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Suppose a series of functions $\sum f_n$ has uniform absolute convergence on $E$ and $(g_n)$ is a uniformly bounded family of functions there, with common bound $M$. The triangle inequality givesThe right-hand side tends uniformly to zero as $q>p\to \infty$. The Uniform Cauchy criterion proves uniform convergence of $\sum f_n{g}_n$. Applying the same estimate to $\sum |f_n{g}_n|$ proves uniform absolute convergence too.

The shared bound over both $n$ and $x$ matters: separate bounds for each $g_n$ can grow too fast. For example, the constant functions $f_n(x)=2^{-n}$ and $g_n(x)=2^n$ have a summable original series but products identically one.

For real-valued functions, the Uniform Cauchy criterion applied to partial sums says that $\sum u_n$ has uniform convergence exactly whenSufficiency uses completeness of the real numbers.

Let $M$ be a common bound for the uniformly bounded family of functions $(g_n)$; increase it if necessary so that $M>0$. The assumed uniform absolute convergence of $\sum f_n$ gives $N$ such thatThe triangle inequality now yieldsThe Uniform Cauchy criterion therefore proves uniform convergence of $\sum f_n{g}_n$. The same estimate with $|f_n{g}_n|$ proves uniform absolute convergence as well.

Use uniform convergence to choose $N$ with $|f_n(x)-f_N(x)|\le 1$ for all $n\ge N$ and $x\in E$; this follows, for example, from the Uniform Cauchy criterion. Since each $f_j$ is a bounded function, put $M_j={\sup }_E|f_j|<\infty$ for $1\le j\le N$. Thenfor every $n$ and $x$. Thus $(f_n)$ is a uniformly bounded family of functions. The finite initial segment is essential: uniform convergence directly controls only a tail of a sequence.

The Uniform Cauchy criterion says that $(f_n:E\to \mathbb{R})$ converges uniformly on $E$ if and only ifFor necessity, use the triangle inequality through the uniform limit. For sufficiency, the condition makes $(f_n(x){)}_n$ a Cauchy sequence for each $x$, so completeness of the real numbers defines a pointwise limit $f(x)$. Apply the condition with $\epsilon /2$ and let $m\to \infty$ to obtain $|f_n(x)-f(x)|\le \epsilon /2<\epsilon$ uniformly in $x$.

A series of functions is uniformly absolutely convergent on a set if the series of their absolute values has uniform convergence there. A sufficient condition is the Weierstrass M-test: if $|f_n(x)|\le M_n$ for every $x$ and ${\sum }_n{M}_n<\infty$, then the tail is bounded uniformly by ${\sum }_{n\ge N}{M}_n\to 0$. This also proves uniform convergence of the original series by the Uniform Cauchy criterion.

If sequences of functions $f_n\to f$ and $g_n\to g$ have uniform convergence on the same domain and each term is a bounded function, their products converge uniformly to $fg$. The two families are uniformly bounded families of functions: one late term and the Uniform Cauchy criterion bound the entire tail, while finitely many early terms have a finite common bound. Write $f_n{g}_n-fg=f_n(g_n-g)+g(f_n-f)$ and take the supremum norm to obtain the displayed estimate. The boundedness hypothesis cannot be dropped: $f_n(x)=x$ and $g_n(x)=1/n$ converge uniformly on $\mathbb{R}$, whereas $x/n$ fails to converge uniformly to zero.