Uniform Convergence

So far we studied convergence of sequence of real numbers and extended it to a more generic setting under metric spaces. We will now consider a sequence of functions and study what convergence means.

Definition. Suppose that $f_n: X \to Y$ is a sequence of functions on a set $X$ to a metric space$(Y, \sigma)$. The sequence $(f_n)$ converges pointwise to a function $f: X \to Y$ if for each $x \in X$,

\[\sigma(f_n(x), f(x)) \to 0 \qquad \text{as} \qquad n \to \infty\]

Alternatively, the sequence is pointwise convergent if

\[(\forall \varepsilon > 0)(\forall x \in X)(\exists N)(\forall n > N) \; \sigma(f_n(x), f(x)) < \varepsilon\]

This is a rather simple definition of convergence but there is a critical shortfall that properties of the functions $f_n$ isn’t carried over to the limit function $f$. For example, consider the sequence of functions $f_n(x) = x^n$ for $0 \le x \le 1$, we have

\[f_n(x) \to f(x) = \begin{cases} 0 & 0 \le x < 1 \\ 1 & x = 1 \end{cases}\]

The functions $f_n$ are all continuous at $x = 1$ but not the limit function $f$ which is not desirable. Therefore, we need a stronger notion of convergence in which we require all the points of $f_n$ to converge to $f$ at the same pace.

Definition. Suppose that $f_n: X \to Y$ is a sequence of functions on a set $X$ to a metric space$(Y, \sigma)$. The sequence $(f_n)$ converges uniformly to a function $f: X \to Y$ if

\[\sup_{x \in X} \,\sigma(f_n(x), f(x)) \to 0 \qquad \text{as} \qquad n \to \infty\]

Alternatively, the sequence is uniformly convergent if

\[(\forall \varepsilon > 0)(\exists N)(\forall n > N) \; \sup_{x \in X} \sigma(f_n(x), f(x)) < \varepsilon\]

The difference is that for pointwise convergence, we can choose a different $N = N(\varepsilon, x)$ for each point but for uniform convergence, we need to choose $N = N(\varepsilon)$ independent of $x$, such that it works for all points. It is obvious that uniform convergence implies pointwise convergence, but the converse is not true. Considering the example above, $f_n(x) = x^n$ is pointwise convergent but not uniformly convergent since

\[\sup_{0 \le x < 1} \vert f_n(x) - f(x) \vert = \lim_{x \to 1} \vert x^n - 0 \vert = 1\]

In case of $X \subseteq \mathbf{R}$ and $Y = \mathbf{R}$, graphical representation of uniform convergence is that for $n > N(\varepsilon)$, $y = f_n(x)$ lies inside the strip between $y = f(x) \pm \varepsilon$.

Uniform Convergence

Definition. A sequence $f_n: X \to Y$ of functions is uniformly Cauchy if

\[(\forall \varepsilon > 0)(\exists N)(\forall m, n > N) \; \sup_{x \in X} \sigma(f_m(x), f_n(x)) < \varepsilon\]

Proposition. [General Principle of Uniform Convergence] Let $(f_n)$ be a sequence of real or complex valued functions. Then $(f_n)$ converges uniformly iff $(f_n)$ is uniformly Cauchy.

Proof.

($\Rightarrow$) Suppose that $f_n \to f$ uniformly. Given $\varepsilon > 0$, there exists $N$ such that for all $n > N$,

\[\sup_{x \in X} \vert f_n(x) - f(x) \vert < {\varepsilon \over 2}\]

Therefore, for $m, n > N$, by triangle inequality,

\[\vert f_m(x) - f_n(x) \vert \le \vert f_m(x) - f(x) \vert + \vert f(x) - f_n(x) \vert < \varepsilon\]

($\Leftarrow$) Since the sequence of functions are uniformly Cauchy, for each $x \in X$, $(f_n(x))$ is a Cauchy sequence that converges to some real or complex value. The pointwise convergence implies that there is a function $f$ such that $f_n \to f$ and we need to show that it converges uniformly.

Given $\varepsilon > 0$. There exists $N$ such that for all $m, n > N$, $\vert f_m(x) - f_n(x) \vert < \varepsilon / 2$. Therefore,

\[\vert f(x) - f_n(x) \vert = \lim_{m \to \infty} \vert f_m(x) - f_n(x) \vert \le { \varepsilon \over 2 } < \varepsilon\]

Since it is true for all $x \in X$,

\[\sup_{x \in X} \vert f(x) - f_n(x) \vert < \varepsilon\]

for $n \ge N$ and $f_n \to f$ uniformly.

