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:

$|f(x)| < M_1$


$|g(x)| < M_2$

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.


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.

  • $\begingroup$ If the functions aren't bounded, though (but are uniformly convergent), then the product doesn't converge, right? I think that's true but I'm having trouble thinking of an example. $\endgroup$ – M T Mar 5 '15 at 21:39
  • $\begingroup$ @Michael At least if $f_n,g_n$ are continuous then they must be uniformly bounded in order to be uniformly convergent on a compact set. $\endgroup$ – Ian Mar 5 '15 at 21:43
  • $\begingroup$ What if $f_n, g_n$ are just uniformly convergent functions? In that general case (without them being bounded, continuous, etc.) then $f_n g_n$ doesn't necessarily converge to $fg$, right? $\endgroup$ – M T Mar 5 '15 at 21:45
  • $\begingroup$ @Michael Even with continuity, by omitting the compactness we can get things like $(x+1/n)(x+1/n)=x^2+2x/n+1/n^2$, which fails to converge uniformly even though the factors do converge uniformly. You could probably compress such a thing into a compact interval by using jumps (and by arbitrarily picking a value at the bad endpoint) and then the convergence would fail again. $\endgroup$ – Ian Mar 5 '15 at 21:59

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