Chapter 3. Differentiation and Integration
1. Differentiation of Integrals
The main goals of this chapter are showing the relation between integration and differentiation called as essential theorem of calculus stating that $F(x) -F(a)= \int_{a}^x \frac{\delta}{\delta y} F(y) dy$ and the condition guaranteeing that there exists $F’$ for an integrable function $F$.
There is an important proposition.
Proposition. Let $f$ be a continuous function on $\mathbb{R}^d$,then for all $x \in \mathbb{R}^d$,
$\underset{m(B)\rightarrow 0 , x \in B}{\lim} \frac{1}{m(B)} \int_B f(y) dy = f(x)$.
pf)
Let fix $\epsilon>0$. For each $x \in \mathbb{R}^d$, there is $\delta>0$ such that $\mid x-y \mid < \delta$ implies $\mid f(x)-f(y) \mid < \epsilon$
If we take a $B$ containing $x$ of radius $\frac{\delta}{2}$, then $\mid y - x \mid < \delta$ for all $y \in B$ and thus
$\mid \frac{1}{m(B)} \int_{B} f(y) dy -f(x)\mid \leq \frac{1}{m(B)} \int_B \mid f(y)-f(x) \mid dy \leq \frac{m(B) \epsilon}{m(B)}= \epsilon$.
To develop the relation, we’ll define the concept called as Hardy-Littlewood Maximal Function
Hardy-Littlewood Maximal Function
Let $f$ be integrable on $\mathbb{R}^d$. Define its maximal function denoted by $f^{\ast}$ such that
$f^{\ast}(x) = \underset{x \in B}{\sup} \frac{1}{m(B)}\int_B \mid f(y) \mid dy , x \in \mathbb{R}^d$.
Check that $\underset{m(B) \rightarrow 0}{\lim} f^{\ast}(x) = \mid f (x) \mid$.
$\def\ud#1#2#3{\underset{#1}{\overset{#2}{#3}}}$
There is another important lemma as follow.
Vitali covering Lemma
Suppose $B = {B_1,B_2,…,B_N}$ is a finite collection of open balls. Then $\exists$ a disjoint sub collection ${B_{j_1},B_{j_2},…,B_{j_k}}$ of $B$ that satisfies $m(\underset{j=1}{\overset{N}{\bigcup}}B_j \leq 3^d \ud{l=1}{k}{\sum}m(B_{j_l}))$
About the Hardy-Littlewood Maximal Function, there are theorems.
Suppose $f$ is integrable on $\mathbb{R}^d$, then
1) $f^{\ast}$ is measurable
2) $f^{\ast}(x)$ is bounded almost everywhere on $x$.
3) $f^{\ast}$ satisfies $m({x\in \mathbb{R}^d ; f^{\ast}(x) > \alpha }) \leq \frac{3^d}{\alpha} \parallel f \parallel_{L_1} , \forall \alpha >0$.
By using the above proposition, we can expand the above proposition into integrable function.
Lebesgue Differentiation Theorem
If $f$ is integrable on $\mathbb{R}^d$, then $\ud{m(B) \rightarrow 0,x \in B}{}{\lim} \frac{1}{m(B)}\int_{B}f(y)dy =f(x)$ almost every where on $x \in \mathbb{R}^d$
pf)
We’ll prove the proposition by showing that for a set $E_{\alpha} = {x \in \mathbb{R}^d ; \ud{m(B) \rightarrow 0 , x \in B}{}{\lim} \mid \frac{1}{m(B)} \int_Bf(y) dy - f(x) \mid > 2\alpha}$, for each $\alpha>0$$, m(E_{\alpha})=0$
Fix $\alpha>0,\epsilon>0$. There exists a continuous function $g$ supported on a compact set such that $\parallel f-g \parallel_{L^1} < \frac{\alpha \epsilon}{3^{d}+1}$
By the above proposition, $\ud{m(B)\rightarrow 0 , x \in B}{}{\lim} \frac{1}{m(B)} \int_{B} g(y) dy = g(x) \forall x$.
For a ball containing $x$,
$\frac{1}{m(B)} \int_{B} f(y) dy - f(x) = \frac{1}{m(B)} \int_B \mid f(y)-g(y) \mid dy + \frac{1}{m(B)}\int_B g(y) dy - g(x) +g(x) -f(x)$
Taking both sides the limit and superemum, we can get following inequality.
