Let $f\in L^1(\mathbb{R}/2\pi\mathbb{Z})$, where $\mathbb{R}/2\pi\mathbb{Z}$ means that $f$ is periodic with period $2\pi$. Let $\sigma_N$ denote the Cesaro mean of the Fourier series of $f$. Suppose that $f$ has a left and right limit at $x$. Prove that as $N$ approaches infinity, $\sigma_N(x)$ approaches $\dfrac{f(x^+)+f(x^-)}{2}$.

We can write $\sigma_N(x)=(f\ast F_N)(x)$, where $F_N$ is the Fejer kernel given by $F_N(x)=\dfrac{\sin^2(Nx/2)}{N\sin^2(x/2)}$. We have $F_N(x)\geq 0$ and $\int_{-\pi}^{\pi}F_N(x)dx=2\pi$.

We want to show that $\left|\dfrac{f(x^+)+f(x^-)}{2}-(f\ast F_N)(x)\right|\rightarrow 0$ as $N\rightarrow\infty$.

How can we go from here?


Note that you can write $$(f*F_N)(x)-\frac{f(x^+)+f(x^-)}{2}=$$$$\int_{-\pi}^{0}f(x-y)F_N(y)\,dy-\frac{f(x^+)}{2}+\int_{0}^{\pi}f(x-y)F_N(y)\,dy-\frac{f(x^-)}{2}=$$$$\int_{-\pi}^{0}(f(x-y)-f(x^+)F_N(y)\,dy+\int_{0}^{\pi}(f(x-y)-f(x^-))F_N(y)\,dy,$$ since $F_N$ is symmetric around $0$, so its integral from $-\pi$ to $0$ is equal to its integral from $0$ to $\pi$, so they are both equal to $\frac{1}{2}$. Now, use that $|f(x-y)-f(x^+)|$ is small for $y$ small and negative, also $|f(x-y)-f(x^-)|$ is small for $y$ small and positive, and the fact that $F_N$ is nonnegative and goes uniformly to $0$ away from $0$.

  • $\begingroup$ I think I almost get it. But when $y$ is away from $0$, how would you bound $|f(x-y)-f(x^+)|$? There's no assumption here that $f$ is bounded. $\endgroup$ – PJ Miller Nov 1 '13 at 1:58
  • $\begingroup$ There you use that $f$ is in $L^1$. If you are away from $0$, then $F_N$ goes to $0$ uniformly, so its maximum goes to $0$. So, you bound the integral by the maximum of $F_N$ times the $L^1$ norm of $f$. $f(x^+)$ is a fixed number, so it does not cause a problem; it can come out of the integral. $\endgroup$ – detnvvp Nov 1 '13 at 5:33

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