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A theorem in my notes claims the following:

If $f\in L_1(\mathbb{T})$, and both $f(x+0) = \lim\limits_{t\to x^+} f(t)$ and $f(x - 0) = \lim\limits_{t\to x^-} f(t)$ exist, then

$$\lim_{n\to\infty}\sigma_n (f) = \frac{[f(x + 0) + f(x - 0)]}{2}.$$

The proof begins with defining

$$I:= \sigma_n(f) - \frac{[f(x + 0) + f(x - 0)]}{2}.$$

(1) Next it is noted that

$$I = \frac1{2\pi}\int_0^\pi K_n (t)\{[f(x + t) - f(x+0)] + [f(x - t) - f(x - 0)]\}\mathrm dt.$$

I just can't see it.

(2) In a previous argument we did show that $$[\sigma_n(f)](x) - f(x) = \frac1{2\pi}\int_{-\pi}^\pi [f(x + t) - f(x)]K_n (t)\mathrm dt.$$

Can someone help me understand how we obtain (1)? I'm not sure if it is a consequence of (2) or if it is unrelated.

Thank you.

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1 Answer 1

up vote 2 down vote accepted

We use the facts that $\frac 1{2\pi}\int_{-\pi}^{\pi}K_n(t)\,dt=1$, $K_n$ is even and $\displaystyle\sigma_n(f)(x)=\dfrac{1}{2\pi}\int_{-\pi}^{\pi}K_n(t)\,f(x+t)\,dt$. Hence $$\begin{align*} \sigma_n(f)-\frac{f(x+0)-f(x-0)}2&=\frac 1{2\pi}\int_{-\pi}^{\pi}K_n(t)\,f(x+t)\,dt -\frac 1{2\pi} \int_{-\pi}^{\pi}K_n(t)\frac{f(x+0)+f(x-0)}2dt\\\ &=\frac 1{2\pi}\int_0^{\pi}K_n(t)\,f(x+t)\,dt+\frac 1{2\pi}\int_{-\pi}^0K_n(t)\,f(x+t)\,dt\\\ &\phantom{=}-\frac 1{2\pi}\int_0^{\pi}K_n(t)\left[f(x+0)+f(x-0)\right]dt\\\ &=\frac 1{2\pi}\int_0^{\pi}K_n(t)\,f(x+t)\,dt+\frac 1{2\pi}\int_0^{\pi}K_n(t)\,f(x-t)\,dt\\\ &\phantom{=}-\frac 1{2\pi}\int_0^{\pi}K_n(t)\left[f(x+0)+f(x-0)\right] dt, \end{align*}$$ which is the expected result (so it's not really a consequence of (2), but the facts used to show (2) are the same that those which were used to show (1), if we add the evenness of $K_n$).

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