LEC ANA1 33

Sequences of functions

Definition 1(Definition).

Suppose $(f_n{)}_{n=1}^{\infty }$ is a sequence of functions defined on a set $E$, and suppose that the sequence of numbers $(f_n(x))$ converges for every $x\in E$. We can then define a function $f$ by

$$f(x)\equiv \underset{n\to \infty }{\lim }{f}_n(x)(x\in E).$$

Under these circumstances, we say that $(f_n)$ converges pointwise (or simply converges) on $E$ to the limit function $f$.

Definition 2(Definition).

Similarly, if $\sum f_n(x)$ converges for every $x\in E$, and if we define

$$f(x)=\sum _{n=1}^{\infty }{f}_n(x)(x\in E),$$

then $f$ is called the sum of the series $\sum f_n$.

We want to determine whether important properties of functions such as continuity and differentiability are preserved under these limit operations. For example, if $(f_n)\to f$ is a sequence of pointwise convergent functions, and every $f_n$ is continuous, is the limit function $f$ continuous? Several easy examples can be constructed with little thought to show that this is definitely not the case. For example, consider the sequence of functions $(f_n)$ defined by $f_n=x^n$ on $[0,1]$. Each $f_n$ is continuous. Now, consider the limit function, $f={\lim }_{n\to \infty }{f}_n$. For any $x\in [0,1)$, $(f_n(x))$ converges to $0$. But at $x=1$, $(f_n(1))\to 1$. So, we get

$$f=\underset{n\to \infty }{\lim }{f}_n=\{\begin{array}{ll}0& x\in [0,1)\\ 1& x=1,\end{array}$$

which is clearly discontinuous. So, pointwise convergence does not preserve properties such as continuity. However, a stronger form of convergence, called uniform convergence, does preserve these properties.

Remark

To say $f$ is continuous at a limit point $x$ means

$$\underset{t\to x}{\lim }f(t)=f(x).$$

Hence, to ask whether the limit of a sequence of continuous functions is continuous is the same as to ask whether

$$\begin{array}{rl}& \underset{t\to x}{\lim }f(t)=\underset{n\to \infty }{\lim }{f}_n(x)\\ & \underset{t\to x}{\lim }(\underset{n\to \infty }{\lim }{f}_n(t))=\underset{n\to \infty }{\lim }(\underset{x\to t}{\lim }{f}_n(t))\end{array}$$

(We have used the definition of $f$ as the pointwise limit of $(f_n)$ and the hypothesis that $f_n$ are continuous). So, we are essentially asking whether the order in which limit processes are carried out is immaterial. As noted before, this is not true for pointwise convergent functions, but is true for uniformly convergent functions.

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Uniform convergence

Recall our definition of uniform continuity and how it relates to vanilla continuity. Something similar will happen here. Notice that the definition of pointwise convergence can be restated like so:

Definition 3(Definition).

Let $f_n:E\to \mathbb{R}$. $(f_n)$ is pointwise convergent if

$$\forall x\in E\forall ϵ>0\exists N\text{ such that }n>N⟹|f_n(x)-f(x)|<ϵ.$$

Compare with the definition of uniform convergence:

Definition 4(Definition).

Let $f_n:E\to \mathbb{R}$. $(f_n)$ is uniformly convergent if

$$\forall ϵ>0\exists N\text{ such that }(\forall x\in E\text{ and }n>N)⟹|f_n(x)-f(x)|<ϵ.$$

$(f_n)$ converges uniformly if and only if

$$\underset{n\to \infty }{\lim }\underset{x\in E}{\sup }|f_n(x)-f(x)|=0.$$

$(f_n)$ converging uniformly to $f$ is denoted by $(f_n)\rightrightarrows f$. (The parentheses denoting that $f_n$ are elements of a sequence may periodically be dropped.)

There exist parallel definitions for $\sum f_n\to f$ and $\sum f_n\rightrightarrows f$.

Results due to uniform convergence

Limit of uniformly converging continuous functions is continuous

Theorem 5(Rudin 7.12).

$f_n\rightrightarrows f$ and each $f_n$ continuous on $E$ $⟹$ $f$ is continuous on $E$.

Proof.

