Ch3. Differentiation and Integration (Ex.01 ~ Ex.11)

(a) Show that \(\{K_\delta\}_{\delta>0}\) is a family of good kernels.

proof

① \(\int_{\mathbb{R}^d} K_\delta(x)\,dx = 1\)

\[ \begin{aligned} \int_{\mathbb{R}^d} K_\delta(x)\,dx &= \int_{\mathbb{R}^d} \delta^{-d} \psi\left(\frac{x}{\delta}\right)\,dx \quad (\text{Let } z = \tfrac{x}{\delta} \Rightarrow dx = \delta^d dz) \\ &= \delta^{-d} \int_{\mathbb{R}^d} \psi(z)\, \delta^d dz = \int_{\mathbb{R}^d} \psi(z)\,dz = 1. \end{aligned} \]

② \(\int_{|x|>1} |K_\delta(x)|\,dx \le A \quad (A>0)\)

\[ \int_{\mathbb{R}^d} |K_\delta(x)|\,dx = \int_{\mathbb{R}^d} \delta^{-d} |\psi(z)|\,dz = \int_{\mathbb{R}^d} |\psi(z)|\,dz < \infty \quad (\because\ \psi \text{ is integrable}) \]

③ \(\lim_{\delta\to 0} \int_{|x|\ge \gamma} |K_\delta(x)|\,dx = 0\) for all \(\gamma>0\).

\(\psi\) is integrable, so there is \(r_\varepsilon>0\) such that \(\int_{|t|\ge r_\varepsilon} |\psi(t)|\,dt < \varepsilon\) for given \(\varepsilon>0\). (Prop. 2.1.12(ii)). Let \(z = \frac{x}{\delta}\), then \(\int_{|x|\ge \gamma} |K_\delta(x)|\,dx = \int_{|z|\ge \gamma/\delta} \delta^{-d} |\psi(z)|\,dz < \varepsilon\).

For given \(\gamma>0\), consider \(\delta>0\) with \(\delta r_\varepsilon < \gamma\).

\(\{x\in\mathbb{R}^d : |x|>\gamma \} \subset \{ x\in\mathbb{R}^d : |z| > \delta r_\varepsilon \}\) (since \(\gamma > \delta r_\varepsilon\)), so

\[ \int_{|x|\ge \gamma} |K_\delta(x)|\,dx = \int_{|z|\ge \gamma/\delta} \delta^{-d} |\psi(z)|\,dz < \varepsilon, \]

and \(\int_{|z|\ge r_\varepsilon} |\psi(z)|\,dz < \varepsilon\).

To sum up, for every \(\varepsilon>0\) there is \(\frac{\gamma}{\varepsilon}>0\) such that \(0<\delta<\frac{\gamma}{\varepsilon}\) implies \(\int_{|x|\ge\gamma} |K_\delta(x)|\,dx < \varepsilon\). So \(\lim_{\delta\to 0} \int_{|x|\ge\gamma} |K_\delta(x)|\,dx = 0\).

This holds for all \(\gamma>0\).

(b) Assume that \(\psi\) is bounded and supported on a bounded set. Show that \(\{K_\delta\}_{\delta>0}\) is an approximation to the identity.

proof

\(|\psi(t)| \le M\) and \(\psi(t) = 0\) for \(|t| > B \quad (M,B>0)\)

① \(|K_\delta(x)| \le M \delta^{-d}\)

When \(\left|\frac{x}{\delta}\right| \le B\),
\[|K_\delta(x)| = |\delta^{-d}\psi(x/\delta)| = \delta^{-d}|\psi(t)| \le M \delta^{-d}\]

When \(\left|\frac{x}{\delta}\right| > B\),
\[|K_\delta(x)| = 0 \le M \delta^{-d}\]

② \(|K_\delta(x)| \le C\,\delta^{-1}\frac{1}{|x|^{d+1}}\)

When \(\left|\frac{x}{\delta}\right| \le B\),
\[ \begin{aligned} |K_\delta(x)| &= \delta^{-d} |\psi(x/\delta)| = \delta^{-(d+1)} |\delta \psi(x/\delta)| \\ &\le \frac{B^{d+1}}{|x|^{d+1}} M \delta = C \delta \frac{1}{|x|^{d+1}} \quad (C = M B^{d+1}). \end{aligned} \]

When \(\left|\frac{x}{\delta}\right| > B\),
\[|K_\delta(x)| = 0 \le C\,\delta\frac{1}{|x|^{d+1}}\]

③ Since \(\int_{\mathbb{R}^d} K_\delta(x)\,dx = 1\), \(K_\delta\) is an approximation to identity.

