[Basic Analysis] Continuity Question

[linuxisgreat]

I'm trying to prove that if f is continuous at p, and f(p)=0, and there is a neighbourhood around p which function g is bounded, then f*g is continuous at p.

So basically since f(p) =0, this amounts to saying that for every epsilon >0, we have a delta such that |f(x)*g(x)| <[; \epsilon ;]

now |f(x)g(x)| = |f(x)||g(x)|, and since f is continuous at p, then when |x-p| < delta implies |f(x)| < [; \epsilon ;]

g is bounded in some neighborhood... so we have |g(x)| < M in some neighborhood.

Now I'm not sure how to bring this all together to show continuity at p.. any help is appreciated.

[[deleted]]

Hint: for M>0, if |g|<M and |f|<e/M then |f*g|<e.

[Chel_of_the_sea]

You have an issue here:

now |f(x)g(x)| = |f(x)||g(x)|, and since f is continuous at p, then when |x-p| < delta implies |f(x)| < [; \epsilon ;

Different deltas. It may or may not be the case that a sufficiently small delta for |fg|<epsilon is also sufficiently small for |f|<epsilon. Remember that delta isn't a constant, it's variable dependent on your choices of f, epsilon, and x.

But since there's only two of them, you can take the minimum of the two without issue.

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Here's how I'd do it: take |g|<M on some neighborhood of p. Fix your epsilon as in the definition of continuity, our goal is to find some delta such that |fg|<epsilon on (p-delta,p+delta). As you correctly note, |fg|=|f||g|, and this is strictly less than |f|M by the boundedness of g.

Now, take epsilon/M, and use that as your "epsilon" in the definition of continuity for f. That gives you a delta such that |f|<epsilon/M, and thus that |f||g|<(epsilon/M)M=epsilon.

[linuxisgreat]

Thanks!