Uniform Limits of Continuous Functions
How to preserve continuity in the limit of a sequence of continuous functions.
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Study Uniform Limits of Continuous Functions
Now we show that a uniformly convergent sequence of continuous functions has to converge to a continuous function.
Theorem 1. Let \(I\subset \mathbb{K}\) and let \((f_n)_{n\in\mathbb{N}}\) be a sequence of continuous functions \(f_n:I\to\mathbb{K}\) that uniformly converges to some \(f:I\to\mathbb{K}\). Then \(f\) is continuous.
Proof: Let \(\varepsilon>0\) and let \(x_0\in I\). Since \((f_n)_{n\in\mathbb{N}}\) converges uniformly to \(f:I\to\mathbb{K}\), there exists some \(N\) such that for all \(n\geq N\) and \(x\in I\) holds \[|f(x)-f_n(x)|<\frac\varepsilon3.\] Since \(f_n\) is continuous on \(I\), there exists some \(\delta>0\) such that for all \(x\in I\) with \(|x-x_0|<\delta\) holds \[|f_n(x_0)-f_n(x)|<\frac\varepsilon3.\] Altogether, we then have \[\begin{aligned} |f(x)-f(x_0)|=&\,|f(x)-f_n(x)+f_n(x)-f_n(x_0)+f_n(x_0)-f(x_0)|\\ \leq &\,|f(x)-f_n(x)|+|f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x_0)|\\ < &\,\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon. \end{aligned}\] Therefore, \(f\) is continuous by the \(\varepsilon\)-\(\delta\) criterion. \(\Box\)
Remark 2. The above result also gives a sufficient characterisation for a sequence of pointwisely convergent functions \((f_n)_{n\in\mathbb{N}}\) that are not uniformly convergent: If \((f_n)_{n\in\mathbb{N}}\) converges to a discontinuous function, then this sequence cannot be uniformly convergent. For instance, consider the sequence \((f_n)_{n\in\mathbb{N}}\) with \(f_n:[0,1]\to\mathbb{R}\), \(x\mapsto x^n\). The pointwise limit is given by \[f(x)=\lim_{n\to\infty}f_n(x)=\begin{cases}0&,\text{ if }x\in[0,1),\\1&,\text{ if }x=1,\end{cases}\] which is discontinuous at \(x=1\) though each \(f_n\) is continuous. Therefore, this sequence cannot be uniformly convergent.
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