(c) Show that \(\lim_{\delta\to 0} \|(f * K_\delta) - f\|_\infty = 0\) for good kernels (\(L^1 * L^\infty \to L^\infty\)).

proof

\[ \begin{aligned} &\|(f * K_\delta) - f\| \\ &= \int_{\mathbb{R}^d} |(f * K_\delta)(x) - f(x)|\,dx \\ &= \int_{\mathbb{R}^d} \left| \int_{\mathbb{R}^d} f(x-y)K_\delta(y)\,dy - f(x) \right| dx \quad (\text{def of } f*K_\delta) \\ &= \int_{\mathbb{R}^d} \left| \int_{\mathbb{R}^d} (f(x-y) - f(x)) K_\delta(y)\,dy \right| dx \\ &\le \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} |f(x-y) - f(x)| |K_\delta(y)|\,dy\,dx. \end{aligned} \]

(\(f\) is measurable on \(\mathbb{R}^d\), so \(f(x-y)\) and \(f(x)\) are measurable on \(\mathbb{R}^{2d}\);
\(K_\delta\) is measurable on \(\mathbb{R}^d\), as well as on \(\mathbb{R}^{2d}\) (Prop 2.3.9, Cor 2.3.7)
\(\Rightarrow\) Tonelli’s theorem)

\[ \begin{aligned} &... \\ &= \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} |f(x-y) - f(x)| |K_\delta(y)|\,dx\,dy. \\ &= \int_{|y|\le \gamma} \int_{\mathbb{R}^d} |f(x-y) - f(x)| |K_\delta(y)|\,dx\,dy \quad (\text{for any } \gamma>0) \;+\; \int_{|y|>\gamma} \int_{\mathbb{R}^d} |f(x-y) - f(x)| |K_\delta(y)|\,dx\,dy. \end{aligned} \]

① \(\int_{|y|\le \gamma} \int_{\mathbb{R}^d} |f(x-y) - f(x)| |K_\delta(y)|\,dx\,dy\)

\[ \begin{aligned} &... \\ &= \int_{|y|\le \gamma} |K_\delta(y)| \left( \int_{\mathbb{R}^d} |f(x-y) - f(x)|\,dx \right) dy \\ &= \int_{|y|\le \gamma} |K_\delta(y)| \|f(x-y) - f(x)\|_1 dy \end{aligned} \]

\(f \in L^1(\mathbb{R}^d)\) so \(\lim_{y\to 0} \|f(x-y) - f(x)\|_1 = 0\) (Prop 2.2.7). Then, for given \(\varepsilon>0\), there is \(\gamma_\varepsilon > 0\) such that \(|y| < \gamma_\varepsilon\) implies \(\|f(x-y) - f(x)\|_1 < \varepsilon\). Replace \(\gamma\) to \(\gamma_\varepsilon\), then

\[ \begin{aligned} \int_{|y|\le \gamma_\varepsilon} \int_{\mathbb{R}^d} |f(x-y) - f(x)|\,|K_\delta(y)|\,dx\,dy < \int_{|y|\le \gamma_\varepsilon} |K_\delta(y)| \cdot \varepsilon \, dy \le \varepsilon \int_{\mathbb{R}^d} |K_\delta(y)|\,dy \le A \cdot \varepsilon \end{aligned} \]

② \(\int_{|y| > \gamma_\varepsilon} \int_{\mathbb{R}^d} |f(x-y) - f(x)| |K_\delta(y)|\,dx\,dy\)

\[ \begin{aligned} &... \\ &= \int_{|y| > \gamma_\varepsilon} |K_\delta(y)| \left( \int_{\mathbb{R}^d} |f(x-y) - f(x)|\,dx \right) dy \\ &\le \int_{|y| > \gamma_\varepsilon} |K_\delta(y)| \left( \int_{\mathbb{R}^d} (|f(x-y)| + |f(x)|)\,dx \right) dy \\ &= \int_{|y| > \gamma_\varepsilon} |K_\delta(y)| \cdot 2\|f\|_1 \, dy \quad (\because \text{translation invariance}) \\ &= 2\|f\|_1 \int_{|y| > \gamma_\varepsilon} |K_\delta(y)| \, dy \end{aligned} \]

Note that \(\lim_{\delta\to 0} \int_{|y| > \gamma_\varepsilon} |K_\delta(y)|\,dy = 0\) for all \(\gamma_\varepsilon > 0\).

