Exercise 16: Prove a pointwise version of Fej\'er's theorem: {\it If $f\in\mathscr R$ and $f(x+)$, $f(x-)$ exist for some $x$, then} $$\lim_{N\rightarrow\infty}\sigma_N(f;x)=\frac{f(x+)+f(x-)}{2}.$$

\def\eps{\varepsilon} Let $\varepsilon>0$ and let $\delta>0$ such that $\big|f(x-t)-f(x+)\big|<\varepsilon/2$ for $-\delta<t<0$ and $\big|f(x-t)-f(x-)\big|<\varepsilon/2$ for $0<t<\delta$. Since $$K_N(x)= \frac{1}{N+1}\cdot\frac{1-\cos(N+1)x}{1-\cos x}$$ is an even function, we have from Exercise 15(b) that $$\frac{1}{2\pi}\int_{-\pi}^0K_N(t)\,dt=\frac{1}{2\pi}\int_0^\pi K_N(t)\,dt=\frac{1}{2}.$$ Hence by the results of Exercise 15, \begin{align*} & \bigg|\sigma_N(f;x)-\frac{f(x+)+f(x-)}{2}\bigg| \\ & \quad\quad\le\frac{1}{2\pi}\int_{-\pi}^0\big|f(x-t)-f(x+)\big|K_N(t)\,dt + \frac{1}{2\pi}\int_0^\pi\big|f(x-t)-f(x-)\big|K_N(t)\,dt \\ & \quad\quad\le\frac{1}{2\pi}\frac{1}{N+1}\frac{2}{1-\cos\delta}\Bigg(\int_{-\pi}^{-\delta}\big|f(x-t)-f(x+)\big|\,dt+\int_\delta^\pi\big|f(x-t)-f(x-)\big|\,dt\Bigg)\;+ \\ & \quad\quad\phantom{\le}\;\;\frac{\varepsilon}{2}\Bigg(\frac{1}{2\pi}\int_{-\delta}^0K_N(t)\,dt+\frac{1}{2\pi}\int_0^\delta K_N(t)\,dt\Bigg) \end{align*} The two integrals in the first term are finite, so we can make it as small as we would like as $N\rightarrow\infty$, and the second term is less than $\varepsilon/2$. Hence $\sigma_N(f;x) \rightarrow\big(f(x+)+f(x-)\big)/2$ as $N\rightarrow\infty$.

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Principles of Mathematical Analysis

Exercise 8.16

Exercise 16: Prove a pointwise version of Fejér’s theorem: If $f \in R$ and $f (x +)$ , $f (x -)$ exist for some $x$ , then

\lim_{N \to \infty} σ_{N} (f; x) = \frac{f (x +) + f (x -)}{2} .

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Let $𝜀 > 0$ and let $δ > 0$ such that $| f (x - t) - f (x +) | < 𝜀 ∕ 2$ for $- δ < t < 0$ and $| f (x - t) - f (x -) | < 𝜀 ∕ 2$ for $0 < t < δ$ . Since

K_{N} (x) = \frac{1}{N + 1} \cdot \frac{1 - \cos (N + 1) x}{1 - \cos x}

is an even function, we have from Exercise 15(b) that

\frac{1}{2 π} \int_{- π}^{0} K_{N} (t) dt = \frac{1}{2 π} \int_{0}^{π} K_{N} (t) dt = \frac{1}{2} .

Hence by the results of Exercise 15,

\begin{array}{l} | σ_{N} (f; x) - \frac{f (x +) + f (x -)}{2} | \\ \leq \frac{1}{2 π} \int_{- π}^{0} | f (x - t) - f (x +) | K_{N} (t) dt + \frac{1}{2 π} \int_{0}^{π} | f (x - t) - f (x -) | K_{N} (t) dt \\ \leq \frac{1}{2 π} \frac{1}{N + 1} \frac{2}{1 - \cos δ} (\int_{- π}^{- δ} | f (x - t) - f (x +) | dt + \int_{δ}^{π} | f (x - t) - f (x -) | dt) + \\ \frac{𝜀}{2} (\frac{1}{2 π} \int_{- δ}^{0} K_{N} (t) dt + \frac{1}{2 π} \int_{0}^{δ} K_{N} (t) dt) \end{array}

The two integrals in the first term are finite, so we can make it as small as we would like as $N \to \infty$ , and the second term is less than $𝜀 ∕ 2$ . Hence $σ_{N} (f; x) \to (f (x +) + f (x -)) ∕ 2$ as $N \to \infty$ .

analambanomenos

2023-08-07 00:00

Exercise 8.16

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