LEC ANA2 8

Reviewed Theorem 143.2.

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Examples for The Arzelà–Ascoli Theorem

Example 168.1(Example).

Let $k:[0,1]\times [0,1]\to \mathbb{R}$ be continuous. Define $T_k:C[0,1]\to C[0,1]$ by

$$(T_{k}f)(x)={\int }_0^{1}k(x,y)f(y)dy.$$

We need to verify the continuity of $T_{k}f$ to justify the codomain in the definition. Let $k_x(y):=k(x,y)$. If $\{x_n\}\to x$, the sequence of functions $\{k_{x_n}\}$ must converge to $k_x$ pointwise due to the continuity of $k$. Also, by the extreme value theorem, $k\le M$ on $[0,1]\times [0,1]$ for some $M>0$, so each $k_{x_n}$ is bounded by the integrable constant function $M$ on $[0,1]$. By the dominated convergence theorem,

$$\begin{array}{rl}\underset{n\to \infty }{\lim }(T_{k}f)(x_n)& =\underset{n\to \infty }{\lim }{\int }_0^{1}k(x_n,y)f(y)dy\\ & ={\int }_0^1\underset{n\to \infty }{\lim }k(x_n,y)f(y)dy\\ & ={\int }_0^{1}k(x,y)f(y)dy\\ & =(T_{k}f)(x).\end{array}$$

So, $T_{k}f$ is continuous for every $f$ ¹.

Consider the collection

$$\mathcal{F}=\overline{\{T_{k}f:\Vert f{\Vert }_{\infty }\le 1\}}\subseteq C[0,1].$$E1

$\mathcal{F}$ is closed, clearly. $\mathcal{F}$ is also bounded ²:

$$\begin{array}{rl}& \left|(T_{k}f)(x)\right|\le {\int }_0^1\left|k(x,y)\right|\left|f(y)\right|dy\le \Vert k{\Vert }_{\infty }\Vert f{\Vert }_{\infty }\forall x\\ & ⟹\Vert T_{k}f{\Vert }_{\infty }\le \Vert f{\Vert }_{\infty }\Vert k{\Vert }_{\infty }.\end{array}$$E2

If $\Vert f{\Vert }_{\infty }\le 1$, then $\Vert T_{k}f\Vert \le \Vert k{\Vert }_{\infty }$. Thus, $\mathcal{F}$ is bounded in $C[0,1]$.

We’re equicontinuity away from applying the Arzelà–Ascoli Theorem. Let $ϵ>0$ and $T_{k}f\in \mathcal{F}$. Choose $\delta$ by the uniform continuity of $k$. If $|x-x^{\prime }|<\delta$,

$$\begin{array}{rl}|T_{k}f(x)-T_{k}f(x^{\prime })|& \le {\int }_0^1\left|k(x,y)-k(x^{\prime },y)\right|dy\\ & \le {\int }_0^1ϵdy\\ & =ϵ.\end{array}$$

We also have to consider functions that are in $\mathcal{F}$ by virtue of us taking the closure in (E1). Let $F\in \mathcal{F}$ be such a function. Then, for $ϵ>0$, there exists $f$ with $\Vert f\Vert \le 1$ such that $\Vert T_{k}f-F\Vert <ϵ$. Then, for $|x-x^{\prime }|<\delta$,

$$\begin{array}{rl}|F(x)-F(x^{\prime })|& \le |F(x)-T_{k}f(x)|+|T_{k}f(x)-T_{k}f(x^{\prime })|+|T_{k}f(x^{\prime })-F(x^{\prime })|\\ & \le 3ϵ.\end{array}$$

so $\mathcal{F}$ is a equicontinuous family. By Arzelà–Ascoli, $\mathcal{F}$ is compact.

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Consequences of completeness

Reviewed The Banach contraction principle. The conclusion holds even if $T^k$ is a contraction for some $k$.

Proposition 168.2(Proposition).

Let $X$ be a complete metric space. Suppose $T^k$ is a contraction. Then $T$ has a unique fixed point.