③ Conclusion

Then, \[\|(f * K_\delta) - f\|_1 \le A\varepsilon + 2\|f\|_1 \int_{|y| > \gamma_\varepsilon} |K_\delta(y)|\,dy\] and

\[\limsup_{\delta\to 0} \|(f * K_\delta) - f\|_1 \le A\varepsilon + 0\]

This holds for all \(\varepsilon>0\), so \(\limsup_{\delta\to 0} \|(f*K_\delta) - f\|_1 = 0\) and thus \[\lim_{\delta\to 0} \|(f*K_\delta) - f\|_1 = 0\] (∵ \(\liminf_{\delta\to 0} \|f*K_\delta - f\|_1 \ge 0\)).

(a) \(-x_n \in E\) for all \(n\)

proof

① \(\dfrac{m(B_n \cap E)}{m(B_n)} \ge \dfrac{2}{3} \quad \forall n \ge N_0\)

Let \(B_n = B_{1/n}(0)\), a ball centered at origin with radius \(\frac{1}{n}\). Since \(\lim_{n\to\infty} m(B_n) = 0\) and ‘0’ is a point of density, \(\lim_{n\to\infty} \frac{m(B_n \cap E)}{m(B_n)} = 1\). Then, there is \(N_0 \in \mathbb{N}\) such that \(1 - \frac{m(B_n \cap E)}{m(B_n)} < \frac{1}{3}\) for \(n \ge N_0\). So \(\frac{m(B_n \cap E)}{m(B_n)} \ge \frac{2}{3}\) holds for all \(n \in \mathbb{N}\) with \(\frac{1}{n} \le \frac{1}{N_0}\).

② \(E_n \cap (-E_n) \neq \varnothing\)

Let \(E_n = B_n \cap E\) when \(n \ge N_0\) and \(E_n = B_n \cap E\) when \(n < N_0\).

\(m(E_n) > \frac{2}{3} m(B_n) = \frac{2}{3} \cdot \frac{2}{n} = \frac{4}{3n}\) and \(m(-E_n) = m(E_n) > \frac{4}{3n}\).

Since \(-E_n = (-B_n) \cap (-E) = B_n \cap (-E)\), both \(E_n\) and \(-E_n\) are included in \(B_n\), with measure greater than \(\frac{4}{3n}\). Since \(m(B_n) = \frac{2}{n}\), the measure of \(E_n \cap (-E_n)\) is greater than \(\frac{8}{3n^2}\), which is positive. So \(E_n \cap (-E_n)\) is not empty.

③ \(-x_n \in E\)

Choose \(x_n \in E_n \cap (-E_n)\) other than 0. Since \(|x_n| < \frac{1}{n}\) (because \(x_n \in E_n \subset B_n\)), \(\lim_{n\to\infty} x_n = 0\).

It is obvious that \(x_n \in E_n \subset E\). Also, \(x_n \in -E_n \subset -E\), so \(-x_n \in E\).

(b) \(2x_n \in E\) for all \(n \in \mathbb{N}\).

proof

① \(\dfrac{m(B_n \cap E)}{m(B_n)} \ge \dfrac{2}{3} \quad \forall n \ge N_0\)

Let \(B_n = B_{1/n}(0)\), a ball centered at origin with radius \(\frac{1}{n}\). Since \(\lim_{n\to\infty} m(B_n) = 0\) and ‘0’ is a point of density of \(E\), \(\lim_{n\to\infty} \dfrac{m(B_n \cap E)}{m(B_n)} = 1\). Then, there is \(N_0 \in \mathbb{N}\) such that \(1 - \dfrac{m(B_n \cap E)}{m(B_n)} < \dfrac{1}{3}\) for \(n \ge N_0\). So \(\dfrac{m(B_n \cap E)}{m(B_n)} \ge \dfrac{2}{3}\) holds for all \(n \in \mathbb{N}\) with \(\dfrac{1}{n} \le \dfrac{1}{N_0}\).

