Let us consider the Fréchet space $C^\infty\Bigl([0,1],\mathbb{R} \Bigr)$ of real-valued, periodic smooth functions.

That is, $f_n \to f$ in $C^\infty\Bigl([0,1],\mathbb{R} \Bigr)$ if $f^{(m)}_n$ converges uniformly on $[0,1]$ to $f^{(m)}$ as $n \to \infty$ and for each $m \in \mathbb{N} \cup \{0\}$.

Now, suppose that there exist two sequences $\{g_n\},\{K_m\} \subset C^\infty\Bigl([0,1],\mathbb{R} \Bigr)$ such that

  1. $K_m(x) \to \sum_{k \in \mathbb{Z}} \delta(x-k)$ as $m \to \infty$ in the sense of distributions.

  2. $\{ g_n * K_m \}$ is convergent in $C^\infty\Bigl([0,1],\mathbb{R} \Bigr)$ as $n \to \infty$ for each fixed $m$. Here $(g_n * K_m)(x):=\int_0^1 g_n(x-y)K_m(y)dy$.

Then, I wonder if $g_n$ itself is convergent in $C^\infty\Bigl([0,1],\mathbb{R} \Bigr)$ as well. Or at least, is there a convergent subsequence of $g_n$?

Edition : I am grateful for the counterexample provided by Iosif Pinelis. It certainly seems that I need further conditions. So, I assume that $g_n$ converges almost everywhere on $[0,1]$ to some $L^1$ function $g$. In this case, is $g$ in fact smooth and $g_n$ converges to $g$ in the Frechet topology defined above?

I may write a new post if this is too much of an addition.

  • $\begingroup$ It looks like the question is equivalent to one on the circle, where all distributions have Fourier expansions with polynomial-growth coefficients, and vice-versa. This might simplify things...? $\endgroup$ Jul 17, 2023 at 17:02
  • $\begingroup$ Just to avoid misunderstandings: What exactly do you mean by $C^\infty([0,1],mathbb{R})$? Do all derivatives extend to a continuous function on the closed interval? $\endgroup$
    – Christian
    Jul 17, 2023 at 17:15
  • $\begingroup$ @paulgarrett yes, it is on the circle. How does the polynomial growth simplify things?? $\endgroup$
    – Isaac
    Jul 17, 2023 at 17:57
  • $\begingroup$ @Christian all derivatives are periodic as well as the original function. So, they are in fact smooth on the real line, just like trigonometric functions. Sorry for the confusion. $\endgroup$
    – Isaac
    Jul 17, 2023 at 17:58
  • $\begingroup$ (1) cannot hold if taken at face value since the $K_m$ were introduced as functions on $[0,1]$. Are you referring to their periodic extensions here? $\endgroup$ Jul 17, 2023 at 18:07

2 Answers 2


$\newcommand{\Z}{\mathbb Z}\newcommand{\R}{\mathbb R}$No (assuming that your $C^\infty\big([0,1],\R\big)$ denotes the set of all smooth periodic functions of period $1$, endowed with the topology specified in your post).

E.g., let
\begin{equation} g_n(x):=\sin2\pi nx \tag{0}\label{0} \end{equation} and \begin{equation*} K_m(x):=\frac m2\,\sum_{k\in\Z}\big(L(m(x-k)-1)+L(m(x-k)+1)\big), \end{equation*} where $L$ is a smooth nonnegative function supported on $[-1,1]$ such that $\int_\R L=1$.

Here is the graph $\{(x,K_m(x))\colon-3.5<x<3.5\}$ for $m=20$ and $L(x)=ce^{-1/(1-x^2)}\,1(|x|<1)$, where $c:=1/\int_{-1}^1 dx\,e^{-1/(1-x^2)}$:

