$\textbf{Exercise 12}$ : Consider the function $f(x)=x^2\sin(\frac{1}{x^2})$ for $x \neq 0$ and $f(0)=0$. Show that $f'$ exists for every $x\in[-1,1]$ but $f' \not \in{L^1[-1,1]}$
$\textit{Proof}$ :
For $x \neq 0$, $f'(x)=2x\sin(\frac{1}{x^2})-\frac{2}{x}\cos(\frac{1}{x^2})$...an algebraic combination of continuous functions defined for all nonzero $x$. To check $f'(0)$, we may look at the difference quotient directly:
$$\lim_{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim_{h \rightarrow 0} \frac{f(h)}{h}=\lim_{h \rightarrow 0} h\sin\left(\frac{1}{h^2}\right)=0$$
So, $f'$ exists everywhere in $[-1,1]$.
Now consider a collection of closed intervals of $(0,1]$ defined by the points $\{\alpha_k\}_{k=1}^\infty$ and $\{\beta_k\}_{k=1}^\infty$ where $\alpha_i \neq \beta_j \quad \forall i,j \in{\mathbb{Z}^+}$ On any one of these closed intervals, $f'$ is continuous, so we may apply the fundamental theorem of calculus which gives
$$\int_\alpha^\beta |f'(x)|dx=x^2\sin\left(\frac{1}{x^2}\right)_{\alpha}^{\beta}$$
Let $\alpha_k$ be s.t. $f(\alpha_k)=\alpha_k$ and $\beta_{k}$ s.t. $f(\beta_{k})=0$ i.e. the roots and local maximum values of $f$. Then, $\alpha_k= \pm \sqrt{\frac{2}{\pi (2k-1)}}$ and $\beta_k = \pm \sqrt{\frac{1}{\pi k}}$. Then,
$$\int_{-1}^1 |f'(t)|dt \geq \int_0^1 |f'(t)|dt \geq \sum_{k=1}^\infty \int_{\sqrt{\frac{2}{\pi (2k-1)}}}^{\sqrt{\frac{1}{\pi k}}}|f'(t)|dt = \sum_{k=1}^\infty x^2\sin\left(\frac{1}{x^2}\right)_{\sqrt{\frac{2}{\pi (2k-1)}}}^{\sqrt{\frac{1}{\pi k}}}$$
$$\geq \sum_{k=1}^\infty\frac{2}{\pi (2k-1)}= \frac{2}{\pi} \sum_{k=1}^\infty \frac{1}{2k-1} \geq \frac{2}{\pi}\sum_{k=2}^\infty\frac{1}{3k} = \frac{2}{3\pi} \sum_{k=2}^\infty \frac{1}{k} = +\infty$$
$\therefore$ $f' \not \in{L^1}$ $\blacksquare$
Monday, December 9, 2013
Thursday, December 5, 2013
Stein 1.26
$\textbf{Exercise 1.26}$ Given the conditions that $A \subseteq E \subseteq B$, $A,B \in{\mathcal{M}}$ where $\mathcal{M}$ is the $\sigma$-algebra of Lebesgue measurable sets, $m(A)=m(B) < \infty$ then $E\in{\mathcal{M}}$
Stategy: We wish to write $E$ as the union or intersection of some measurable sets.
$\textit{Proof}$ : Consider $B-A=B \cap A^c$. Since $A,B\in{\mathcal{M}}$, then by the $\sigma$-algebra structure of $\mathcal{M}$, $A^c\in{\mathcal{M}}$ ($\mathcal{M}$ closed under compliments) and $B \cap A^c\in{\mathcal{M}}$ ($\mathcal{M}$ closed under countable intersections). We may write $B=A \cup (B-A)$ and so $A \cap (B-A) = \emptyset$. The measure of a union of disjoint measurable sets is the sum of their measures (this is theorem 3.2) so, $m(A)+m(B-A)=m(B)$. Because of our assumption that $m(A)=m(B) < \infty$, then $m(B-A)=0$. Since $A \subseteq E \subseteq B$ then $E-A \subseteq B-A$. By the monotinicity of $m_*$, $m_*(E-A) \leq m_*(B-A)=0$. So, $m_*(E-A)=0 \Longrightarrow E-A \in{\mathcal{M}}$ since sets of outer measure 0 are measurable. Again using the $\sigma$-algebra structure of $\mathcal{M}$, $E= (E-A) \cup A$...a union of 2 measurable sets. $\therefore$ $E \in{\mathcal{M}} \quad \blacksquare$
Stategy: We wish to write $E$ as the union or intersection of some measurable sets.