② \(E_n \cap \tfrac{1}{2} E_n \neq \varnothing\).

Let \(E_n = B_n \cap E\). Then
\(m(E_n) = m(E_n \cap B_n) > \dfrac{2}{3n}\).

\(\tfrac{1}{2}E_n = \tfrac{1}{2}B_n \cap \tfrac{1}{2}E = B_{2n} \cap \tfrac{1}{2}E\), and
\(m(\tfrac{1}{2}E_n) = \tfrac{1}{2} m(E_n) > \dfrac{2}{3n}\) (∵ \(m(E_n) > \dfrac{4}{3n}\)).

\(E_n \cap B_{2n} = E_{2n}\), so
\(m(E_{2n}) > \dfrac{2}{3} m(B_{1/(2n)}(0)) = \dfrac{2}{3} \cdot \dfrac{1}{2n} = \dfrac{1}{3n}\) (∵ ①).

Both \(E_n\) and \(\tfrac{1}{2}E_n\) are included in \(B_n\), so \(E_n\) and \(\tfrac{1}{2}E_n\) intersect in a set of measure greater than \(\dfrac{1}{3n}\), which is positive. So \(E_n \cap (\tfrac{1}{2}E_n)\) is not empty and contains numbers other than 0.

③ \(2x_n \in E\)

Choose \(x_n \in E_n \cap (\tfrac{1}{2}E_n)\) other than 0. \(x_n \in B_n\), so \(|x_n| < \dfrac{1}{n}\) and thus \(\lim_{n\to\infty} x_n = 0\).

\(x_n \in E_n\) implies \(x_n \in E\), and \(x_n \in \tfrac{1}{2}E_n\) implies \(2x_n \in E\). So, \(x_n \in E\) and \(2x_n \in E\).

(i) \(f^*(x) \ge \dfrac{C}{|x|^d}\) for some \(C>0\) and \(|x|>1\)

proof

① \(|f| \neq 0\), so \(\int_{\mathbb{R}^d} |f| > 0\) and there is a ball \(B_r \subset \mathbb{R}^d\) such that \(\int_{B_r} |f| > 0\).

②

Choose \(x \in \mathbb{R}^d\) such that \(|x| \ge 1\). Let \(\delta = r + |x|\), then \(B_r \subset B_\delta(x)\) and

\[ \begin{aligned} f^*(x) &\ge \frac{1}{m(B_\delta(x))} \int_{B_\delta(x)} |f(y)|\,dy \quad (\text{def of maximal function}) \\ &= v_d^{-1} (r + |x|)^{-d} \int_{B_\delta(x)} |f(y)|\,dy \quad (v_d : \text{volume of unit ball})\\ &\ge v_d^{-1}(r + |x|)^{-d} \int_{B_r} |f(y)|\,dy \quad (\because B_r \subset B_\delta(x))\\ &\ge v_d^{-1} a^d |x|^{-d} \int_{B_r} |f(y)|\,dy \quad (\text{let } r + |x| \le a|x|, a>0)\\ &\equiv \frac{C}{|x|^d} \quad (C = v_d^{-1} a^d \int_{B_r} |f(y)|\,dy), \quad (|x| \ge 1) \end{aligned} \]

(ii) \(f^*\) is not integrable on \(\mathbb{R}^d\)

proof

In exercise 2-10, we’ve shown that \(\dfrac{1}{|x|^b}\) is integrable iff \(b>d\). Thus \(\dfrac{1}{|x|^d}\) is not integrable; \(\int_{|x|\ge 1} \dfrac{1}{|x|^d} dx = \infty\).

