A sequence of functions $k_n\in L^1(\mathbb T),n\in \mathbb N,$ is called good kernel if it satisfies the following conditions:

$1)$ $$\frac{1}{2π} \int_{-π}^{π}k_n=1, \forall n \in \mathbb N$$ $2)$ $\exists M $ finite constance s.t. $$\lVert k_n\rVert_1\leq M,\forall n\in \mathbb N$$ $ 3)$ $\forall ε>0:$ $$\ \frac{1}{2π}\int_{π\ge \rvert x\lvert>ε}\rvert k_n(x)\lvert dx\rightarrow0.$$

Theorem: If $k_n$ is good kernel & $f\in C(\mathbb T )$ then $f*k_n\rightarrow f$ uniformly.

My question is: Is the opposite also true?

In other words, if $f\in C(\mathbb T )$ & $f*k_n\rightarrow f$ uniformly, does that implies that $k_n$ is a good kernel?

Note: $\mathbb T$ is the torus and can be represented as the interval $[-π,π]$ of 2π-$ \text {periodic} $ functions.

My approach:

The third condition is not always true,

Notice that if $k_n$ is a good kernel then $k_n*f\rightarrow f $ uniformly.

we define the functions $e_n(x):=e^{2πinx}$ & the kernel $F_n(x):=k_n(x) + e_n(x),$

Its clear that $|e_n(x)|=1$ , but : $$F_n*f(x)=k_n*f(x) + e_n*f(x)$$ $$\text {and}$$ $$e_n*f(x)=\int_{-π}^{π}f(t)e^{2πin(x-t)}dt=e^{2πinx} \hat f (n)\rightarrow 0$$

By the Riemann-Lebesgue lemma,uniformly $\forall x\in \mathbb T$

Hence $F_n*f\rightarrow f$ uniformly.

Is it correct?

  • 1
    $\begingroup$ Noting the fact that $C_c$ is dense in $L^p, \forall 1 \le p < \infty$ and combining it with Young's ineqauilty, at least it's true that such a sequence must consist of functions of $L^1$. Anyway, I think your argument by modification of a good kernel, by the characters seems ok to me. $\endgroup$
    – Brozovic
    Jul 31, 2020 at 15:06


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