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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.

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  • $\begingroup$ 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). $\endgroup$
    – hardmath
    Jan 23 at 3:39

2 Answers 2

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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.

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  • $\begingroup$ 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 $sup|f_n(x)g_n(x)-f(x)g(x)|$? $\endgroup$
    – Abake
    Jan 21 at 19:31
  • $\begingroup$ Your definition is right. Here $f(x)g(x) = 0$ $\endgroup$
    – ZAF
    Jan 21 at 19:32
  • $\begingroup$ how are we able to conclude that $f(x)g(x) = 0$? $\endgroup$
    – Abake
    Jan 21 at 19:33
  • $\begingroup$ @Abake I think there is some confusion here; I can settle it, but first tell me what you think $E$ is. $\endgroup$ Jan 21 at 19:33
  • $\begingroup$ $f_{n}$ 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 $x \in E$. $\endgroup$
    – ZAF
    Jan 21 at 19:36
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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 $0$.

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  • $\begingroup$ 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 $|f_n(x)g_n(x) -f(x)g(x)| \leq M|f_n(x) - f(x)g(x)|$? $\endgroup$
    – Abake
    Jan 21 at 19:29
  • $\begingroup$ @Abake what is $g$? The limit of the sequence $g_n$? It doesn't need to converge for $f_ng_n \to 0$. $\endgroup$ Jan 21 at 19:32
  • $\begingroup$ Yes g is the limit function of the sequence. i thought it was needed in order to calculate $sup|f_n(x)g_n(x) - f(x)g(x)|$ $\endgroup$
    – Abake
    Jan 21 at 19:36

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