Proof.

$T^k$ must have a unique fixed point, say $x$.

$$\begin{array}{rl}d(x,Tx)& =d(T^{k}x,T(T^{k}x))\\ & =d(T^{k}x,T^k(Tx))\\ & \le cd(x,Tx).\end{array}$$

Since $c<1$ $d(x,Tx)=0$, so $x=Tx$. It is clear that $T$ cannot have any other fixed points, since a fixed point of $T$ is also a fixed point of $T^k$.□

Proposition 168.3(Proposition).

Suppose $X$ is compact. $T:X\to X$ such that $d(T{x}_1,T{x}_2)<d(x_1,x_2)$ for all $x_1\ne x_2$. Then there exists a unique fixed point.

Proof.

$x\mapsto d(x,Tx)$ is a continuous function:

$$\begin{array}{rl}d(x,Tx)& \le d(x,y)+d(y,Ty)+d(Ty,Tx)\\ d(x,Tx)-d(y,Ty)& \le 2d(x,y).\end{array}$$

Thus there exists $x_0\in X$ such that $d(x_0,T{x}_0)={\inf }_{x\in X}d(x,Tx)$. Let $y=T{x}_0$. Then $d(y,Ty)<d(x_0,T{x}_0)$. Thus, $x_0=T{x}_0$.□

Proposition 168.4(Proposition).

Let $V\subseteq B(S)$ be a closed NLS containing all constant functions of $B(S)$. Let $T:V\to V$ (need not be linear) satisfy:

1. $T(f)\le T(g)$ if $f\le g$ (these are pointwise comparisons).
2. There exists $0<\beta <1$ such that for all $f\in V$ and constant functions $c$, $T(f+c)\le (Tf)+\beta c$.

Then $T$ has a unique fixed point.

Proof.

It suffices to show $T$ is a contraction with $\beta$. Fix $f,g\in V$. Let $c=\Vert f-g{\Vert }_{\infty }$. Then,

$$\begin{array}{r}f\le g+c\\ g\le f+c\end{array}$$

So,

$$\begin{array}{rl}(Tf)& \le T(g+c)\\ & \le (Tg)+\beta c\\ & \\ Tf-Tg& \le \beta c=\beta \Vert f-g{\Vert }_{\infty }.\end{array}$$

Similarly, $Tg-Tf\le \beta \Vert f-g{\Vert }_{\infty }$. So, $\Vert Tf-Tg{\Vert }_{\infty }\le \beta \Vert f-g{\Vert }_{\infty }$, and we’re done.□

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Baire Category theorem

Definition 168.5(Definition).

$X$ is a metric space. $S\subseteq X$ is nowhere dense if $(\overline{S}{)}^{\circ }=\varnothing$.

Definition 168.6(Definition).

$F\subseteq X$ is said to be meagre or of category I if $F$ is contained in a countable union of closed nowhere dense sets. If $F\subseteq X$ is not of category I, it is said to be of category II.

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Note

There are two definitions of a meagre set: the one given above, and this: $F\subseteq X$ is meagre if $F$ is equal to a countable union of nowhere dense sets. These are equivalent; the one given above is more convenient to work with because closed sets are easier to manipulate in proofs.

Example 168.7(Example).

1. Note that “meagre/Category I” is a property relative to the ambient topological space. $\mathbb{Z}$ is of first category in $\mathbb{R}$, but $\mathbb{Z}$ is not of first category in $\mathbb{Z}$, because nowhere dense sets do not exist in $\mathbb{Z}$ when it is regarded as the whole space. When we say a space is of category x, we mean it is of category x in itself.
2. $\mathbb{Q}$ is of first category in $\mathbb{R}$, and of first category in $\mathbb{Q}$. In fact, any countable metric space having no isolated points is of first category in itself.

Theorem 168.8(Baire Category Theorem).

Let $X$ be a complete metric space. Let $\{U_n{\}}_{n=1}^{\infty }$ be a collection of open and dense subsets of $X$. Then, $U={\bigcap }_{n=1}^{\infty }{U}_n$ is dense in $X$.