Thus,

\[ \int_{\mathbb{R}^d} f^*(x)\,dx \ge \int_{|x|\ge 1} f^*(x)\,dx \ge C \int_{|x|\ge 1} \frac{1}{|x|^d} dx = \infty. \]

(iii) if \(f\) is supported on the unit ball \(B_1\), with \(\int_{B_1} |f| = 1\), then \(m(\{x : f^*(x) > \alpha\}) \ge \dfrac{C}{\alpha}\) for some \(C>0\) and all sufficiently small \(\alpha\). (Let \(F_\alpha = \{x : f^*(x) > \alpha\}\); let \(B_1 = \{ |x| \le 1\}\) w.l.o.g.)

proof

① \(x \in B_1\) : \(\dfrac{1}{m(B_1)} \int_{B_1} |f(y)|\,dy = \dfrac{1}{v_d}\), so \(\sup_{B \subset \mathbb{R}^d} \dfrac{1}{m(B)} \int_B |f(y)|\,dy \ge \dfrac{1}{v_d}\) and \(f^*(x) \ge \dfrac{1}{v_d}\).

② \(|x|>1\) : let \(B\) be a ball containing \(x\).

1. \(B \cap B_1 = \varnothing\) : \(\int_B |f| = 0\) so \(\dfrac{1}{m(B)} \int_B |f(y)|\,dy = 0\).

2. \(B \cap B_1 \ne \varnothing\).

Let \(B'\) be the smallest ball such that \(x \in B'\) and \(B_1 \subset B'\). The radius of \(B'\) is \(\tfrac{1}{2}(|x|+1)\). Then \[ \begin{aligned} \dfrac{1}{m(B')} \int_{B'} |f(y)|\,dy &= \dfrac{1}{\left(\frac{|x|+1}{2}\right)^d v_d} \int_{B'} |f(y)|\,dy \\ &\ge \left\{\left(\frac{|x|+1}{2}\right)^d v_d\right\}^{-1} \int_{B_1} |f(y)|\,dy \quad (\because B_1 \subset B') \\ &= \left(\frac{|x|+1}{2}\right)^{-d} v_d^{-1} = \dfrac{2^d}{(|x|+1)^d v_d}. \end{aligned} \]

③ \(f^* \ge\)

Then, \(f^*(x) \ge \dfrac{1}{v_d}\) (∵①) and \(f^*(x) \ge \dfrac{2^d}{(|x|+1)^d v_d}\) for \(|x|>1\) (∵②).

④ \(\{x : f^*(x) > \alpha\} \supset \left\{ x : \dfrac{2^d}{(|x|+1)^d v_d} > \alpha \right\}\) for \(0<\alpha<\dfrac{1}{v_d}\).

Consider two cases for \(x \in \left\{ x : \dfrac{2^d}{(|x|+1)^d v_d} > \alpha \right\}\).

When \(|x|>1\), we’ve shown that \(f^*(x) \ge \dfrac{2^d}{(|x|+1)^d v_d}\), so \(f^*(x) > \alpha\) (∵③). When \(|x| \le 1\), \(f^*(x) \ge \dfrac{1}{v_d} > \alpha\) (∵①), so \(f^*(x) > \alpha\).

For the both case, \(x \in \{x : f^*(x) > \alpha\}\).

⑤ \(m(F_\alpha) \ge \dfrac{C'}{\alpha}\) for some \(0 < C' < (2 - (\sqrt[v_d]{\alpha})^2)^d\) and all \(0 < \alpha < \dfrac{1}{v_d}\).

\(\{ x : f^*(x) > \alpha \} = \left\{ x : \dfrac{2^d}{(|x|+1)^d v_d} > \alpha \right\} = \left\{ x : |x| < \dfrac{2}{(\sqrt[v_d]{\alpha})} - 1 \right\}\).

So, \(m(F_\alpha) \ge v_d \left( \dfrac{2}{(\sqrt[v_d]{\alpha})} - 1 \right)^d\).

Note that \(v_d \left( \dfrac{2}{(\sqrt[v_d]{\alpha})} - 1 \right)^d \ge \dfrac{C'}{\alpha}\) \((C' > 0)\)

\(\Longleftrightarrow \left( \dfrac{2}{(\sqrt[v_d]{\alpha})} - 1 \right) \ge \left( \dfrac{C'}{v_d \alpha} \right)^{1/d}\) (and \(2 - (\sqrt[v_d]{\alpha})^2 > 0\))

\(\Longleftrightarrow \dfrac{2 - C'^{1/d}}{(\sqrt[v_d]{\alpha})} \ge 1 \quad \Longleftrightarrow \quad 0 < C'^{1/d} \le 2 - (\sqrt[v_d]{\alpha})^2\).