$\textit{Proof}$ : Consider $B-A=B \cap A^c$. Since $A,B\in{\mathcal{M}}$, then by the $\sigma$-algebra structure of $\mathcal{M}$, $A^c\in{\mathcal{M}}$ ($\mathcal{M}$ closed under compliments) and $B \cap A^c\in{\mathcal{M}}$ ($\mathcal{M}$ closed under countable intersections). We may write $B=A \cup (B-A)$ and so $A \cap (B-A) = \emptyset$. The measure of a union of disjoint measurable sets is the sum of their measures (this is theorem 3.2) so, $m(A)+m(B-A)=m(B)$. Because of our assumption that $m(A)=m(B) < \infty$, then $m(B-A)=0$. Since $A \subseteq E \subseteq B$ then $E-A \subseteq B-A$. By the monotinicity of $m_*$, $m_*(E-A) \leq m_*(B-A)=0$. So, $m_*(E-A)=0 \Longrightarrow E-A \in{\mathcal{M}}$ since sets of outer measure 0 are measurable. Again using the $\sigma$-algebra structure of $\mathcal{M}$, $E= (E-A) \cup A$...a union of 2 measurable sets. $\therefore$ $E \in{\mathcal{M}} \quad \blacksquare$
Wednesday, December 4, 2013
Stein 3.32
$\textbf{Exercise 32}$ : Let $f: \mathbb{R} \longrightarrow \mathbb{R}$. Prove that $f$ is Lipschitz continuous i.e.
$\exists M>0$ s.t.
$$\frac{|f(x)-f(z)|}{|x-z|} \leq M \quad \forall x,z\in{\mathbb{R}} \quad (\star)$$
iff $f$ is absolutely continuous $\textit{and}$ $|f'(x)| \leq M$ for a.e. $x$.
$\textit{Proof}$
"$(\Longrightarrow)$" Assume that $f$ has the above Lipschitz condition.
Remark: we should first establish that $f$ is absolutely continuous and then used the fact that an absolutely continuous function has a first derivative almost everywhere. This first derivative is then bounded by the Lipschitz constant $M$.
Recall that $f$ is $\textit{absolutely continuous}$ if $\forall \epsilon>0$, $\exists \delta(\epsilon)>0$ s.t. for any collection of disjoint intervals $\{(a_i,b_i)\}_{i=1}^n$ defined by a partition that $\sum_{i=1}^n(b_i-a_i) \leq \delta \Longrightarrow \sum_{i=1}^n|f(b_i)-f(a_i)| \leq \epsilon$. With the Lipschitz condition $(\star)$ let $\delta=\frac{\epsilon}{M}$. Then, $|f(b_i)-f(a_i)|\leq M |b_i-a_i|=M \frac{\epsilon}{M}=\epsilon$. Let $|b_i-a_i| \leq \frac{\delta}{n}$. Then, $|f(b_i)-f(a_i)|\leq \frac{\epsilon}{n}$ and so $$\sum_{i=1}^n(b_i-a_i) \leq n \cdot \frac{\delta}{n}=\delta \Longrightarrow \sum_{i=1}^n|f(b_i)-f(a_i)| \leq n \cdot \frac{\epsilon}{n} =\epsilon$$ So, $f$ is absolutely continuous.
