\endgroup$ – M Lemm May 17 '14 at 0:40
  • 2
    $\begingroup$ Actually I don't. $\endgroup$ – Christian Remling May 17 '14 at 0:42

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.