Continuity

The example of $f_n(x) = x^n$ shows that continuity is not preserved with pointwise convergence, but it is not the case for uniform convergence.

Proposition. [Uniform Convergence and Continuity] Let $(X, \rho)$ and $(Y, \sigma)$ be metric spaces. Let the sequence of functions $f_n: X \to Y$ converge to $f$ uniformly on $X$. If $c$ is a point of $X$ at which each $f_n$ is continuous, then $f$ is continuous at $c$.

Proof.

We have $\delta > 0$ such that

\[\sigma(f_n(x), f_n(c)) < {\varepsilon \over 3} \qquad \text{for} \qquad \rho(x, c) < \delta\]

and $N$ such that

\[\sigma(f_n(x), f(x)) < {\varepsilon \over 3} \qquad \text{for} \qquad n > N\]

therefore

\[\sigma(f(x), f(c)) \le \sigma(f(x), f_n(x)) + \sigma(f_n(x), f_n(c)) + \sigma(f_n(c), f(c)) < \varepsilon\]

so $f$ is continuous at $c$.

Uniform convergence is a sufficient condition but not a necessary condition for the limit function to be continuous. Consider the sequence of functions $f_n: [0, 1] \to \mathbf{R}$ formed by joining $(0, 0), (1/n, n), (2/n, 0), (1, 0)$, i.e.

Uniform Convergence and Continuity

We have $f_n(x) \to f(x) = 0$ as $n \to \infty$ in which $f_n$ and $f$ are continuous but the convergence is not uniform.

Also, the uniform convergence implies continuity means that we have

\[\lim_{x \to c} \lim_{n \to \infty} f_n(x) = \lim_{x \to c} f(x) = f(c) = \lim_{n \to \infty} \lim_{x \to c} f_n(x)\]

so it can be used to justify the inversion of the order of repeated limits.

Integrability

Again, pointwise convergence is not enough to preverse integrability. Suppose that $(r_n)$ is a sequence consisting all the rational numbers in $[0, 1]$ and consider the sequence of $f_n$ on $[0, 1]$ defined by

\[f_n(x) = \begin{cases} 1 & x = r_1, ..., r_n \\ 0 & \text{otherwise} \end{cases}\]

then

\[f_n(x) \to f(x) = \begin{cases} 1 & x \in \mathbb{Q} \\ 0 & x \not\in \mathbb{Q} \end{cases}\]

In this case, we have all $f_n$ being Riemann integrable since there are only finite number of points mapped to $1$ but $f$ being not integrable.

We also have $g_n(x) = nxe^{-nx^2} \to g(x) = 0$ being an example that even the limit function is integrable but the sequence of integrals need to converge to it.

Proposition. [Uniform Convergence and Integrability] Let $(f_n)$ be a sequence of real functions integrable over the finite interval $[a, b]$. If $f_n \to f$ uniformly on $[a, b]$, then $f$ is integrable over $[a, b]$ and

\[\int_a^b f_n \to \int_a^b f\]

Proof.

Given $\varepsilon > 0$. There exists $N$ such that

\[\sup_{a \le x \le b} \vert f_n(x) - f(x) \vert < \varepsilon \qquad \text{for} \qquad n \ge N\]

For $a \le x \le b$, we have

\[f_N(x) - \varepsilon < f(x) < f_N(x) + \varepsilon\]

so for any dissection $D$ of $[a, b]$,

\[s_D(f_N) - \varepsilon(b - a) \le s_D(f) \le S_D(f) \le S_D(f_N) + \varepsilon(b - a)\]

Thus,

\[S_D(f) - s_D(f) \le S_D(f_N) - s_D(f_N) + 2 \varepsilon (b - a)\]

and $f_N$ is integrable so

\[S_D(f) - s_D(f) \le \varepsilon (1 + 2 (b - a))\]

and $f$ is integrable.

Moreoever, for $n > N$,

\[\left| \int_a^b f_n(x) - \int_a^b f(x) \right| = \int_a^b \vert f_n(x) - f(x) \vert \le \sup_{a \le x \le b} \vert f_n(x) - f(x) \vert (b - a) \to 0\]

Note that the proposition is only proved for finite interval, and is false for an infinite interval.

References