So choose some \(C'\) with \(0 < C' \le (2 - (\sqrt[v_d]{\alpha})^2)^d\) and \(0 < \alpha < \dfrac{1}{v_d}\). Then, \(m(F_\alpha) \ge \dfrac{C'}{\alpha}\) holds.

(a) Show that \(f\) is integrable

proof
\[ \begin{aligned} \int_{\mathbb{R}} |f(x)|\,dx &= 2\int_{0}^{1/2} \frac{1}{x\bigl(\log \tfrac{1}{x}\bigr)^2}\,dx \quad (\text{by symmetry}) \\ &= 2\int_{0}^{1/2} \frac{1}{x\bigl(\log \tfrac{1}{x}\bigr)^2}\,dx \quad (\text{let } t = \log \tfrac{1}{x},\ dt = -\tfrac{1}{x}dx) \\ &= 2\int_{\log 2}^{\infty} \frac{1}{t^2}\,dt = 2\left[ -\frac{1}{t} \right]_{\log 2}^{\infty} = 2\cdot(-\log \tfrac{1}{2}) = 2\log 2 < \infty. \end{aligned} \] So \(f\) is integrable.

(b) Show that \(f^*(x) \ge C (|x|\log \tfrac{1}{|x|})^{-1}\) for some \(C>0\) and all \(|x| < \dfrac{1}{2}\)

proof
Let \(B = B_{|x|}(x)\): a ball of radius \(|x|\) centered at \(x\) (either \((0,2x_1)\) or \((2x_1,0)\) in the picture). Then \[ \begin{aligned} \frac{1}{m(B)} \int_B |f(y)|\,dy &= \frac{1}{2|x|} \int_{0}^{2|x|} \frac{1}{y\bigl(\log \tfrac{1}{y}\bigr)^2}\,dy \\ &\ge \frac{1}{2|x|} \int_{|x|}^{2|x|} \frac{1}{y\bigl(\log \tfrac{1}{y}\bigr)^2}\,dy \quad (\text{restrict interval}) \\ &= \frac{1}{2|x|} \int_{\log \tfrac{1}{2|x|}}^{\log \tfrac{1}{|x|}} \frac{1}{t^2}\,dt \quad (\text{let } t = \log \tfrac{1}{y},\ dt = -\tfrac{1}{y}dy) \\ &= \frac{1}{2|x|} \left[ -\frac{1}{t} \right]_{\log \tfrac{1}{2|x|}}^{\log \tfrac{1}{|x|}}\\ &= \frac{1}{2|x|} \left( -\frac{1}{\log \tfrac{1}{|x|}} + \frac{1}{\log \tfrac{1}{2|x|}} \right) \;\\ &\ge\; \frac{1}{2}\,\frac{1}{|x|\log \tfrac{1}{|x|}} \end{aligned} \] for all sufficiently small \(|x|\). Hence \(f^*(x) \ge \dfrac{1}{2} \dfrac{1}{|x|\log \tfrac{1}{|x|}}\), so for example \(C = \dfrac{1}{2}\) works.

(c) Show that \(f^*\) is not locally integrable

proof
For \(0<\delta<\dfrac{1}{2}\), \[ \begin{aligned} \int_{0}^{\delta} f^*(x)\,dx &\ge \int_{0}^{\delta} \frac{1}{2x\log \tfrac{1}{x}}\,dx = \frac{1}{2}\int_{0}^{\delta} \frac{1}{x\log \tfrac{1}{x}}\,dx \quad (\text{use the lower bound from (b)}) \\ &= \frac{1}{2}\int_{\log \tfrac{1}{\delta}}^{\infty} \frac{1}{t}\,dt \quad (\text{let } t = \log \tfrac{1}{x},\ dt = -\tfrac{1}{x}dx) \\ &= \frac{1}{2}\left[ \log t \right]_{\log \tfrac{1}{\delta}}^{\infty} = \infty. \end{aligned} \] Therefore there exists \(\delta \in (0,\tfrac{1}{2})\) such that \(\int_{0}^{\delta} f^*(x)\,dx = \infty\), so \(f^*\) is not locally integrable.