$\ud{m(B) \rightarrow 0, x \in B}{}{\lim \sup} \mid \frac{1}{m(B)} \int_B f(y) dy - f(x) \mid \leq (f-g)^{\ast}(x) + \mid g(x) - f(x) \mid$
Let define $F_{\alpha} = {x \in \mathbb{R}^d ; (f-g)^{\ast}(x) > \alpha}0$ and $G_{\alpha} = {x \in \mathbb{R}^d ; \mid g(x) - f(x) \mid > \alpha}$ then $E_{\alpha} \subset F_{\alpha} \cup G_{\alpha}$
We know that $m(F_\alpha) \leq \frac{3^d}{\alpha} \parallel f -g \parallel_{L^1}$ and $m(G_{\alpha}) \leq \frac{1}{\alpha} \parallel f- g \parallel_{L^1}$ by above theorem and Tchevysev’s inequality.
Therefore, $m(E_{\alpha}) \leq m(F_{\alpha})+m(G_{\alpha}) \leq \frac{3^d+1}{\alpha} \parallel f- g \parallel < \epsilon$.
This conclusion could be expanded into a set of locally integrable function. The locally integrable function means that for any compact set $K$, $f(x) \chi_{K}$ is integrable. Notice that it is more general set than the set of integrable function. For example, a constant function $C$ is locally integrable function but not integrable functions on $\mathbb{R}$.
Corollary. If $f$ is a locally integrable function, then $\underset{m(B) \rightarrow 0}{\lim} \frac{1}{m(B)}\int_B f(y)dy = f(x)$ almost everywhere on $x$.
We say that $x$ is a point of Lebesgue density of $E$, if $\ud{m(B) \rightarrow 0 , x \in B}{}{\lim}\frac{m(B \cap E)}{m(B)} = 1$.
Suppose $E$ is measurable, then
- Almost every point $x \in E$ is a point of Lebesgue density of $E$
- Almost every point $x \notin E$ is not a point of Lebesgue density of $E$
pf)
$\frac{m(B\cap E)}{m(B)}=\frac{1}{m(B)} \int_B \chi_{E}(y) dy \rightarrow \chi_E(x)$ as $m(B) \rightarrow 0$.
Let’s define the Lebegue set $\mathcal{L}f$ as a set of $\overline{x}$ such that $\ud{m(B) \rightarrow 0,\bar{x} \in B}{}{\lim} \frac{1}{m(B)}\int{B}\mid f(y)- f(\bar{x}) \mid dy =0$
Then we can know that if $f$ is continuous at $\bar{x}$, then $\bar{x}$ is in $\mathcal{L}f$. If $\bar{x} \in \mathcal{L}{f}$, then $\ud{m(B) \rightarrow 0,\bar{x} \in B}{}{\lim} \frac{1}{m(B)}\int_{B} f(y) dy =f(\bar{x})$
By corollary, if $f$ is locally integrable on $\mathbb{R}^d$, then almost every point belongs to $\mathcal{L}_f$.
3.2 Good kernels and Approximation to the identity
There is a concept called by kernels. By using the kernel and convolution, we can handle the functions in much more good conditions.
Good Kernels
Let \(\{k_{\delta}\}_{\delta>0}\) be a collection of integrable functions, we say \(k_{\delta}\) are good kernels if for \(\delta>0\)
- $\int k_{\delta}(x) dx = 1$
- $\int \mid k_{\delta}(x) \mid dx \leq A$ for some $A >0$ independent of $\delta$
- For every $\eta >0$, $\int_{\mid x \mid \geq \eta} \mid k_{\delta}(x) \mid dx \rightarrow 0$ as $\delta \rightarrow 0$.
There is a weaker concept than the Good kernels.
Approximation to the Identity
Let \(\{k_{\delta}\}_{\delta>0}\) be a collection of integrable functions, we say \(k_{\delta}\) are approximations to the idenity if for \(\delta>0\)
- \[\int k_{\delta}(x) dx = 1\]
\(\mid k_{\delta}(x) \mid \leq A \delta^{-d}\) for some \(A >0\) independent of \(\delta\) for all \(x \in \mathbb{R}^d\)
- \(\mid k_{\delta}(x) \mid \leq \frac{A \delta}{\mid x \mid ^{d+1}}\) for some \(A >0\) independent of \(\delta\) for all \(x \in \mathbb{R}^d\)
As a pointwise limit of $k_{\delta}$ as $\delta \rightarrow \infty$, we define Dirac Delta Function, denoted by $\delta(x)$ by $\delta(x) = \begin{cases}\infty \mbox{ if }x =0 \\ 0 \mbox{ otherwise } \end{cases}$ and $\int \delta(x) dx =1$.
