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.
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.
There is an answer to this question in the book of Folland:
Folland, Gerald B. Real analysis: modern techniques and their applications. Vol. 40. John Wiley and Sons, 1999.
8.15 Theorem. Suppose $|\phi(x)| ≤ C(1 + |x|)^ {-n-\varepsilon} $for some $C, \varepsilon>0$ (which implies that $\phi \in L^1$) and $\int_{R^n} \phi (x) dx =a.$ If $f\in L^p$ $(1 \leq p \leq \infty),$ then $f * \phi_{t}(x) \to f(x)$ as $t \to 0$ for every $x$ in the Lebesgue set of $f$ — in particular: for almost every $x,$ and for every $x$ at which $f$ is continuous. Here $\phi_t(x):=t^{-n}\phi(t^{-1}x).$
