Up till now, when we talked about limits of sequences we talked about sequences of numbers. A very useful concept in analysis is a sequence of functions. For example, a solution to some differential equation might be found by finding only approximate solutions. Then the actual solution is some sort of limit of those approximate solutions.
When talking about sequences of functions, the tricky part is that there are multiple notions of a limit. Let us describe two common notions of a limit of a sequence of functions.
Limits of sequences of numbers are unique, and so if a sequence \(\{ f_n \}_{n=1}^\infty\) converges pointwise, the limit function \(f\) is unique. It is common to say that \(f_n \colon S \to \R\)converges pointwise to \(f\) on \(T \subset S\) for some \(f \colon T \to \R\text{.}\) In that case we mean \(f(x) = \lim_{n\to\infty} f_n(x)\) for every \(x \in T\text{.}\) In other words, the restrictions of \(f_n\) to \(T\) converge pointwise to \(f\text{.}\)
When \(x = 1\) or \(x=-1\text{,}\) then \(x^{2n} = 1\) for all \(n\) and hence \(\lim_{n\to\infty}f_n(x) = 1\text{.}\) For all other \(x\text{,}\) the sequence \(\bigl\{ f_n(x) \bigr\}_{n=1}^\infty\) does not converge.
The subtle point here is that while \(\frac{1}{1-x}\) is defined for all \(x \not=1\text{,}\) and \(f_n\) are defined for all \(x\) (even at \(x=1\)), convergence only happens on \((-1,1)\text{.}\) Therefore, when we write
Let \(f_n(x) \coloneqq \sin(nx)\text{.}\) Then \(f_n\) does not converge pointwise to any function on any interval. It may converge at certain points, such as when \(x=0\) or \(x=\pi\text{.}\) It is left as an exercise that in any interval \([a,b]\text{,}\) there exists an \(x\) such that \(\sin(xn)\) does not have a limit as \(n\) goes to infinity. See Figure 6.2.
Before we move to uniform convergence, let us reformulate pointwise convergence in a different way. We leave the proof to the reader—it is a simple application of the definition of convergence of a sequence of real numbers.
Let \(f_n \colon S \to \R\) and \(f \colon S \to \R\) be functions. Then \(\{ f_n \}_{n=1}^\infty\) converges pointwise to \(f\) if and only if for every \(x \in S\) and every \(\epsilon > 0\text{,}\) there exists an \(N \in \N\) such that
\begin{equation*}
\abs{f_n(x)-f(x)} < \epsilon \quad \text{for all } n \geq N .
\end{equation*}
The key point is that \(N\) can depend on \(x\text{,}\) not just on \(\epsilon\text{.}\) For each \(x\text{,}\) we can pick a different \(N\text{.}\) If we could pick one \(N\) for all \(x\text{,}\) we would have what is called uniform convergence.
Let \(f_n \colon S \to \R\) and \(f \colon S \to \R\) be functions. The sequence \(\{ f_n \}_{n=1}^\infty\)converges uniformly to \(f\) if for every \(\epsilon > 0\text{,}\) there exists an \(N \in \N\) such that for all \(n \geq N\text{,}\)
\begin{equation*}
\abs{f_n(x) - f(x)} < \epsilon \qquad \text{for all } x \in S.
\end{equation*}
In uniform convergence, \(N\) cannot depend on \(x\text{.}\) Given \(\epsilon > 0\text{,}\) we must find an \(N\) that works for all \(x \in S\text{.}\) See Figure 6.3 for an illustration. Uniform convergence implies pointwise convergence, and the proof follows by Proposition 6.1.5:
Let \(\{ f_n \}_{n=1}^\infty\) be a sequence of functions \(f_n \colon S \to \R\text{.}\) If \(\{ f_n \}_{n=1}^\infty\) converges uniformly to \(f \colon S \to \R\text{,}\) then \(\{ f_n \}_{n=1}^\infty\) converges pointwise to \(f\text{.}\)
The functions \(f_n(x) \coloneqq x^{2n}\) do not converge uniformly on \([-1,1]\text{,}\) even though they converge pointwise. To see this, suppose for contradiction that the convergence is uniform. For \(\epsilon \coloneqq \nicefrac{1}{2}\text{,}\) there would have to exist an \(N\) such that \(x^{2N} = \abs{x^{2N} - 0} < \nicefrac{1}{2}\) for all \(x \in
(-1,1)\) (as \(f_n(x)\) converges to 0 on \((-1,1)\)). But that means that for every sequence \(\{ x_k \}_{k=1}^\infty\) in \((-1,1)\) such that \(\lim_{k\to\infty} x_k = 1\text{,}\) we have \(x_k^{2N} < \nicefrac{1}{2}\) for all \(k\text{.}\) On the other hand, \(x^{2N}\) is a continuous function of \(x\) (it is a polynomial). Therefore, we obtain a contradiction
However, if we restrict our domain to \([-a,a]\) where \(0 < a < 1\text{,}\) then \(\{ f_n \}_{n=1}^\infty\) converges uniformly to 0 on \([-a,a]\text{.}\) Note that \(a^{2n} \to 0\) as \(n \to \infty\text{.}\) Given \(\epsilon > 0\text{,}\) pick \(N \in \N\) such that \(a^{2n} < \epsilon\) for all \(n \geq N\text{.}\) If \(x \in [-a,a]\text{,}\) then \(\abs{x} \leq a\text{.}\) So for all \(n \geq N\) and all \(x \in [-a,a]\text{,}\)
For bounded functions, there is another more abstract way to think of uniform convergence. To every bounded function we assign a certain nonnegative number that measures the “distance” of the function from the constant function \(0\text{.}\) This number allows us to “measure” how far two functions are from each other. We then translate a statement about uniform convergence into a statement about a certain sequence of real numbers converging to zero.