enter image description here

Then your condition 1 holds. Also, for $$h_{n,m}:=g_n*K_m$$ we have \begin{equation*} h_{n,m}(x)=\int_{x-1}^x dt\,K_m(x-t)g_n(t). \end{equation*} Therefore and because the function $K_m$ and all its derivatives vanish on $\Z$, for all $j=0,1,\dots$ we have \begin{equation*} h^{(j)}_{n,m}(x)=\sum_{k\in\Z}I_{n,m,k,j}(x)=\sum_{k=-1}^2I_{n,m,k,j}(x) \tag{10}\label{10} \end{equation*} for $m\ge2$, where \begin{equation*} \begin{aligned} I_{n,m,k,j}(x):=\frac{m^j}2\int_{x-1}^x & dt\, \sin2\pi nt\, \\ &\times\big(L^{(j)}(m(x-t-k)-1)+L^{(j)}(m(x-t-k)+1)\big), \end{aligned} \end{equation*} so that $I_{n,m,k,j}(x)=0$ if $k\in\Z\setminus\{-1,0,1,2\}$. Writing now $dt\, \sin2\pi nt=-d\dfrac{\cos2\pi nt}{2\pi n}$ and integrating by parts, we get \begin{equation*} |I_{n,m,k,j}(x)|\le C/n, \end{equation*} where $C$ is a positive real number not depending on $n$ or $x$. So, in view of \eqref{10}, $g_n*K_m=h_{n,m}$ converges to $0$ (as $n\to\infty$) in $C^\infty\big([0,1],\R\big)$. So, your condition 2 holds as well.

However, in view of \eqref{0}, $g_n$ does not converge in $C^\infty\big([0,1],\R\big)$. $\quad\Box$

  • $\begingroup$ Again, thank you very much for your amazing answer.. I in fact added further conditions that $g_n$ converges a.e. to some integrable function $g$. Then, perhaps the situation will change? Could you provide any insight? $\endgroup$
    – Isaac
    Jul 17, 2023 at 19:16
  • 1
    $\begingroup$ @Isaac : The answer would still be no: Replace $g_n(x):=\sin2\pi nx$ by (say) $g_n(x):=\frac1n\,\sin2\pi nx$. $\endgroup$ Jul 17, 2023 at 19:36
  • $\begingroup$ I see. At least, I can see that $g=0$ is smooth. I am more curious about regularity of $g$ than convergence to it. Perhaps, it may be possible that $g$ is necessarily smooth? $\endgroup$
    – Isaac
    Jul 17, 2023 at 20:13
  • 1
    $\begingroup$ @Isaac : I am not sure what you mean by $g$ here. $\endgroup$ Jul 17, 2023 at 20:37
  • 2
    $\begingroup$ @Isaac : I think the answer would still be no, but this will probably require a separate question and a separate answer. $\endgroup$ Jul 17, 2023 at 23:46

The Fourier series version is a variant of @IosifPinelis' argument:

Let's grant ourselves that we're talking about functions on the circle, and that smooth functions are the same as Fourier series with rapidly decaying coefficients. Somewhat more subtly, distributions have Fourier expansions with moderate-growth (slower than some polynomial) coefficients, and the evaluation map is the obvious restriction-and-extension of Parseval's theorem….

Convolution is (conveniently?) converted to multiplication: $$(\sum_m a_me^{2\pi imx})*(\sum_n b_ne^{2\pi inx})\;=\; \sum_\ell a_\ell b_\ell e^{2\pi i\ell x}. $$

In that context, perhaps even without using some of those facts, take $K_n(x)=\sum_{\lvert\ell\rvert\le n} 1\cdot e^{2\pi i\ell x}$ and $g_n(x)=e^{2\pi i(n+1)x}$, or maybe $n\cdot e^{2\pi i (n+1)x}$. Since the Dirac comb (= periodic $\delta$) has Fourier expansion $\sum_n 1\cdot e^{2\pi inx}$ (convergent in the sense of distributions, and, in fact, in a Sobolev space…), $K_n\to \delta$. Since $K_n*g_n=0$ for all $n$, it converges even in the strongest possible sense to the smooth function $0$. On the other hand, the $g_n$'s are not a Cauchy sequence in the topology on smooth functions… and by tweaking the constant, they're not Cauchy in $L^2$, etc.

EDIT: but/and, yes, various hypotheses on the $g_n$ do allow some conclusions. One significant mechanism is the "basic" fact that separately-continuous bilinear functionals on pairs of Frechet spaces (and maybe LF... or other colimits) are auto-magically jointly continuous. Thus, if the $g_n$'s have a limit, e.g., in some Sobolev space, and $K_n$'s have a limit in a Sobolev space (where periodic $\delta$ lives), then $K_n*g_n$'s limit must be the same as $\lim K_n * \lim g_n$, etc. In the example I gave, of $g_n=e^{2\pi inx}$, while this is not Cauchy in $L^2$, it is Cauchy in any negative-index $L^2$ Sobolev space.


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