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 $\eta_kf\notin X$.

  • $\begingroup$ @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. $\endgroup$
    – username
    Commented Jun 30, 2014 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.) $\endgroup$ Commented Jun 30, 2014 at 0:29

1 Answer 1


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 $-\epsilon_k \varphi(x)$, $\varphi\in C_0^{\infty}$, $\int\varphi = 1$.

  • $\begingroup$ I see. Why do you need to shift $\varphi$, though? $\endgroup$
    – username
    Commented May 17, 2014 at 0:40
  • 2
    $\begingroup$ Actually I don't. $\endgroup$ Commented May 17, 2014 at 0:42

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