Concept of Uniform Convergence

Section 2.1 Concept of Uniform Convergence

Simple examples illustrate that pointwise limit of a sequence of continuous functions can fail to be continuous, and that even pointwise limit of a sequence of continuously differentiable functions can also fail to be continuous. The key cause is that the convergence may not be uniform, namely, for each \(\epsilon >0\) and each \(x\in E\text{,}\) there exists \(N\) which may depend on \(\epsilon\) as well as on \(x\) such that when \(n\ge N\text{,}\) we have \(|f_n(x)-f(x)| < \epsilon\text{;}\) but the minimum \(N\) needed for \(x\in E\) may not be uniform over \(x\in E\text{.}\) The following definition defines the notion of uniform convergence.

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Definition 2.1.1. Uniform Convergence.

Let \(\{ f_n(x)\}\) be a sequence of functions defined on a set \(E\text{.}\) Then \(\{ f_n(x)\}\) converges uniformly on \(E\) to a limit function \(f(x)\) on \(E\) if, for every \(\epsilon >0\text{,}\) there exists \(N\in \mathbb N\) such that \(|f_n(x)-f(x)| < \epsilon\) for all \(n\ge N\) and for all \(\; x\in E\text{.}\)

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Remark 2.1.2.

Note that one needs to specify the set \(E\) when discussing the notion of uniform convergence. For example, the sequence of functions \(f_{n}(x)=x^{n}\) is converging pointwise to \(f(x):=0\) on \([0,1)\text{,}\) but not uniformly on \([0,1)\text{.}\) However, for any fixed \(\; 0 < \delta < 1\text{,}\) \(f_{n}(x) \to 0\) uniformly over \([0,\delta]\text{.}\)

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Example 2.1.3. Examine the notion of pointwise and uniform convergence.

We continue to work with \(f_{n}(x)=x^{n}\) and consider it on \([0, 1]\text{.}\) It still converges pointwise, but the limiting function

\begin{equation*} f(x)=\begin{cases} 0 \amp 0\le x \lt 1 \\ 1 \amp x=1\\ \end{cases} \end{equation*}

fails to be continuous at \(x=1\text{.}\) This is due to the failure of uniform convergence on \([0, 1]\text{.}\)

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Here we still have

\begin{equation*} \lim_{n\to \infty} \int_{0}^{1}x^{n}\, dx = \int_{0}^{1} f(x)\, dx\text{.} \end{equation*}

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If we modify \(f_{n}(x)\) into
\begin{equation*} g_{n}(x)=(n+1) x^{n}(1-x)\text{.} \end{equation*}
Then \(g_{n}(x)\to g(x):=0\) pointwise on \([0, 1]\text{,}\) although the convergence is still not uniform, as \(\max_{[0,1]}g_{n}(x)=\left(\frac{n}{n+1}\right)^{n}\to e^{-1}\) as \(n\to \infty\text{.}\) When we examine \(\int_{0}^{1}g_{n}(x)\, dx\text{,}\) we find it equal to \((n+2)^{-1}\to 0\) as \(n\to \infty\text{.}\)

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However, if we modify \(g_{n}(x)\) into \(h_{n}(x) :=(n+2)g_{n}(x)\text{,}\) we continue to have \(h_{n}(x) \to 0\) pointwise on \([0, 1]\text{.}\) The convergence is still not uniform, as \(\max_{[0,1]}h_{n}(x)= (n+2)\left(\frac{n}{n+1}\right)^{n} \to \infty\text{.}\) And we find that
\begin{equation*} \int_{0}^{1}h_{n}(x) = 1 \not \to \int_{0}^{1} 0\, dx\text{!} \end{equation*}

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Note that in all the cases above, once we fix some \(0 < \delta < 1\text{,}\) the relevant sequences of functions converge to \(0\) uniformly over \([0,\delta]\text{,}\) so they fail to converge uniformly only over a neighborhood of \(x=1\text{.}\) To examine question (b) raised earlier, we just need to answer whether the following holds:

\begin{gather*} \forall \; \epsilon>0, \exists \; 0 < \delta < 1 \text{ and } N \text{ such that for all } n\ge N,\\ \left| \int_{\delta}^{1} k_{n}(x)\, dx \right| < \epsilon, \end{gather*}

where \((k_{n}(x))\) stands for one of the sequences \(\{ f_{n}(x)\}, \{g_{n}(x)\}, \{h_{n}(x)\}\) above. Through direct examination we find that this holds for \(\{f_{n}(x)\}, \{g_{n}(x)\}\text{,}\) but fails for \(\{h_{n}(x)\}\text{.}\)

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Remark 2.1.4.

In most contexts we confine to \(\mathbb R\) or \(\mathbb C\)-valued functions defined on a set \(E\text{.}\) The notion of uniform convergence can be defined as along as there is a quantitative way to describe how close \(f_{n}(x)\) is to \(f(x)\text{.}\) If \(Y\) is a metric space with \(d\) as the metric, and \(f_{n}, f: E \mapsto Y\text{,}\) then the notion of uniform convergence of \(f_{n}\) makes sense. More specifically, \(f_{n}\to f\) uniformly over \(E\text{,}\) if for any \(\epsilon > 0\text{,}\) there exists \(N\) such that for all \(n\ge N\text{,}\) \(\sup_{x\in E}d(f_{n}(x), f(x)) < \epsilon\text{.}\) This notion does not require \(E\) to be a metric space.

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On the other hand, the notion of uniform continuity also requires a mechanism of quantitatively describing how close \(x, y\) are in \(E\text{,}\) and it would make sense if both \(E\) and \(Y\) are metric spaces.

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