Question Given $K_n=\cases 0$ elsewhere , $n- n^2|x|$ for $x<|\frac 1n|$ , $f$ is continuous, $2\pi$ periodic $\Bbb R \to \Bbb C$ .
$f_n(x)=\int _{-\pi}^ \pi f(t)Kn(x-t)$
prove that $f_n\to^uf$.
Thoughts I know it has something to do with Fourier series and Fejer kernel. Since this is not the same kernel - I don't really know how to handle it.
Since $K_n(x-t)=0$ when $|x-t|>1/n$, the integral can be taken only over $|x-t|<1/n$:
$$ f_n(x) = \int_{x-1/n}^{x+1/n} f(t)\,K_n(x-t)\,dt $$ By continuity of $f$, for every $\epsilon$ there is $N$ such that $|f(t)-f(x)|<\epsilon$ when $|t-x|\le 1/N$. So, for every $n\ge N$ we have $$\begin{split} |f_n(x)-f(x)| &= \left|\int_{x-1/n}^{x+1/n} (f(t)-f(x))\,K_n(x-t)\,dt \right| \\&\le \int_{x-1/n}^{x+1/n} \left|f(t)-f(x)\right|\,K_n(x-t)\,dt \\ & \le \epsilon \int_{x-1/n}^{x+1/n} \,K_n(x-t)\,dt \\&=\epsilon \end{split}$$ where the first and last steps use the fact that the integral of $K_n$ is $1$. The second step is the integral triangle inequality.