(1) \(\delta(x+y) \le |y|\) whenever \(x \in F\)

proof \(\delta(x+y) \le |x+y-x| = |y|\) for \(x \in F\). Note that \(\delta(x) = 0\) for \(x \in F\).

(2) \(\delta(x+y) = o(|y|)\) a.e. \(x \in F\) ; \(\displaystyle \lim_{|y|\to 0} \frac{\delta(x+y)}{|y|} = 0 \text{ a.e. } x \in F\)

proof ① \(\delta\) is uniformly continuous

For given \(\gamma > 0\), there is \(z \in F\) such that \(|y-z| < \delta(y) + \gamma\) (::\(\,\)def of \(\delta\)).

\(\delta(x) \le |x-z| \le |x-y| + |y-z| < |x-y| + \delta(y) + \gamma\). So, \(\delta(x) - \delta(y) < |x-y| + \gamma\). Switching \(x\) and \(y\) yields \(\delta(y) - \delta(x) < |x-y| + \gamma\).

Then, \(|\delta(x)-\delta(y)| < |x-y| + \gamma\) holds for all \(\gamma > 0\), so \(|\delta(x)-\delta(y)| \le |x-y| \ \ (x,y \in \mathbb{R})\).

Then, for every \(\varepsilon > 0\), \(|\delta(x)-\delta(y)| < \varepsilon\) holds if \(|x-y| < \varepsilon\) satisfies. Thus, \(\delta\) is (uniformly) continuous on \(\mathbb{R}\).

② \(\delta\) is of bounded variation

Let \(a = t_0 < t_1 < \cdots < t_N = b\) be a partition of \([a,b]\) (with \(t_i - t_{i-1} = \dfrac{b-a}{N}\)). Then
\(\displaystyle \sum_{i=1}^N \left| \delta(t_i) - \delta(t_{i-1}) \right| \le \sum_{i=1}^N |t_i - t_{i-1}| = b-a\). So \(\delta\) is of bounded variation.

③ Conclusion

\(\delta(x)\) is non-negative, so \(0\) is a minimum of \(\delta\). Since \(\delta(x) = 0\) for all \(x \in F\), \(\delta\) achieves minimum at \(x \in F\).

\(\delta\) is of bounded variation, so \(\delta'\) exists a.e. and \(\delta'(x) = 0\) a.e. \(x \in F\). Thus,
\[ \begin{aligned} \lim_{|y|\to 0} \frac{|\delta(x+y) - \delta(x)|}{|y|} &= \lim_{|y|\to 0} \frac{|\delta(x+y)|}{|y|} = 0 \quad \text{a.e. } x \in F \end{aligned} \]

(i) Show that \(f\) is of bounded variation iff \(a>b\)

proof

① \(a>b \Rightarrow f\) is bv.

\(f'(x)=a x^{a-1}\sin(x^{-b})+x^a\cos(x^{-b})(-b)x^{-b-1} = a x^{a-1}\sin(x^{-b})-b x^{a-b-1}\cos(x^{-b})\)

\[ \begin{aligned} \int_0^1 |f'(x)|dx &\le a\int_0^1 |x^{a-1}\sin(x^{-b})|dx + b\int_0^1 |x^{a-b-1}\cos(x^{-b})|dx \\ &\le a\int_0^1 x^{a-1}dx + b\int_0^1 x^{a-b-1}dx \\ &= a\int_0^1 x^{-(a-1)}dx + b\int_0^1 x^{-(a-b+1)}dx \end{aligned} \]

\(-(a-1)=-a+1<1\), so \(\int_0^1 x^{a-1}dx<\infty\) (∵ ex 2-10).

\(a-b>0\), so \(\int_0^1 x^{a-b-1}dx<\infty\) for the same reason.

This leads to \(\int_0^1 |f'(x)|dx<\infty\), so \(f'\) is integrable and \(f\) is absolutely continuous.

Thus, \(f\) is of bounded variation (p.127).

② \(b\ge a \Rightarrow f\) is NOT bv.