Now, $f$ has a bounded first derivative, which is defined a.e. This can be seen as follows: $\forall x,h \in{\mathbb{R}}$, $$\left|\dfrac{f(x+h)-f(x)}{(x+h)-x}\right| \leq M$$ This statement is true for all $x$ and $h$, so it is therefore true in the limit $$\lim_{h \rightarrow 0}\left|\dfrac{f(x+h)-f(x)}{h}\right| =|f'(x)| \leq M$$ "$(\Longleftarrow)$" Assume that $|f'(x)| \leq M$ and $f$ is absolutely continuous. By Theorem 3.11, an absolutely continuous function $f$ has the properties that $f'$ exists a.e., $f'\in{L^1}$ and $f$ can be represented with the integral of its derivative: $$f(b)-f(a)=\int_a^bf'(t)dt \quad \forall a,b \in{\mathbb{R}}$$ Using the triangle inequality, the assumption $|f'(t)| \leq M$, and the linearity of the Lebesgue integral, $$|f(b)-f(a)| = \left|\int_a^bf'(t)dt\right| \leq \int_a^b|f'(t)|dt \leq \int_a^b M dt = M\int_a^b dt=M|b-a|$$ $\therefore$ $f$ is Lipschitz continuous. $\blacksquare$
$\textit{Proof}$
"$(\Longrightarrow)$" Assume that $f$ has the above Lipschitz condition.
Remark: we should first establish that $f$ is absolutely continuous and then used the fact that an absolutely continuous function has a first derivative almost everywhere. This first derivative is then bounded by the Lipschitz constant $M$.
Recall that $f$ is $\textit{absolutely continuous}$ if $\forall \epsilon>0$, $\exists \delta(\epsilon)>0$ s.t. for any collection of disjoint intervals $\{(a_i,b_i)\}_{i=1}^n$ defined by a partition that $\sum_{i=1}^n(b_i-a_i) \leq \delta \Longrightarrow \sum_{i=1}^n|f(b_i)-f(a_i)| \leq \epsilon$. With the Lipschitz condition $(\star)$ let $\delta=\frac{\epsilon}{M}$. Then, $|f(b_i)-f(a_i)|\leq M |b_i-a_i|=M \frac{\epsilon}{M}=\epsilon$. Let $|b_i-a_i| \leq \frac{\delta}{n}$. Then, $|f(b_i)-f(a_i)|\leq \frac{\epsilon}{n}$ and so $$\sum_{i=1}^n(b_i-a_i) \leq n \cdot \frac{\delta}{n}=\delta \Longrightarrow \sum_{i=1}^n|f(b_i)-f(a_i)| \leq n \cdot \frac{\epsilon}{n} =\epsilon$$ So, $f$ is absolutely continuous.
Now, $f$ has a bounded first derivative, which is defined a.e. This can be seen as follows: $\forall x,h \in{\mathbb{R}}$, $$\left|\dfrac{f(x+h)-f(x)}{(x+h)-x}\right| \leq M$$ This statement is true for all $x$ and $h$, so it is therefore true in the limit $$\lim_{h \rightarrow 0}\left|\dfrac{f(x+h)-f(x)}{h}\right| =|f'(x)| \leq M$$ "$(\Longleftarrow)$" Assume that $|f'(x)| \leq M$ and $f$ is absolutely continuous. By Theorem 3.11, an absolutely continuous function $f$ has the properties that $f'$ exists a.e., $f'\in{L^1}$ and $f$ can be represented with the integral of its derivative: $$f(b)-f(a)=\int_a^bf'(t)dt \quad \forall a,b \in{\mathbb{R}}$$ Using the triangle inequality, the assumption $|f'(t)| \leq M$, and the linearity of the Lebesgue integral, $$|f(b)-f(a)| = \left|\int_a^bf'(t)dt\right| \leq \int_a^b|f'(t)|dt \leq \int_a^b M dt = M\int_a^b dt=M|b-a|$$ $\therefore$ $f$ is Lipschitz continuous. $\blacksquare$
Monday, December 2, 2013
Stein 2.19
$\textbf{Exercise 2.19}$ Let $f\in{L^1(\mathbb{R}^d)}$. For each $\alpha>0$ define $E_\alpha=\{x : |f(x)|>\alpha\}$. Prove that
$$\int_{\mathbb{R}^d}|f(x)|dx = \int_0^\infty m(E_\alpha)d\alpha$$
Strategy: this proof uses some techniques similar to the proof of corollary 3.8. The first trick is relate function value to measure of an interval: $|f(x)|=m([0,|f(x)|))$.
