## Basic Analysis: Uniform Convergence

Once again, let $D\subseteq \mathbf{R}$ be a subset. Suppose we now have a sequence of functions $f_n : D\to\mathbf{R}$, where n = 1, 2, 3, … ,

such that for each x in D, the sequence $(f_n(x))$ converges to some real value. We’ll denote this value by f(x), thus defining a function fD → R. We then say that the sequence of functions fn(x) converges to f(xpointwise.

The big question is: if each fn(x) is continuous at x=a, is it true that f(x) is also continuous at x=a?

It may surprise you that the answer is no. [ Historical note: Cauchy actually believed the answer is yes and provided an erroneous proof. This shouldn’t be held against Cauchy since analysis was still in its infancy in the 19th century. On the contrary, one should use this as an indication of how counter-intuitive results in analysis can be. ]

Example

Consider the sequence of functions $f_n : [0,1]\to\mathbf{R}, \quad f_n(x) = x^n.$

If 0≤x<1, then clearly fn(x) → 0 as n approaches infinity. On the other hand, for x=1, the sequence is a constant sequence (1, 1, 1, …). Thus fn(x) converges to f(x) pointwise, where f takes any element in [0,1) to 0, and 1 to 1. ## Uniform Convergence

In order to ensure that f is continuous, we need a further condition.

Definition. The sequence of functions fn(x) is said to converge uniformly to f(x) if:

• for each ε>0, there exists N such that whenever n>N and x is in D, we have |fn(x)-f(x)|<ε.

Pictorially, we have: [ For large n, the function fn(x) lies between f(x)-ε and f(x)+ε. ]

Let’s consider the difference between pointwise convergence and uniform convergence. In the former case, we have:

• For each x in D and ε>0, there exists N such that for all n>N, we have |fn(x)-f(x)|<ε.

On the other hand, the latter says:

• For each ε>0, there exists N such that for all x in D and n>N, we have |fn(x)-f(x)|<ε.

Using logical quantifiers, the two statements become: $\forall x,\forall\epsilon, \exists N, P(N,x,\epsilon)$ and $\forall \epsilon, \exists N, \forall x, P(N,x,\epsilon)$,

where P(Nx, ε) says that whenever n>N, we have |fn(x)-f(x)|<ε. Written in this manner, it’s clear that uniform convergence implies pointwise convergence.

Main Therorem. If fn(x) converges to f(x) uniformly, and each fn(x) is continuous at x=a, then so is f(x).

Proof. Let ε>0.

• Since fn(x) → f(x) uniformly, there exists N such that for all x in D and n>N, we have |fn(x)-f(x)|<ε/3. Fix any n>N.
• Since fn(x) is continuous at x=a, there exists δ>0 such that for all x in D satisfying |xa|<δ, we have |fn(x)-fn(a)|<ε/3, which gives: $|f(x)-f(a)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(a)|+|f_n(a)-f(a)|<\frac\epsilon 3+\frac\epsilon 3+\frac\epsilon 3=\epsilon.$ ♦ ## Alternate Definition

Before going through some examples, we consider an alternate definition of uniform convergence which is somewhat conceptually clearer.

Consider the set V of all functions D → R. This forms a real vector space via pointwise addition and scalar multiplication: i.e. $(f+g)(x) := f(x) + g(x), \quad (c\cdot f)(x) = c\cdot f(x), \ c\in\mathbf{R}.$

Now we define the norm of f to be $||f|| := \sup\{|f(x)| : x\in D\}$ which is either a non-negative real number or +∞.  It’s not hard to show that $||f+g|| \le ||f||+||g||$, $||fg|| \le ||f||\cdot||g||$ and $||c\cdot f|| = |c|\cdot ||f||$.

[ Indeed, to prove the first inequality, just note that for any x in D, we have |(f+g)(x)| = |f(x)+g(x)| ≤ |f(x)|+|g(x)| ≤ RHS. The same proof holds for the second inequality. ]

Alternate Definition. The sequence $f_n(x) \to f(x)$ uniformly if:

• for every ε>0, there exists N such that when n>N, we have $||f_n - f||<\epsilon.$

Note that this definition is quite similar to that of convergent sequences, which makes it convenient to work with. [ Side note: the acute reader may suspect that one can study analysis on the above function space V and he’d be right! Indeed, this is the primary motivation behind functional analysis. ]

Examples

1. Let’s go back to the example $f_n(x) = x^n$ on D=[0,1]. Since the functions converge to a discontinuous function, the above theorem tells us the convergence is not uniform, but let’s verify this directly. Indeed, the function $f_n-f$ takes xn on [0,1) and 0 at x=1. Thus, $||f_n-f||=1$ for all n, which shows that the convergence is not uniform.
2. On the other hand, if we take $f_n(x) = x^n$ on D=[0, 1/2], then they converge to f(x) = 0 uniformly. Indeed, we have $||f_n - f|| = ||f_n|| = 1/2^n$ for each n, so $\lim_{n\to\infty} ||f_n - f|| = 0$. As expected, since each fn(x) is continuous, so is the resulting converged function f(x).
3. If we now take $f_n(x) = x^n$ on D=[0, 1), then they converge to f(x) = 0, albeit not uniformly. Indeed, it’s easy to see that $||f_n - f|| = 1$ for each nIn other words, one can have pointwise convergence of a sequence of continuous functions to a continuous function, even if the convergence is not uniform.
4. Consider the Riemann zeta function $f(x)=\zeta(x) = 1^{-x} + 2^{-x} + 3^{-x} + \ldots$ on (1, ∞). We know this converges at every x>1, but is the function continuous? Indeed, it is: via the following useful trick. First, write $f_n(x) = 1^{-x} + 2^{-x} + \ldots + n^{-x}$, so that $\lim_{n\to\infty} f_n(x) \to \zeta(x)$ pointwise. Suppose we wish to show that ζ(x) is continuous at x=a>1.
• Pick any b, 1<b<a.
• Let gn (resp. g) be the restriction of fn (resp. f) to [b, ∞).
• Now $||g_n - g||=(n+1)^{-x} + (n+2)^{-x} + \ldots \le (n+1)^{-b} + (n+2)^{-b} + \ldots$ which is a convergent series since b>1.
• Hence $||g_n-g|| \to 0$, which shows that gn converges uniformly to g.
• Thus means g is continuous, in particular f is continuous at a. Hence, f is continuous at any a>1.

