Given 2 sequences of functions $f_n$ and $g_n$ on an interval $[a, b]$.
$f_n$ is uniformly convergent to $f$.
$g_n$ is uniformly convergent to $g$.
And there exists 2 real numbers $M_1$ and $M_2$ such that:
$|f(x)| < M_1$
and
$|g(x)| < M_2$
for any $x \in [a,b]$.
How would I prove the the product $f_n \cdot g_n$ converges uniformly to $f\cdot g$?
I think the product does not converge uniformly, but having trouble proving it.
Best Answer
Hint: recall the proof for sequences of real numbers: if $x_n \to x$ and $y_n \to y$ then $x_n y_n \to xy$, which goes like:
\begin{align}|x_n y_n - xy| & = |x_n y_n - x y_n + x y_n - xy| \\ & = |y_n (x_n - x) + x (y_n - y)| \\ & \leq |y_n| |x_n - x| + |x| |y_n - y|. \end{align}
Now you must show $|y_n|$ is bounded as an intermediate lemma, after which the result is immediate from definitions.
For the context of uniform convergence you'll have the same thing for each $x$:
$$|f_n(x) g_n(x) - f(x) g(x)| \leq |f_n(x)| |g_n(x) - g(x)| + |g(x)| |f_n(x) - f(x)|.$$
Then for each $x$ you have that $|f_n(x)|$ is bounded for the same reason you have this on real numbers. So we immediately get pointwise convergence. Now take a supremum over $x$ on both sides and see what happens.