Choose \(P_n: 0<\cdots<x_k<\cdots<x_2<x_1<1\) with \(x_k=\left(\frac{k\pi}{2}\right)^{-1/b}\ (k\in\mathbb{N}\ \text{s.t.}\ (\frac{k\pi}{2})^b\le1,\ k\le n)\)

\[ \begin{aligned} T_f(0,1) &\ge\sum_{k=1}^n |f(x_k)-f(x_{k-1})| \\ &=\sum_{k=1}^n\left|\left(\frac{k\pi}{2}\right)^{-a/b}\sin\frac{k\pi}{2}-\left(\frac{(k-1)\pi}{2}\right)^{-a/b}\sin\frac{(k-1)\pi}{2}\right| \\ &\ge\sum_{k=1}^n 2\left(\frac{\pi}{2}\right)^{-a/b}\left(\frac{1}{2k-1}\right)^{a/b} \end{aligned} \]

for some \(m\in\mathbb{N}\).

So, \[T_f(0,1)\ge\sum_{k=1}^\infty |f(x_k)-f(x_{k-1})| =2\left(\frac{\pi}{2}\right)^{-a/b}\sum_{k=1}^\infty\left(\frac{1}{2k-1}\right)^{a/b}\]

\(\sum_{n=1}^\infty \frac{1}{n^p}=\infty\) if \(p\le1\), and \[\sum_{k=1}^\infty\left(\frac{1}{2k-1}\right)^{a/b}\ge\sum_{k=1}^\infty\left(\frac{1}{2k}\right)^{a/b} =\left(\frac12\right)^{a/b}\sum_{k=1}^\infty\left(\frac{1}{k}\right)^{a/b}\], and \(\frac{a}{b}\le1\)

So, \[2\left(\frac{\pi}{2}\right)^{-a/b}\sum_{k=1}^\infty\left(\frac{1}{2k-1}\right)^{a/b}=\infty.\]

∴ \(f\) is not of bounded variation.

(ii) Let \(b=a\). Show that \(f\) follows \(|f(x)-f(y)|\le A|x-y|^\alpha\) \((0<\alpha<1)\), the Lipschitz condition of exponent \(\alpha\), and \(f\) is not of bounded variation.

proof

① \(|f(x+h)-f(x)|\le 2(x+h)^a\) & \(|f(x+h)-f(x)|\le 2a h\cdot \frac1x\)

For \(x\in[0,1]\) and \(h\in[0,1]\) s.t. \(x+h\le1\),
\[|f(x+h)-f(x)|\le |f(x+h)|+|f(x)|\le (x+h)^a+x^a\le 2(x+h)^a\]

Also, \(f'(x)=a x^{a-1}\sin(x^{-a})-a x^{-1}\cos(x^{-a})\), so
\[|f'(x)|\le a x^{a-1}|\sin(x^{-a})|+a x^{-1}|\cos(x^{-a})| \le a(x^{a-1}+x^{-1})\le 2a x^{-1}\ (x\in(0,1))\]

Then, \(|f(x+h)-f(x)|=|f'(c)\cdot h|\) (: MVT, \(c\in(x,x+h)\)) \(\le 2a h\cdot \frac1c\le 2a h\cdot \frac1x\).

② \(|f(x+h)-f(x)|\le A\cdot h^{\frac{a}{a+1}}\)

1. \(h\le x^{a+1}\iff h^{\frac1{a+1}}\le x\iff x^{-1}\le h^{-\frac1{a+1}}\)

\(|f(x+h)-f(x)|\le 2a h\cdot x^{-1}\le 2a h\cdot h^{-\frac1{a+1}} =2a\cdot h^{\frac{a}{a+1}}\).

2. \(h>x^{a+1}\iff x<h^{\frac1{a+1}}\)

\(|f(x+h)-f(x)|\le 2(x+h)^a\le 2(h^{\frac1{a+1}}+h)^a \le 2(2h^{\frac1{a+1}})^a\) (: \(0\le h\le1\)) \(=2^{a+1}\cdot h^{\frac{a}{a+1}}\).

\(2^{a+1}>2a\) for all \(a>0\), thus
\(|f(x+h)-f(x)|\le 2^{a+1}\cdot h^{\frac{a}{a+1}} =2^{a+1}\cdot h^\alpha\ (\alpha=\frac{a}{a+1})\).

③ not bounded variation

Since \(b>a\) does not hold, \(f\) is not bounded variation (∵ (i)).