Let \(f \colon S \to \R\) be a bounded function. Define
\begin{equation*}
\snorm{f}_S \coloneqq
\sup \bigl\{ \abs{f(x)} : x \in S \bigr\} .
\end{equation*}
We call \(\snorm{\cdot}_S\) the uniform norm. Sometimes other notation 2
The notation nor terminology is not completely standardized. The norm is also called the sup norm or infinity norm, and in addition to \(\snorm{f}_u\) and \(\snorm{f}_S\) it is sometimes written as \(\snorm{f}_{\infty}\) or \(\snorm{f}_{\infty,S}\text{.}\)
First suppose \(\lim_{n\to\infty} \norm{f_n - f}_S = 0\text{.}\) Let \(\epsilon > 0\) be given. There exists an \(N\) such that for \(n \geq N\text{,}\) we have \(\snorm{f_n - f}_S < \epsilon\text{.}\) As \(\norm{f_n-f}_S\) is the supremum of \(\abs{f_n(x)-f(x)}\text{,}\) we see that for all \(x \in S\text{,}\) we have \(\abs{f_n(x)-f(x)} \leq \snorm{f_n - f}_S < \epsilon\text{.}\)
On the other hand, suppose \(\{ f_n \}_{n=1}^\infty\) converges uniformly to \(f\text{.}\) Let \(\epsilon > 0\) be given. Then find \(N\) such that for all \(n \geq N\text{,}\) we have \(\abs{f_n(x)-f(x)} < \epsilon\) for all \(x \in S\text{.}\) Taking the supremum over \(x \in S\text{,}\) we see that \(\snorm{f_n - f}_S \leq \epsilon\text{.}\) Hence \(\lim_{n\to\infty} \snorm{f_n-f}_S = 0\text{.}\)
Sometimes it is said that \(\{ f_n \}_{n=1}^\infty\) converges to \(f\) in uniform norm instead of converges uniformly if \(\snorm{f_n-f}_S \to 0\text{.}\) The proposition says that the two notions are the same thing for bounded functions.
Let \(f_n \colon [0,1] \to \R\) be defined by \(f_n(x) \coloneqq \frac{nx+ \sin(nx^2)}{n}\text{.}\) We claim \(\{ f_n \}_{n=1}^\infty\) converges uniformly to \(f(x) \coloneqq x\text{.}\) Let us compute:
Let \(f_n \colon S \to \R\) be bounded functions. The sequence is Cauchy in the uniform norm or uniformly Cauchy if for every \(\epsilon > 0\text{,}\) there exists an \(N \in \N\) such that for all \(m,k \geq N\text{,}\)
Let \(f_n \colon S \to \R\) be bounded functions. Then \(\{ f_n \}_{n=1}^\infty\) is Cauchy in the uniform norm if and only if there exists an \(f \colon S \to \R\) and \(\{ f_n \}_{n=1}^\infty\) converges uniformly to \(f\text{.}\)
First suppose \(\{ f_n \}_{n=1}^\infty\) is Cauchy in the uniform norm. Let us define \(f\text{.}\) Fix \(x\text{.}\) The sequence \(\bigl\{ f_n(x) \bigr\}_{n=1}^\infty\) is Cauchy because
The sequence \(\{ f_n \}_{n=1}^\infty\) converges pointwise to \(f\text{.}\) To show that the convergence is uniform, let \(\epsilon > 0\) be given. Find an \(N\) such that for all \(m, k \geq N\text{,}\) we have \(\norm{f_m-f_k}_S < \nicefrac{\epsilon}{2}\text{.}\) In other words, for all \(x\text{,}\) we have \(\abs{f_m(x)-f_k(x)} < \nicefrac{\epsilon}{2}\text{.}\) For any fixed \(x\text{,}\) take the limit as \(k\) goes to infinity. Then \(\abs{f_m(x)-f_k(x)}\) goes to \(\abs{f_m(x)-f(x)}\text{.}\) Consequently for all \(x\text{,}\)
Next, we prove the other direction. Suppose \(\{ f_n \}_{n=1}^\infty\) converges uniformly to \(f\text{.}\) Given \(\epsilon > 0\text{,}\) find \(N\) such that for all \(n \geq N\text{,}\) we have \(\abs{f_n(x)-f(x)} < \nicefrac{\epsilon}{4}\) for all \(x \in S\text{.}\) Therefore, for all \(m, k \geq N\) and all \(x\text{,}\)