Remark that $(f \ast \delta) (x) = \int f(y) \delta(x-y)dy = f(x)$ and therefore, $(f \ast k_{\delta})(x) \rightarrow f$ as $\delta \rightarrow 0$ pointwisely for a bounded function $f$.
The following two functions are approximation to identity.
Example. (Poisson Kernel)
Let $P_{y}$ be the Poisson kernel defined by $P_{y}(x)=\frac{1}{x}\frac{y}{x^2+y^2}, x \in \mathbb{R}$.
Example. (Fejer Kernel)
Let $F_N, N \in \mathbb{N}$ be the Fejer kernel defined by $F_N(x) = \begin{cases} \frac{1}{2 \pi N} \frac{\sin^2(Nx/2)}{\sin^2(x/2)} \quad \mbox{ if } \mid x \mid \leq \pi \\ 0 \qquad \qquad \qquad \mbox{ o.w. }\end{cases}$
Now, we’ll expand the result $(f\ast k_{\delta}) \rightarrow f$ into integrable function and uniform continuous.
Lemma
Suppose \(f\) is integrable and \(x \in \mathcal{L}_f\). Let \(A(r)=\frac{1}{r^d} \int_{\mid y \mid \leq r} \mid f(x-y)-f(x) \mid dy\) whenever \(r>0\). Then \(A(r)\) is continuous for \(r>0\) and \(A(r) \rightarrow 0\) as \(r \rightarrow 0\). Moreover, \(A(r)\) is bounded.
By using the lemma, we can prove a following theorem.
Theorem (Pointwise Convergence)
Let \(f\) be integrable and let \(\{k_{\delta}\}\) be an approximation to the identity. Then \((f \ast k_{\delta})(x) \rightarrow f(x)\) as \(\delta \rightarrow 0\) for \(x \in \mathcal{L}_f\). The limit hold for almost every \(x\).
pf)
Fix $x \in \mathcal{L}_f$.
A direct computation gives $\mid (f\ast k_{\delta})(x) -f(x) \mid \leq \int \mid f(x-y) -f(x) \mid k_{\delta}(y) \mid dy$
Let define $F_{\delta}(x,y) = \mid f(x-y) - f(x) \mid \mid k_{\delta}(y)\mid$.
$\mid (f \ast k_{\delta})(x) -f(x) \mid = \int_{\mid y \mid \leq r}F_{\delta}(x,y)dy + \ud{k=0}{\infty}{\sum} \int_{2^k \delta \leq \mid y \mid \leq 2^{k+1}\delta}F_{\delta}(x,y) dy$
Since $\mid K_{\delta}(x) \mid \leq c \delta^{-d}$, we have $\int_{\mid y \mid \leq \delta} F_{\delta}(x,y) dy \leq c A(\delta)$.
\[\begin{align*}\int_{2^k \delta \leq \mid y \mid \leq 2^{k+1}\delta}F_{\delta}(x,y) dy &\leq \frac{c \delta}{(2^k \delta)^{d+1}} \int_{\mid y \mid \leq 2^{k+1}}\mid f(x-y)-f(x) \mid dy \\ &= \frac{c2^d}{2^k } A(2^{k+1}\delta)\end{align*}\]$\Rightarrow \mid (f \ast k_{\delta})(x) -f(x) \mid \leq cA(\delta) + c2^d \sum_{k=0}^{\infty}2^{-k}A(2^{k+1}\delta)$
Let $\epsilon>0$ be given, $A(r)$ is bounded.
Thus, there exists $N \in \mathbb{N}$ such that $\sum_{k\geq N}2^{-k} < \frac{\epsilon}{3Mc2^d}$
For this $N$, $A(r) \rightarrow 0$ as $r \rightarrow 0$.
We can fix $\eta>0$ such that if $r < \eta$, $A(r) < \frac{\epsilon}{3Mc2^d}$.
Thus $\delta < \frac{\eta}{2^N}$, then $A(2^{k+1}\delta) < \frac{\epsilon}{3Nc2^d}$ and $A(\delta) < \frac{\epsilon}{3C}$
Therefore, $(f\ast k_{\delta})(x) -f(x) \leq cA(\delta)+c 2^d \ud{k=0}{N-1}{\sum} 2^{-k}A(2^{k}\delta) +c 2^d \ud{k\geq N}{\infty}{\sum} 2^{-k}A(2^{k}\delta) < \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3} = \epsilon.$
The theorem could be expanded into uniform continuity.
