16.4: Uniform Convergence of Function Sequences

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Uniform Convergence of Function Sequences

For this discussion, we will only consider functions with \(g_n\) where

\[\mathbb{R} \rightarrow \mathbb{R} \nonumber \]

Definition: Uniform Convergence

The sequence (Section 16.2) \(\left.\left\{g_{n}\right\}\right|_{n=1} ^{\infty}\) converges uniformly to function \(g\) if for every \(\varepsilon > 0\) there is an integer \(N\) such that \(n \geq N\) implies

\[\left|g_{n}(t)-g(t)\right| \leq \epsilon \label{16.11} \]

for all \(t \in \mathbb{R}\).

Obviously every uniformly convergent sequence is pointwise (Section 16.3) convergent. The difference between pointwise and uniform convergence is this: If \(\left\{g_{n}\right\}\) converges pointwise to \(g\), then for every \(\varepsilon> 0\) and for every \(t \in \mathbb{R}\) there is an integer \(N\) depending on \(\varepsilon\) and \(t\) such that Equation \ref{16.11} holds if \(n≥N\). If \(\left\{g_{n}\right\}\) converges uniformly to \(g\), it is possible for each \(\varepsilon>0\) to find one integer \(N\) that will do for all \(t \in \mathbb{R}\).

Example \(\PageIndex{1}\)

\[g_{n}(t)=\frac{1}{n}, t \in \mathbb{R} \nonumber \]

Let \(\varepsilon > 0\) be given. Then choose \(N=\left\lceil\frac{1}{\varepsilon}\right\rceil\). Obviously,

\[\left|g_{n}(t)-0\right| \leq \epsilon, \quad n \geq N \nonumber \]

for all \(t\). Thus, \(g_n(t)\) converges uniformly to \(0\).

Example \(\PageIndex{2}\)

\[g_{n}(t)=\frac{t}{n}, t \in \mathbb{R} \nonumber \]

Obviously for any \(\varepsilon > 0\) we cannot find a single function \(g_n(t)\) for which Equation \ref{16.11} holds with \(g(t)=0\) for all \(t\). Thus \(g_n\) is not uniformly convergent. However we do have:

\[g_{n}(t) \rightarrow g(t) \text{ pointwise }\nonumber \]

Conclusion

Uniform convergence always implies pointwise convergence, but pointwise convergence does not guarantee uniform convergence.