Let$x\in E$. We’ll show continuity of $f$ at $x$. Let $ϵ>0$. Pick $N$ such that $n\ge N$ $⟹$ $|f_n(y)-f(y)|<ϵ$ for all $y\in E$. Let $\delta >0$ such that $d(x,t)<\delta ⟹|f_N(t)-f_N(x)|<ϵ$. Now, if $d(x,t)<\delta$, we have

$$\begin{array}{rl}|f(x)-f(t)|& \le |f(x)-f_N(x)|+|f_N(x)-f_N(t)|+|f_N(t)-f(t)|\\ & <ϵ+ϵ+ϵ\end{array}$$ □

Limit of uniformly converging integrable functions is integrable (and more)

Theorem 6(Rudin 7.16).

$f_n\rightrightarrows f$ and each $f_n\in \mathcal{R}$ on $[a,b]$ $⟹$ $f\in \mathcal{R}$ on $[a,b]$ and

$${\int }_a^{b}fdx=\underset{n\to \infty }{\lim }{\int }_a^b{f}_{n}dx.$$

Proof.

Let${ϵ}_n={\sup }_{x\in [a,b]}|f(x)-f_n(x)|$. Since $f_n$ converges uniformly to $f$, $({ϵ}_n)\to 0$. Now,

$$\begin{array}{r}{f}_n(x)-{ϵ}_n\le f(x)\le f_n(x)+{ϵ}_n\end{array}$$

Consider a partition $P$ of $[a,b]$. Then, the first inequality above gives us

$$\begin{array}{r}L(P,f_n(x)-{ϵ}_n)\le L(P,f(x))\le \underline{{\int }_a^b}f(t)dt.\end{array}$$

The above is true for every partition $P$. So, we can take the supremum or infimum on the left (doesn’t matter which, both are equal):

$${\int }_a^b{f}_n(t)-{ϵ}_{n}dt\le \underline{{\int }_a^b}f(t)dt.$$

Applying the same idea on the other side gives us

$$\begin{array}{r}{\int }_a^b{f}_n(t)dt-{ϵ}_n(b-a)\le \underline{{\int }_a^b}f(t)dt\le \overline{{\int }_a^b}f(t)dt\le {\int }_a^b{f}_n(t)dt+{ϵ}_n(b-a)\end{array}$$

Thus,

$$\overline{{\int }_a^b}f(t)dt-\underline{{\int }_a^b}f(t)dt\le 2{ϵ}_n(b-a).$$

Thus, $f$ is integrable on $[a,b]$.□

A not so neat but nevertheless useful result on differentiation

Theorem 7(Rudin 7.17).

Suppose

• $f_n$ are differentiable on $[a,b]$,
• $f_n^{\prime }\rightrightarrows \text{ a function, say }g\text{ on }[a,b]$, and
• there exists some $x_0\in [a,b]$ such that $f_n(x_0)$ converges.

Then, $f_n\rightrightarrows \text{a function }f$ and $f^{\prime }(x)=g(x)$ for all $x\in [a,b]$.

Another way to understand uniform convergence

Let $X$ be a metric space. Define

$$\mathcal{B}(X)\equiv \{f:X\to \mathbb{R}|f(X)\text{ is a bounded set}\}.$$

Define a norm on $\mathcal{B}(X)$ by

$$\Vert f\Vert =\underset{x\in X}{\sup }|f(x)|.$$

Note that this turns $\mathcal{B}(X)$ into a normed linear space over $\mathbb{R}$, since all the properties of the norm are satisfied by our definition:

1. $\Vert \alpha f\Vert =|\alpha |\Vert f\Vert$ for all $\alpha \in \mathbb{R}$ and $f\in \mathcal{B}(X)$. This is easy to see, since if a function is scaled by $\alpha$, its maximum value is also scaled by $\alpha$.
2. $\Vert f\Vert \ge 0$ with $\Vert f\Vert =0$ if and only if $f=0$.
3. $\Vert f+g\Vert \le \Vert f\Vert +\Vert g\Vert$.