Suppose \(f_n \colon S \to \R\) are functions that converge uniformly to \(f \colon S \to \R\text{.}\) Suppose \(A \subset S\text{.}\) Show that the sequence of restrictions \(\{ f_n|_A \}_{n=1}^\infty\) converges uniformly to \(f|_A\text{.}\)
Suppose \(\{ f_n \}_{n=1}^\infty\) and \(\{ g_n \}_{n=1}^\infty\) defined on some set \(A\) converge to \(f\) and \(g\) respectively pointwise. Show that \(\{ f_n+g_n \}_{n=1}^\infty\) converges pointwise to \(f+g\text{.}\)
Suppose \(\{ f_n \}_{n=1}^\infty\) and \(\{ g_n \}_{n=1}^\infty\) defined on some set \(A\) converge to \(f\) and \(g\) respectively uniformly on \(A\text{.}\) Show that \(\{ f_n+g_n \}_{n=1}^\infty\) converges uniformly to \(f+g\) on \(A\text{.}\)
Find an example of a sequence of functions \(\{ f_n \}_{n=1}^\infty\) and \(\{
g_n \}_{n=1}^\infty\) that converge uniformly to some \(f\) and \(g\) on some set \(A\text{,}\) but such that \(\{ f_ng_n \}_{n=1}^\infty\) (the multiple) does not converge uniformly to \(fg\) on \(A\text{.}\) Hint: Let \(A \coloneqq \R\text{,}\) let \(f(x)\coloneqq g(x) \coloneqq x\text{.}\) You can even pick \(f_n = g_n\text{.}\)
Suppose there exists a sequence of functions \(\{ g_n \}_{n=1}^\infty\) uniformly converging to \(0\) on \(A\text{.}\) Now suppose we have a sequence of functions \(\{ f_n \}_{n=1}^\infty\) and a function \(f\) on \(A\) such that
Let \(\{ f_n \}_{n=1}^\infty\text{,}\)\(\{ g_n \}_{n=1}^\infty\) and \(\{ h_n \}_{n=1}^\infty\) be sequences of functions on \([a,b]\text{.}\) Suppose \(\{ f_n \}_{n=1}^\infty\) and \(\{ h_n \}_{n=1}^\infty\) converge uniformly to some function \(f \colon [a,b] \to \R\) and suppose \(f_n(x) \leq g_n(x) \leq h_n(x)\) for all \(x \in [a,b]\text{.}\) Show that \(\{ g_n \}_{n=1}^\infty\) converges uniformly to \(f\text{.}\)
Let \(f_n \colon [0,1] \to \R\) be a sequence of increasing functions (that is, \(f_n(x) \geq f_n(y)\) whenever \(x \geq y\)). Suppose \(f_n(0) = 0\) and \(\lim\limits_{n \to \infty} f_n(1) = 0\text{.}\) Show that \(\{ f_n \}_{n=1}^\infty\) converges uniformly to \(0\text{.}\)
Consider a sequence of functions \(f_n \colon [0,1] \to \R\) so that there is a sequence of distinct numbers \(x_n \in [0,1]\) such that for all \(n\text{,}\)
Fix a continuous \(h \colon [a,b] \to \R\text{.}\) Let \(f(x) \coloneqq h(x)\) for \(x \in [a,b]\text{,}\)\(f(x) \coloneqq h(a)\) for \(x < a\) and \(f(x) \coloneqq h(b)\) for all \(x > b\text{.}\) First show that \(f \colon \R \to \R\) is continuous. Now let \(f_n\) be the function \(g\) from Exercise 5.3.7 with \(\epsilon = \nicefrac{1}{n}\text{,}\) defined on the interval \([a,b]\text{.}\) That is,
\begin{equation*}
f_n(x) \coloneqq \frac{n}{2} \int_{x-1/n}^{x+1/n} f .
\end{equation*}
Show that \(\{ f_n \}_{n=1}^\infty\) converges uniformly to \(h\) on \([a,b]\text{.}\)
Prove that if a sequence of functions \(f_n \colon S \to \R\) converge uniformly to a bounded function \(f \colon S \to \R\text{,}\) then there exists an \(N\) such that for all \(n \geq N\text{,}\) the \(f_n\) are bounded.
Suppose there is a single constant \(B\) and a sequence of functions \(f_n \colon S \to \R\) that are bounded by \(B\text{,}\) that is \(\abs{f_n(x)} \leq B\) for all \(x \in S\text{.}\) Suppose that \(\{ f_n \}_{n=1}^\infty\) converges pointwise to \(f \colon S \to \R\text{.}\) Prove that \(f\) is bounded.
