uniform convergence on a sequence of two functions

I have the following real analysis question.

Let $$f_n , g_n$$ be two sequences of functions such that $$f_n \rightarrow 0$$ uniformly and $$g_n$$ is bounded by $$M$$: $$\forall x: |g_n(x)| \leq M$$. Prove that $$f_ng_n \rightarrow 0$$ uniformly.

In my class, when asked to prove the uniform convergence of a given sequence of functions $$f_n$$, we have been showing that $$\sup|f_n(x) - f(x)| \rightarrow 0$$. My question is: how can we use the bound on $$g_n$$ in this approach.

If we have $$|f_n(x)g_n(x) - f(x)g(x)|$$, I'm unsure what we can do about the g(x) function.

Thanks so much.

• Note that $g(x)$ doesn't appear in the problem setup. I don't think you were actually asked to do anything "about the $g(x)$ function" (it was never defined). Jan 23 at 3:39

we have that $$f_{n} \to 0$$ uniformly on $$E$$. Then $$\sup|f_{n}(x)| \to 0$$. Thus for $$\varepsilon > 0$$ there exists some $$N>0$$ such that if $$n\geq N \implies |f_{n}(x)| \leq \sup|f_{n}(x)|< \varepsilon$$ for all $$x \in E$$. Now, take $$\varepsilon > 0$$, and $$\tilde{\varepsilon} = \frac{\varepsilon}{M}$$. Then there exist some $$N > 0$$ such that if $$n \geq N \implies |f_{n}(x)| < \tilde{\varepsilon}$$ for all $$x \in E$$. Then $$|f_{n}(x)g_{n}(x)| = |f_{n}(x)||g_{n}(x)| \leq |f_{n}(x)|M < \tilde{\varepsilon}M = \varepsilon$$ for all $$n\geq N$$, for all $$x \in E$$.

Then $$\sup|f_{n}(x)g_{n}(x)| \to 0$$

So $$f_{n}g_{n} \to 0$$ uniformly.

• Thanks for the quick reply. Do i have the definition wrong? can we just say $sup|f_n(x)g_n(x)| \rightarrow 0$? i thought it was $sup|f_n(x)g_n(x)-f(x)g(x)|$? Jan 21 at 19:31
• Your definition is right. Here $f(x)g(x) = 0$
– ZAF
Jan 21 at 19:32
• how are we able to conclude that $f(x)g(x) = 0$? Jan 21 at 19:33
• @Abake I think there is some confusion here; I can settle it, but first tell me what you think $E$ is. Jan 21 at 19:33
• $f_{n}$ converges uniformly to $f$ if and only if $sup|f_{n}(x) - f(x)| \to 0$. As we have that $sup|f_{n}(x)g_{n}(x) - 0| = sup|f_{n}(x)g_{n}(x)| \to 0$. We can conclude that $f(x)g(x) = 0$ for all $x \in E$.
– ZAF
Jan 21 at 19:36

Hint: $$|f_n(x)g_n(x)| \leq |f_n(x)|M$$. If you know that $$f_n \to 0$$ uniformly, then you know $$|f_n(x)| < \varepsilon$$ for every $$x$$, if $$n$$ is large enough. Just pick $$\varepsilon' = \varepsilon/M$$

Edit: The OP seems a little confused about uniform convergence itself, so I'll give a brief explanation.

I suppose you are taking a course on Real Analysis, and so you must have already met notions of function convergence such as point-wise convergence. Now, you are working with uniform convergence. When does $$f_n \to f$$ uniformly? Well, when, for any $$\varepsilon$$ (no matter how tiny), you can find a number $$N$$ after which all members in the sequence $$f_n$$ are $$\varepsilon$$-close to $$f$$ in the sense that $$\forall x: |f_n(x)-f(x)| < \varepsilon$$.

How does uniform convergence look like? Specially, what does it mean for $$f_n \to 0$$?

If $$|f_n(x)-0| < \varepsilon$$, then it is as if the function $$f$$ is entirely contained inside a "street" (or "margin", "strip", suit yourself here) of length $$\varepsilon$$ centered at the origin.

One example of such $$f_n$$ is $$f_n(x) = 1/n$$.

Ok, having said that, suppose now you have an additional sequence $$g_n$$ such that $$\forall x: |g_n(x)| < M$$, for some positive $$M$$. These $$g_n$$ don't need to converge in any sense. For example, picture a "street" of length $$1$$ and, inside it, lots of functions that just wiggle randomly without leaving the strip.

Well, these guys are still bounded, so multiplying them by $$1/n$$ (for example) must slowly bring the resulting sequence $$f_ng_n$$ to ever smaller "streets" (of length $$\varepsilon$$). In the limit, this process brings $$f_ng_n$$ to $$0$$.

Formally, since $$f_n \to 0$$, we can pick $$N$$ s.t., for every $$n \geq N$$ and every $$x$$, we have $$|f_n(x)| < \varepsilon/M$$

Thus, for every $$n \geq N$$ and every $$x$$, $$|f_ng_n(x) - 0| = |f_n(x)g_n(x)| = |f_n(x)||g_n(x)| \leq |f_n(x)|M < \varepsilon$$.

This proves that $$f_ng_n$$ converge to $$0$$ uniformly. The important thing to note here is that "f" as in our definition of unif. convergence here is $$0$$.

• Thank you for the response. I see that we can bound like that but I'm unsure of what to do with g(x). if we are given $g_n(x)$ is bounded does that tell us anything about g(x)? can we bound it as well? Becuase wouldn't we now have $|f_n(x)g_n(x) -f(x)g(x)| \leq M|f_n(x) - f(x)g(x)|$? Jan 21 at 19:29
• @Abake what is $g$? The limit of the sequence $g_n$? It doesn't need to converge for $f_ng_n \to 0$. Jan 21 at 19:32
• Yes g is the limit function of the sequence. i thought it was needed in order to calculate $sup|f_n(x)g_n(x) - f(x)g(x)|$ Jan 21 at 19:36