$\textit{Proof}$: Since $f$ integrable, the following equalities hold in the non-extended sense. Again relating integration and measure, we have $m([0,|f(x)|))=\int_\mathbb{R}\chi_{[0,|f(x)|)}$ $$\int_{\mathbb{R}^d}|f(x)|dx=\int_{\mathbb{R}^d}m[0,|f(x)|)dx =\int_{\mathbb{R}^d} \left(\int_\mathbb{R}\chi_{[0,|f(x)|)}(\alpha)d\alpha\right)dx \quad (\star)$$ with $\alpha$ acting as a dummy variable in the above right integral.
Observe that for $\alpha\in{\mathbb{R}}$ (or even $\alpha>0$) $$ \chi_{[0,|f(x)|)}(\alpha) = \left\{ \begin{array}{lr} 1 & : 0<\alpha< f(x) \\ 0 & : \alpha > |f(x)| \\ \end{array} \right. $$ $$ \chi_{E_\alpha}(x) = \left\{ \begin{array}{lr} 1 & : x\in{E_\alpha} \Longleftrightarrow \alpha< f(x) \\ 0 & : x\not\in{E_\alpha} \Longleftrightarrow f(x) \leq \alpha \\ \end{array} \right. $$ Note that since $f\in{L^1}$ then $f$ is measurable, implying by definition of measurable functions that the set $E_\alpha$ is measurable, which is iff $\chi_{E_\alpha}$ is measurable.
By Tonelli's theorem, and $\alpha>0$, $(\star)$ becomes: $$=\int_\mathbb{R} \left( \int_{\mathbb{R}^d}\chi_{E_\alpha}(x)dx \right) d \alpha =\int_\mathbb{R}m(E_\alpha)d\alpha = \int_0^\infty m(E_\alpha)d \alpha \quad \blacksquare$$
$\textit{Proof}$: Since $f$ integrable, the following equalities hold in the non-extended sense. Again relating integration and measure, we have $m([0,|f(x)|))=\int_\mathbb{R}\chi_{[0,|f(x)|)}$ $$\int_{\mathbb{R}^d}|f(x)|dx=\int_{\mathbb{R}^d}m[0,|f(x)|)dx =\int_{\mathbb{R}^d} \left(\int_\mathbb{R}\chi_{[0,|f(x)|)}(\alpha)d\alpha\right)dx \quad (\star)$$ with $\alpha$ acting as a dummy variable in the above right integral.
Observe that for $\alpha\in{\mathbb{R}}$ (or even $\alpha>0$) $$ \chi_{[0,|f(x)|)}(\alpha) = \left\{ \begin{array}{lr} 1 & : 0<\alpha< f(x) \\ 0 & : \alpha > |f(x)| \\ \end{array} \right. $$ $$ \chi_{E_\alpha}(x) = \left\{ \begin{array}{lr} 1 & : x\in{E_\alpha} \Longleftrightarrow \alpha< f(x) \\ 0 & : x\not\in{E_\alpha} \Longleftrightarrow f(x) \leq \alpha \\ \end{array} \right. $$ Note that since $f\in{L^1}$ then $f$ is measurable, implying by definition of measurable functions that the set $E_\alpha$ is measurable, which is iff $\chi_{E_\alpha}$ is measurable.