Theorem (Uniform Convergence)
Let $f$ be integrable and let ${k_{\delta}}$ be an approximation to the identity. Then for each $\delta>0$, the function $(f \ast k_{\delta})(x) = \int f(x-y)k_{\delta}(y)dy$ is integrable and $\parallel (f \ast k_{\delta})-f \parallel \rightarrow 0$ as $\delta \rightarrow 0$
pf)
Note that $\mid (f \ast k_{\delta})(x) - f(x) \mid \leq \int \mid f(x-y) -f(x) \mid k_{\delta}(y) \mid dy$.
By Tonelli theorem, $\parallel (f \ast k_{\delta}) - f \parallel \leq \int \mid k_{\delta}(y) \int \mid f(x-y)-f(x) \mid dx dy$.
Let $\epsilon>0$ be given. $f_{h}(x) = f(x-h) \rightarrow f$.
We can find $\eta >0 $ such that if $\mid y \mid < \eta$, then $\parallel f- f_y \parallel < \frac{\epsilon}{2A}$ and $\int \mid k_{\delta}(x) \mid dx \leq A$.
Remark that $\int_{\mid x\mid \geq \eta}\mid k_{\delta}(x) \mid dx \rightarrow 0$ as $\delta \rightarrow 0$.
For this $\eta >0$, we can find $r >0$ such that if $\delta < r$,then $\int_{\mid x \mid \geq \eta} \mid k_{\delta}(x) \mid dx < \frac{\epsilon}{4 \parallel f \parallel}$
Moreover, if $\delta < r$, then
\[\begin{align*} \parallel (f \ast k_{\delta})- f \parallel & \leq \int_{\mid y \mid < \eta} \mid k_{\delta}(y) \mid (\int \mid f(x-y)-f(x) \mid dx )dy +\int_{\mid y \mid \geq \eta} \mid k_{\delta}(y) \mid (\int \mid f(x-y)-f(x) \mid dx )dy \\ & < \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon \end{align*}\]3. Differentiability
Now, we’ll see the conditions to hold the equation $F(b)-F(a) = \int_{a}^b F’(x) dx$. To see them, we use another concept called as Bounded Variation.
Bounded Variation
Let $F: [a,b] \rightarrow \mathbb{R}$
- $P = {x_0,x_1,…,x_n} = {x_j ; j =0,1,..,N}$; partition of $[0,1]$ if $a = x_0 < x_1 < … < x_N = b$
- $\tilde{P} = {x_j ; 1,2,…,M}$ ; Refinement of $P$ if $\tilde{P}$ is a partition of $[a,b]$ and $P \subset \tilde{P}$.
- For a partition $P$ of $[a,b]$, $V(F,P) = \ud{j=1}{N}{\sum} \mid F(x_j)-F(x_{j-1}) \mid =: \ud{j=1}{N}{\sum} \mid \Delta F_j \mid$ is called by Variation of $F$ on $P$.
- $F$ is of Bounded Variation on $[a,b]$ if the set ${V(F,P) ; P \mbox{ is a partition of }[a,b]}$ is bounded about $P$.
- $T_F(a,b) = \sup {V(F,P) ; P \mbox{ is a partition of }[a,b]}$; Total Variation of $F$ on $[a,b]$
- $BV[a,b]$ ; the set of all functions on $[a,b]$ that are of bounded variation
Remark that $V(F,P) \leq V(F,\tilde{P})$ and if $F \in BV[a,b]$, then $F$ is bounded at $[a,b]$.
Let’s see the details of the Bounded Variation.
- Let $F:[a,b] \rightarrow \mathbb{R}$ be monotone, then $F \in BV[a,b]$ and $T_F(a,b) = \mid F(b)-F(a)$
- Let $F: [a,b]\rightarrow \mathbb{R}$ be continuous and differentiable on $[a,b]$, then $F \in BV[a,b]$
- Let $F,G \in BV[a,b]$ and $c \in \mathbb{R}$, then
- $F+G, cF,FG \in BV[a,b]$
- $T_{F+G}[a,b] \leq T_F[a,b] + T_G[a,b]$
- $T_{cF}[a,b] = \mid c \mid T_{F}[a,b]$
- Let $F \in BV[a,b]$ and $c \in (a,b)$,then $F\in BV[a,c] \cap BV[c,d]$ and $T_{F}[a,b] = T_F[a,c]+ T_{F}[c,b]$
- Let $F:[a,b] \rightarrow \mathbb{R},F \in BV[a,b]$ iff there are two increasing functions $G,H$ such that $F=G-H$.
- Let $F:[a,b] \rightarrow \mathbb{R}$ be continuous $F \in BV[a,b]$ iff $F$ is difference of two continuous increasing functions $F = G-H$.