Proof.

It suffices to show$B_0\cap U\ne \varnothing$ for any closed ball $B_0\subseteq X$ of radius $\delta$. Clearly, $B_0^{\circ }\cap U_1$ is open, and hence contains a closed ball $B_1$ of radius less than $\delta /2$. Iteratively, choose a closed ball $B_n$ of radius $\delta /2^n$ such that $B_n\subseteq B_{n-1}^{\circ }\cap U_n$. $\{B_n{\}}_{n=1}^{\infty }$ is a contracting sequence of nonempty closed subsets. Since $X$ is complete, ${\bigcap }_{n=1}^{\infty }{B}_n=\{x\}$ for some $x\in X$. It follows that $x\in U$.□

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Corollary 168.9(Corollary).

If $X$ is complete, then $X$ is of second category. In other words, a complete metric space cannot be expressed as a countable union of closed nowhere dense subsets.

Proof.

Let$\{F_n{\}}_{n=1}^{\infty }$ be a collection of closed nowhere dense subsets of $X$. Then, $\{F_n^c{\}}_{n=1}^{\infty }$ is a collection of open dense subsets of $X$. By Theorem 8, ${\left({\bigcup }_{n=1}^{\infty }{F}_n\right)}^c={\bigcap }_{n=1}^{\infty }{F}_n^c$ is dense in $X$. Thus, ${\bigcup }_{n=1}^{\infty }{F}_n\ne X$.□

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Exercise 168.10(Exercise).

Show that there does not exist $f:\mathbb{R}\to \mathbb{R}$ that is only continuous on $\mathbb{Q}$.

Recall Thomae’s function as an example of a function which is continuous only on the irrationals; something must go pear shaped when we try to achieve the same for the rationals. First, observe that the set of discontinuities $B$ of any function from $\mathbb{R}$ to $\mathbb{R}$ is a countable union of closed sets (aka an $F_{\sigma }$ set):

$$B=\bigcup _{n=1}^{\infty }\{x\in \mathbb{R}:o(f,x)\ge 1/n\};$$

see Theorem 195.8 and Theorem 195.9¹. Thus, if a function from $\mathbb{R}$ to $\mathbb{R}$ which is continuous only on $\mathbb{Q}$ existed, the irrationals would be a countable union of closed sets ${\bigcup }_{i=1}^{\infty }{C}_i$, but since the irrationals do not contain an interval, neither can any of the $C_i$. Therefore, each of the $C_i$ would be nowhere dense, and the irrationals would be a meagre set. It would follow that $\mathbb{R}$ is meagre, contradicting Corollary 9.

[!Exercise]
For $f:[0,1]\to \mathbb{R}$, define

$$(D^{+}f)(a)=\underset{x\to a^{+}}{\mathrm{limsup}}\frac{f(x)-f(a)}{x-a}.$$

Prove that for each $a\in [0,1]$ the set $\{f\in C[0,1]:D^{+}f(a)=\infty \}$ is a dense $G_{\delta }$ subset. (A set is said to be $G_{\delta }$ if it is a countable intersection of open sets.)

I thought of considering the sets $S_n=\{f\in C[0,1]:D^{+}f(a)>n\}$ for $n\in \mathbb{N}$, but i do not think these are open.

Footnotes

1. Alternatively, you could observe that the convergence $\{k_{x_n}\}\to k_x$ is actually uniform, and use Theorem 155.7 to conclude that $T_{k}f$ is continuous. ↩ ↩²

2. Incidentally, since $T_k$ is linear, (E2) implies $$\begin{gathered} \Vert T_{k}f-T_{k}g{\Vert }_{\infty }=\Vert T_k(f-g){\Vert }_{\infty } \\ \le \Vert f-g{\Vert }_{\infty }\Vert k{\Vert }_{\infty } \end{gathered}$$. So, $T_k$ is actually uniformly continuous! ↩