# If $k_n$ is good kernel & $f\in C(\mathbb T )$ then $f*k_n\rightarrow f$ uniformly opposite way

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?

• 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. Jul 31, 2020 at 15:06