We have seen which functions satisfy the Bounded variations.
Now, we’ll see how the conditions are connected with the differentiability. We need additional lemmas.
Let $F:[a,b] \rightarrow \mathbb{R}$ be a monotone function. Then $F$ has at most countably many discontinuity.
Let $G$ be a real-valued continuous function and $E$ be the set of points $x$ such that $G(x+h) > G(x)$ for some $h= h_x>0$
- If $E$ is non-empty,then it must be open and hence $E=\underset{j \in \mathbb{N}}{\bigcup} (a_j,b_j)$
- If $(a_j,b_j)$ is finite set, then $G(b_j)-G(a_j)=0$
- The set $E$ is either $\phi$ or open.
Theorem If $F$ is continuous and of bounded variation on $[a,b]$, then $F$ is differentiable almost everywhere.
Corollary
If $f$ is increasing and continuous, then $F’$ exists almost everywhere.
Moreover, $F’$ is measurable, nonnegative and $\int_a^b F’(x)dx \leq F(b) -F(a) \leq 2 \max F$.
Particulary, if $F$ is bounded on $\mathbb{R}$, $F’$ is integrable.
There is an important concept about continuity.
Absolutely Continuous Function
A function $F$ defined on $[a,b]$ is said to be absolutely continuous, if for any $\epsilon>0$, $\exists \delta>0$ such that for $a\leq a_1<b_1\leq a_2 <b_2 \leq…\leq a_N < b_N = b$, $\ud{j=1}{N}{\sum}(b_j -a_j) < \delta$ implies $\ud{j=1}{N}{\sum}\mid F(b_j)-F(a_j)\mid <\epsilon$
There are properties about the absolutely continuous function.
Let $F,G$ be a absolutely continuous function on $[a,b]$. Then
- $F$ is uniformly continuous.
- $F \in BV[a,b]$
- $F+G,FG$ are also absolutely continuous.
- $\mid F \mid^{a}$ are absolutely continuous for $a \in \mathbb{R}$.
- When $f$ is integrable, then $F(x) = \int_{-\infty}^x f(y) dy$ is absolutely continuous.
Lemma. Let $E$ be a set of finite measure, and let $B$ be a Vitali covering of $E$. For any $\delta >0$, we can find finite many balls $B_j\in B, j=1,2,…,N$ such that $\ud{j=1}{N}{\sum}m(B_j) \geq m(E)-\delta$
Corollary Let $E$ be a set of finite measure and let $B$ be a Vitali covering of $E$. For any $\delta >0$, we can find finitely many balls $B_j$ such that $m(E-\ud{j=1}{N}{\bigcup} B_j)<2 \delta$
Theorem
If $F$ is absolutely continuous on $[a,b]$, then $\exists F’(x)$ a.e. $x$. Moreover, if $F’(x) =0$ a.e. $x$, then $F$ is constant.
By using the above lemma and theorem, we can derive the conclusion about the relation between integration and differentiation
Theorem
If $F$ is absolutely continuous on $[a,b]$, then $F’$ exists almost everywhere, and is integrable. Moreover, $F(x)-F(a)=\int_{a}^{x}F’(y)dy, a\leq x \leq b$. Conversely, if $f$ is integrable on $[a,b]$, then there exists an absolutely continuous function $F$ such that $F’=f$ almost everywhere.
The remaining part is differentiation about Jump function.
Let ${x_n}$ denote the points at which $F$ is discontinuous and let $\alpha_n=F(x_n^+)-F(x_n^-)>0$ denote the jump of $F$ at $x_n$.
For each $n \in \mathbb{N}$, we can find $0 < \theta_n <1$ such that $F(x_n)=F(x_n^-)+\alpha_n \theta_n$.
Then define the jump function associated to $F$ by $J_F(x) = \underset{n \in \mathbb{N}}{\sum} \alpha_n j_n(x)$ where $j_n(x) = \begin{cases}0 \mbox{ if } x< x_n \ \theta_n \mbox{ if } x= x_n \ 1 \mbox{ if } x > x_n\end{cases}$.
Lemma. Suppose that $F: [a,b] \rightarrow \mathbb{R}$ is increasing
- $J_F(x)$ is discontinuous at the points ${x_n}$ and has a jump at $x_n$ equal to that of $F$
- $F-J_F$ is increasing and continuous.
Theorem
If $J_F$ is the jump function, than $J_F(x)$ exists and vanishes almost everywhere on $x$.
Theorem
If $F$ is of bounded variation on $[a,b]$, then $F$ is differentiable almost everywhere.