Denote $K(x)=\frac{1}{\sqrt{|x|}\cdot(1+x^2)}$ and $K_n(x)=nK(xn), n\in\mathbb{N}$. For $f\in L^1(R),$ define $T_nf(x)=\int_R K_n(x-y)f(y)\, dy.$ Prove or disprove that for every $f\in L^1(R), \sup_n |T_nf(x)|<\infty$ for Lebesgue almost every $x$.
I tried doing this: $||T_nf(x)||_{L^1}\leq\int_R\int_R |K_n(x-y)f(y)|\, dy\, dx=||K_n||_{L^1}||f||_{L^1}=||K_1||_{L^1}||f||_{L^1}<\infty.$ So $|T_nf(x)|<\infty$ a.e for each n. Then I don't know how to prove $\sup_n |T_nf(x)|<\infty$ a.e.