By Tonelli's theorem, and $\alpha>0$, $(\star)$ becomes: $$=\int_\mathbb{R} \left( \int_{\mathbb{R}^d}\chi_{E_\alpha}(x)dx \right) d \alpha =\int_\mathbb{R}m(E_\alpha)d\alpha = \int_0^\infty m(E_\alpha)d \alpha \quad \blacksquare$$
Stein 3.7
$\textbf{Exercise 3.7}$ Prove that if a measurable subset $E$ of $[0,1]$ satisfies
$$m(E \cap I) \geq \alpha m(I) \quad (\star)$$
for some $\alpha > 0$ and all intervals $I \in{[0,1]}$ then $m(E) = 1$
Strategy: Instead of appealing directly to corollary 1.5 , we consider $\chi_E$. If we can then show that $\chi_E=1$ a.e. for $E\in{[0,1]}$ then this implies that $m(E)=1$.
$\textit{Proof}$ : Since $I$ is an interval, then $m(I)>0$, so in $(\star)$ above, we can divide through to get $$0<\alpha \leq \frac{m(E \cap I)}{m(I)}$$ Consider $\chi_E$, which is of course integrable on [0,1]. Then, the Lebesgue differentiation theorem gives us that $$\lim_{x\in{I},m(I)\rightarrow0}\frac{1}{m(I)} \int_I\chi_E(x)dx=\chi_E \quad a.e. x\in{[0,1]}$$ Again, one of the key themes in these types of problems is to relate integration and measure. We may write $\int \chi_E = \int_E 1 = m(E)$. Further, we can write $\int_I \chi_E = \int_{I \cap E} 1 = m(E \cap I)$. So, our above expression becomes $$\lim_{x\in{I},m(I)\rightarrow0}\frac{1}{m(I)} \int_{(I \cap E)} 1 dx= \lim_{x\in{I},m(I)\rightarrow0}\frac{m(E \cap I)}{m(I)} = \chi_E \quad a.e. x\in{[0,1]}$$ $$\lim_{x\in{I},m(I)\rightarrow0}\frac{m(E \cap I)}{m(I)} \geq \alpha > 0$$ So, $$\chi_E > 0 \quad a.e.$$ However, characteristic functions take on only values of 0 and 1. So, $$\chi_E>0 \Longrightarrow \chi_E = 1 \quad a.e. \Longrightarrow \int \chi_E = 1 = m(E) \quad \blacksquare$$
Strategy: Instead of appealing directly to corollary 1.5 , we consider $\chi_E$. If we can then show that $\chi_E=1$ a.e. for $E\in{[0,1]}$ then this implies that $m(E)=1$.
$\textit{Proof}$ : Since $I$ is an interval, then $m(I)>0$, so in $(\star)$ above, we can divide through to get $$0<\alpha \leq \frac{m(E \cap I)}{m(I)}$$ Consider $\chi_E$, which is of course integrable on [0,1]. Then, the Lebesgue differentiation theorem gives us that $$\lim_{x\in{I},m(I)\rightarrow0}\frac{1}{m(I)} \int_I\chi_E(x)dx=\chi_E \quad a.e. x\in{[0,1]}$$ Again, one of the key themes in these types of problems is to relate integration and measure. We may write $\int \chi_E = \int_E 1 = m(E)$. Further, we can write $\int_I \chi_E = \int_{I \cap E} 1 = m(E \cap I)$. So, our above expression becomes $$\lim_{x\in{I},m(I)\rightarrow0}\frac{1}{m(I)} \int_{(I \cap E)} 1 dx= \lim_{x\in{I},m(I)\rightarrow0}\frac{m(E \cap I)}{m(I)} = \chi_E \quad a.e. x\in{[0,1]}$$ $$\lim_{x\in{I},m(I)\rightarrow0}\frac{m(E \cap I)}{m(I)} \geq \alpha > 0$$ So, $$\chi_E > 0 \quad a.e.$$ However, characteristic functions take on only values of 0 and 1. So, $$\chi_E>0 \Longrightarrow \chi_E = 1 \quad a.e. \Longrightarrow \int \chi_E = 1 = m(E) \quad \blacksquare$$
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