Proof of triangle inequality

$$\begin{array}{rl}L& \equiv \underset{x\in X}{\sup }|f(x)+g(x)|\\ R& \equiv \underset{x\in X}{\sup }|f(x)|+\underset{x\in X}{\sup }|g(x)|\end{array}$$

From the triangle inequality for real numbers, we know that

$$\begin{array}{r}|f(x)+g(x)|\le |f(x)|+|g(x)|\forall x\in X\end{array}$$

We know that $|f(x)|+|g(x)|\le \sup |f(x)|+\sup |g(x)|=\Vert f\Vert +\Vert g\Vert$. This is true for all $x$. Thus, $\Vert f\Vert +\Vert g\Vert$ is an upper bound for $|f(x)+g(x)|$. Thus, we have

$$\sup |f(x)+g(x)|\le \Vert f\Vert +\Vert g\Vert .$$

Note the order in which the suprema were taken in the above proof.

Therefore, we get a metric $d(f,g)=\Vert f-g\Vert$.

Claim 8(Claim).

$f_n\rightrightarrows f$ $⟺$ $f_n\to f$ in $\mathcal{B}(X)$, i.e, $\forall ϵ>0$, $\exists N$ such that $n>N$ $⟹$ $\Vert f_n-f\Vert <ϵ$.

Proof.

$$\Vert f_n-f\Vert \le ϵ⟺\underset{x\in X}{\sup }|f_n(x)-f(x)|\le ϵ⟺|f_n(x)-f(x)|\le ϵ\forall x\in X$$ □

Remark

Consider $\mathcal{B}([0,1])$. Let $\mathcal{C}([0,1])$ be the set of all continuous functions on $[0,1]$. Now, since continuous functions on compact sets are bounded, $\mathcal{C}([0,1])\subset \mathcal{B}([0,1])$. Since the limit of uniformly converging continuous functions is continuous the limit of every convergent sequence $(f_n)$ in $\mathcal{C}([0,1])$ is also in $\mathcal{C}([0,1])$, i.e, $\mathcal{C}([0,1])$ is a closed subset of $\mathcal{B}([0,1])$!

Criteria for uniform convergence

The Cauchy Criterion

Theorem 9(Theorem).

The sequence of functions $(f_n)$ defined on $E$ converges uniformly on $E$ if and only if for every $ϵ>0$ there exists an integer $N$ such that $m\ge N$, $n\ge N$, $x\in E$ implies

$$|f_n(x)-f_m(x)|\le ϵ.$$

Proof.

Let$N$ be such that $n>N$ implies $|f_n(x)-f(x)|<ϵ$ for all $x\in E$. Thus, $\forall n,m>N$,

$$\begin{array}{rl}|f_n(x)-f_m(x)|& \le |f_n(x)-f(x)|+|f(x)-f_m(x)|\\ & \le ϵ+ϵ\end{array}$$

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Conversely, the Cauchy criterion for sequences of real numbers tells us that the sequence $(f_n(x))$ converges for every fixed $x$. Thus, $(f_n)$ is pointwise convergent to some function $f$. We need to prove that this convergence is uniform. Choose $N$ such that $m,n\ge N$ implies $|f_n(x)-f_m(x)|\le ϵ$ for all $x\in E$. Now, keep $n$ and $x$ fixed, and consider the sequence obtained by incrementing $m$: $|f_n(x)-f_{m+1}(x)|,|f_n(x)-f_{m+2}(x)|,\dots$. All of these are less than $ϵ$. Thus, their limit must also be less than $ϵ$. So, we have

$$|f_n(x)-f(x)|\le ϵ\forall n\ge N,\forall x\in E.$$ □

The M-test

Theorem 10(Theorem).

Suppose $(f_n)$ is a sequence of functions defined on $E$, and suppose

$$|f_n(x)|\le M_n(x\in E,n=1,2,3,\dots ).$$

Then, $\sum M_n$ converges $⟹$ $\sum f_n$ converges uniformly on $E$.

Proof.

If$\sum M_n$ converges, then, for arbitrary $ϵ>0$, from the Cauchy criterion for series,

$$\left|\sum _{i=n}^m{f}_i(x)\right|\le \sum _{i=n}^m{M}_i\le ϵ(x\in E),$$

provided $m$ and $n$ are large enough. Uniform convergence follows from the Cauchy criterion for uniform convergence. Note the implicit use of the triangle inequality above.□