I'm studying Fourier analysis and have a question about approximate identities.

Let $k_{\epsilon}$ be an approximate identity on $L^{1}(\mathbf{T})$. We know that $k_{\epsilon}*f\to f$ in $L^{1}$ as $\epsilon\to 0$.

Question: Can we construct a $k_{\epsilon}$ such that for every $f\in L^{1}$, $k_{\epsilon}*f$ converges(not necessarily $f$) everywhere ?

I have asked this question on math.stackexchange but I didn't get an answer.

  • $\begingroup$ $k_\varepsilon =0$? And what do you mean by "everywhere"? $\endgroup$ Commented Sep 5, 2016 at 15:36
  • 1
    $\begingroup$ But $k_{\epsilon}=0$ is not an approximate identity... "everywhere" simply means "for every $x$". $\endgroup$
    – yun
    Commented Sep 5, 2016 at 15:44
  • 1
    $\begingroup$ But I didn't require $k_{\epsilon}*f$ converge to $f$ pointwise. The limit is not necessarily $f$. $\endgroup$
    – yun
    Commented Sep 5, 2016 at 15:58
  • $\begingroup$ Do you really want/need this convergence to work for all $f\in L^1$ or merely for some dense subspace of "nice" integrable functions? $\endgroup$
    – Yemon Choi
    Commented Sep 5, 2016 at 17:09
  • $\begingroup$ I hope it work for all $f\in L^1$. But if it is too strong, I hope there are some weaker results. $\endgroup$
    – yun
    Commented Sep 5, 2016 at 17:27

2 Answers 2


No. Given any $k_{\epsilon}$, we can easily build an $f$ so that $I(\epsilon):=k_{\epsilon}*f(0)$ oscillates roughly between $-1$ and $1$ as $\epsilon\to 0$. My $f$ will only take the two values $\pm 1$.

Just take $f=1$ everywhere except on a tiny interval centered at $0$, and take this interval so small that $I(1)\simeq 1$. Next, take $\epsilon_2\ll 1$ so small that $k_{\epsilon_2}$ is essentially supported by that tiny interval that is left, and let $f=-1$ there, except on an even smaller interval centered at $0$. Again, this will be chosen so small that $I(\epsilon_2)\simeq -1$ etc.


Not exactly an answer, but too long for a comment.

Assume there is such approximation to the identity. We may define a linear form $u\colon L^1(T)\to C$ by $$u(f)=\lim_{\varepsilon\to0} f*k_{\varepsilon}(0).$$

This linear form will be continuous when restricted to $L^\infty(T)$ because for $f\in L^\infty(T)$ $$|u(f)|=\lim_{\varepsilon\to0} |f*k_{\varepsilon}(0)|= \lim_{\varepsilon\to0} \Bigl|\frac{1}{2\pi}\int_{T}f(t)k_\varepsilon(-t)\,dt \Bigr|\le \Vert f\Vert_\infty.$$ Also for a continuous $f$ we will have $u(f)=f(0)$.

A continuous linear form in $L^\infty(T)$ is associated to a finitely additive measure $\mu$ on the measurable sets of $T$ by $$u(f)=\int_{T}f(t)\mu(dt)$$ If $\mu$ were countably additive it will be the Dirac delta (this is the only countably additive measure that gives integral $f(0)$ when $f$ is continuous), but this is not true because if $M$ is of Lebesgue measure $0$ we have $u(\chi_M)=u(0)=0$ and there is no problem assuming that $0\in M$.

These finitely additive measures defined on the measurable sets of $T$ vanishing on sets of measure zero but not countably additive can be proved to exists, but they are similar to non-measurable sets (I will call them phantoms). So they can not be so easily defined.

The hyperplane $u^{-1}(0)$ of $L^1(T)$, is
definable in terms of a sequence of ordinals. I think that in Solovay's model of ZFC, it can be proved that, in this case, it is closed. This means $u$ is continuous. But this is clearly false because then there will exists a function $g\in L^\infty(T)$ such that $$u(f)=\frac{1}{2\pi}\int_{T}f(t)g(t)\,dt.$$

So it is consistent with ZFC that there is no such approximation of the identity.

But I have forgotten now the details to show that in Solovay's model an hyperplane definable from a sequence of ordinals is closed. Perhaps it was that a projective hyperplane was closed. Perhaps somebody here can close my argument.


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