Let $f : \mathbb R \to \mathbb C$ be a $2\pi$ periodic function ; the $f$ is determined by its values on $[-\pi , \pi]$ . Let $\sigma_n(f)(x):=\dfrac 1n(S_0(f)(x)+...+S_n(f)(x)) , \forall x \in [-\pi , \pi] , \forall n \in \mathbb N$ , where $S_n(f)(x):=\sum_{k=-n}^n \hat f (k) e^{ikx} , \forall x \in [-\pi , \pi] , \forall n \in \mathbb N $, where $\hat f (k):=\dfrac 1 {2\pi}\int_{-\pi}^\pi f(t)e^{-ikt}dt , \forall k \in \mathbb Z$ .

Simplifying we can also see that $\sigma_n(f)=f * F_n$ , where $F_n(x):=\dfrac 1{2n\pi}\dfrac {\sin^2 (nx/2)}{\sin^2(x/2)} , \forall x \in [-\pi, \pi] ,\forall n \in \mathbb N$ .

Now I know that $F_n$'s are approximate identity , hence for every $1\le p<\infty , f \in L^p[-\pi , \pi] \implies ||f - \sigma_n(f)||_p \to 0$ as $n \to \infty$ and also $f \in C[-\pi ,\pi] \implies \sigma_n(f)$ converges to $f$ uniformly on $[-\pi , \pi]$ . My question is the following :

Let $f \in L^1[-\pi , \pi]$ and $f$ be continuous at some $x \in [-\pi , \pi]$ , then is it true that $\lim _{n\to \infty} \sigma_n(f)(x)=f(x) $ ?


1 Answer 1


Yes, we have pointwise convergence at continuity points for every sufficiently nice approximate identity. Given $\varepsilon > 0$, pick $\delta > 0$ with $\lvert y-x\rvert \leqslant \delta \implies \lvert f(y) - f(x)\rvert \leqslant \varepsilon$. Then split the integral into two parts, on $\lvert t\rvert \leqslant \delta$, the factor $f(x-t) - f(x)$ is small, and on the remaining part, the approximate identity is small:

\begin{align} \lvert (f\ast F_n)(x) - f(x)\rvert &= \Biggl\lvert \int_{-\pi}^{\pi} \bigl(f(x-t) - f(x)\bigr)F_n(t)\,dt\Biggr\rvert \\ &\leqslant \int_{-\delta}^{\delta} \lvert f(x-t) - f(x)\rvert \,\lvert F_n(t)\rvert\,dt + \int_{\delta < \lvert t\rvert \leqslant \pi} \bigl(\lvert f(x-t)\rvert + \lvert f(x)\rvert\bigr)\lvert F_n(t)\rvert\,dt \\ &\leqslant \varepsilon \lVert F_n\rVert_{L^1} + c_n\bigl(\lVert f\rVert_{L^1} + 2\pi \lvert f(x)\rvert\bigr) \end{align}

where $c_n = \sup \{ \lvert F_n(t)\rvert : \lvert t\rvert > \delta\}$. If $F_n$ is such that $c_n\to 0$, and $\lVert F_n\rVert_{L^1}$ is bounded, like that is the case for the Fejér kernels, this yields $\limsup \lvert (f\ast F_n)(x) - f(x)\rvert \leqslant \varepsilon K$ where $K$ is a bound for the norms of $F_n$ (for the Fejér kernels, we have $\lVert F_n\rVert_{L^1} = 1$ for all $n$). Since $\varepsilon > 0$ was arbitrary, it follows that $(f\ast F_n)(x) \to f(x)$ if $f$ is continuous at $x$.

  • $\begingroup$ Completely understood . Thanks a lot $\endgroup$
    – user456828
    Sep 3, 2017 at 13:47
  • $\begingroup$ One doubt ... why $c_n \to 0$ ? $\endgroup$
    – user456828
    Sep 3, 2017 at 14:51
  • $\begingroup$ For $\lvert t\rvert \geqslant \delta$, we have $$0 \leqslant F_n(t) \leqslant \frac{1}{2n\pi \sin^2 \frac{\delta}{2}}$$ here. $\endgroup$ Sep 3, 2017 at 14:54
  • $\begingroup$ Yes yes , thank you $\endgroup$
    – user456828
    Sep 3, 2017 at 14:56

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