Is there a way to read a summability property of a family of functions from the behavior of coefficients of the corresponding Fourier transform?

Question

If $(K_N)_{N\in\mathbb{N}}$ (or indexed by any other directed set) is a family of functions in $L^1(\mathbb{T})$ defined on the 1-torus $\mathbb{T}$ such that: $$\forall p\in[1,+\infty), \forall f\in L^p(\mathbb{T}), \|K_N*f-f\|_p\to0, N\to+\infty$$ is there a way to read this property from the family of its Fourier transforms?

Some examples:

• $F_N(t):=\frac {1}{N+1}\left (\frac{\sin\left(\frac{N+1}{2}t\right)}{\sin\left(\frac{t}{2}\right)}\right)^2$ satisfies this property for $N\to+\infty$ and $\mathcal{F}(F_N)(n)=(1-\frac{|n|}{N})\chi_{[-N,N]}(n);$
• $P_r(t):= \frac{1-r^2}{|1-r e^{it}|^2}$ satisfies this property for $r\to 1^-$ and $\mathcal{F}(P_r)(n)=r^{|n|};$
• $K_s(t):=\sum_{n\in\mathbb{Z}}e^{-sn^2}e^{int}$ satisfies this property for $s\to0^+$ and obviously $\mathcal{F}(K_s)(n)=e^{-sn^2}$;
• $D_N(t):=\frac{\sin\left(\left(N+\frac{1}{2}\right)t\right)}{\sin\left(\frac{t}{2}\right)}$ satisfies this property for $N\to +\infty$ and $1<p<+\infty$ but not for $p=1$ and $\mathcal{F}(D_N)(n)=\chi_{[-N,N]}(n)$.

Now, in all these examples, the family of Fourier transforms converges boundedly to $1$ as the index goes to the proper limit, but I can't see any obvious (sufficient) condition where we can read why for the first three examples the condition is satisfied while it is not for the last one. Any ideas?

Answer 1

In fact

(i) $||f-K_N*f||_2\to0$ for every $f\in L^2$ if and only if (ii) $\sup_{n,N}|\widehat{K_N}(n)|<\infty$ and $\lim_{N\to\infty}\widehat{K_N}(n)=1$ for all $n$.

(i) implies (ii): Define $T_N:L^2\to L^2$ by $T_Nf=K_n*f$. Then $||T_N||=\sup_n|\widehat{K_N}(n)|$ and uniform boundedness implies that $\sup_N||T_N||<\infty$. If $e_n(t)=e^{int}$ then $T_Ne_n=\widehat{K_N}(n)e_n$, hence $\lim_N\widehat{K_N}(n)=1$.

(ii) implies (i): (ii) shows that $||K_N*P-P||_2\to0$ for every trigonometric polynomial $P$, and (i) follows since $||T_N||$ is bounded.

I don't believe there is any such simple characterization for $p\ne2$, just because if $T_N:L^p\to L^p$ is defined by $T_Nf=K_N*f$ nobody knows exactly what $||T_N||$ is in general (this would be a characterization of "Fourier multipliers" for $L^p$, which does not exist). Of course if $1\le p<\infty$ the same argument gives this:

$||K_N*f-f||_p\to0$ for every $f\in L^p$ if and only if $||T_N||$ is bounded and $\lim_N\widehat{K_N}(n)=1$ for every $n$,

and you can often deduce a yes or no for a specific $K_N$ from that, noting that $$\sup_n|\widehat{K_N}(n)|\le||T_N||\le||K_N||_1.$$

Of course if $p=\infty$ it's clear that $||T_N||=||K_N||_1$, but there cannot be any such result for $p=\infty$ because $||K_N*f-f||_\infty\to0$ is impossible, since $K_N*f$ is continuous (the proof fails because the trigonometric polynomials are not dense in $L^\infty$). Note however

$K_N*f\to f$ uniformly for every $f\in C(\Bbb T)$ if and only if $||K_N||_1$ is bounded and $\lim_N\widehat{K_N}(n)=1$ for all $n$;

I believe that's the only other nice clean result in this direction that exists.

This is enough information to settle many of the special cases you mention. For example, yes for the Fejer kernel and the Poisson kernel because the $L^1$ norm is bounded; no for the Dirichlet kernel operating on $C(\Bbb T)$ because the $L^1$ norm is not bounded. But for example nothing above settles the case of the Dirichlet kernel on $L^p$, $1<p<\infty$; that one happens to be a yes, but that's a theorem, not something we can just see by looking at the coefficients.