## Basic Properties of Uniform Convergence

Let’s figure out some basic properties, specifically suppose $f_n : D\to \mathbf{R}$ and $g_n : D\to\mathbf{R}$ are sequences of functions which converge uniformly to $f, g:D\to\mathbf{R}$ respectively. What about the sum $f_n+g_n$, product $f_n g_n$ and reciprocal $1/g_n$?

1. The case of sum is easy.

Indeed, $f_n + g_n \to f+g$ uniformly. To see why, write: $||(f_n+g_n)-(f+g)|| = ||(f_n-f)+(g_n-g)|| \le ||f_n-f|| + ||g_n-g||.$

Now, for any $\epsilon>0$, $||f_n - f|| < \frac\epsilon 2$ and $||g_n-g|| < \frac\epsilon 2$ for large n. Thus, we also have $||(f_n+g_n)-(f+g)|| < \epsilon$ for large n.

2. The case of product is true with some additional conditions.

Indeed, let’s write: $||f_n g_n - fg || = ||(f_n-f)g_n + f(g_n - g)|| \le ||f_n - f||\cdot ||g_n|| + ||f||\cdot ||g_n - g||.$

Now, if ||f|| and ||g|| are both finite, then in particular, we have $||g_n|| \le ||g_n - g|| + ||g|| < ||g||+1$ for large n. It then follows that we can write $||f||, ||g|| < B$ for some positive bound B and $||f_n g_n - fg|| \le B(||f_n - f|| + ||g_n - g||)$. Hence, it also follows that $f_n g_n \to fg$ converges uniformly.

What if either ||f|| or ||g|| is infinite? A counter-example is easy: $f_n, g_n:\mathbf{R} \to\mathbf{R}, \ f_n(x) = x, g_n(x) = \frac 1 n.$

Then $f_n(x) \to x$ and $g_n(x) \to 0$ uniformly. However, $||f_n g_n - 0|| = \sup\{\frac {|x|} n: x\in\mathbf{R}\} = \infty$.

3. The case of reciprocal is true with some additional conditions.

Let’s consider when $1/g_n \to 1/g$ uniformly, assuming that g(x) and gn(x) are never 0. Taking the difference gives: $||\frac{1}{g_n} - \frac{1}{g}|| = ||\frac{g_n - g}{g_n g}|| \le ||g_n -g|| \cdot ||\frac{1}{g_n}||\cdot ||\frac 1 g||.$

Hence, suppose there is a positive B such that $||1/g_n|| < B$ for all n. This means $|g_n(x)| > 1/B$ for all n, and $x\in D$. For some N, we have $||g_n - g|| < 1/(2B)$, i.e. $|g_n(x) - g(x)| < 1/(2B)$ for all n>N and $x\in D$. Thus: $|g(x)| \ge |g_n(x)| - |g_n(x)-g(x)| > 1/B - 1/(2B) = 1/(2B),$ when $x\in D$, i.e. $||1/g||\le 2B$.

Thus, $||\frac 1 {g_n}-\frac 1 g|| \le ||g_n - g||\cdot B\cdot(2B)$ for all n and it follows that $1/g_n \to 1/g$ converges uniformly.

What if such a B doesn’t exist, i.e. the infimum of all $||1/g_n||$ is zero? Consider the example of $g_n: (0, 1]\to\mathbf{R}, g_n(x)=(1+\frac 1 n)x$ $\$. Then $g_n(x)$ converges to g(x)=x uniformly since $||g_n-g|| = ||\frac x n||=\frac 1 n$. On the other hand, $||\frac 1 {g_n}-\frac 1 g|| = ||\frac n {(n+1)x}-\frac 1 x|| = ||\frac 1 {(n+1)x}||=+\infty,$

so $1/g_n \to 1/g$ pointwise, but not uniformly.

Summary. Suppose $f_n \to f$ and $g_n \to g$ uniformly, where each $f_n, g_n, f, g:D\to\mathbf{R}$. Then:

• $f_n+g_n\to f+g$ uniformly as well;
• if $||f||, ||g||<\infty$, then $f_n g_n \to fg$ uniformly;
• if there is a positive B such that $||g_n||>B$ for all n, then $1/g_n \to 1/g$ uniformly.
This entry was posted in Notes and tagged , , , , , , . Bookmark the permalink.

### 1 Response to Basic Analysis: Uniform Convergence

1. l3spam says:

Simple and clean explanation. THANKS!