Uniform Product of Continuous Functions is Continuous

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Let $X$ be a metric space.

Let $\struct {\mathbb K, \norm{\,\cdot\,}}$ be a valued field.

Let $\sequence {f_n}$ be a sequence of bounded continuous mappings $f_n: X \to \mathbb K$.

Let the product $\ds \prod_{n \mathop = 1}^\infty f_n$ converge uniformly to $f$.

Then $f$ is continuous.

Proof 1

Let $n_0 \in \N$ be such that the sequence of partial products of $\ds \prod_{n \mathop = n_0}^\infty f_n$ converges uniformly.

By the Uniform Limit Theorem, $\ds \prod_{n \mathop = n_0}^\infty f_n$ is continuous.

Because $f_1, \ldots, f_{n_0 - 1}$ are continuous, so is $\ds \prod_{n \mathop = 1}^\infty f_n$.


Proof 2

Follows directly from:

Partial Products of Uniformly Convergent Product Converge Uniformly
Uniform Limit Theorem


Also see