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Let $f$ and $g$ be functions defined on $\mathbb R$. Assume that $f$ is continuous at $0$ and $g$ is discontinuous at $0$. Prove that if $f(0) \neq 0$, then $f*g$ is not continuous at $0$. Does $f(0)=0$ imply that $f*g$ is continuous at 0?

The definition of continuous states that function $f(x)$ is continuous at $x_0$ if and only if for each $\epsilon>0$, $\exists \delta$ such that $x \in dom(f)$ and $|x-x_0|<\delta$ imply that $|f(x)-f(x_0)|<\epsilon$

I'm not sure how to do the first part of the question because this is the first time I have had to use multiplication in the continuous definition. However, for the second part, I believe that it does imply that because it will be continuous getting closer to zero and pull $g$ down to zero as well. But again, I'm not sure how to prove this with multiplication.

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  • $\begingroup$ I'm not sure you can prove the first part because if f is constant f(g(x)) is always continuous even if g isn't continuous. Are there any other conditions that are missing? $\endgroup$ – lordoftheshadows Oct 10 '16 at 23:44
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    $\begingroup$ * is the product? The composition? $\endgroup$ – Martín Vacas Vignolo Oct 10 '16 at 23:44
  • $\begingroup$ I think it is just the product, not composition. $\endgroup$ – AndroidFish Oct 11 '16 at 0:04
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Hints: Here are two strategies.

  1. (Contrapositive) If $fg$ is continuous at $0$ and $f$ is continuous and non-zero, then $g = (fg)/f$ is continuous at $0$ (by a theorem you've probably seen recently).

  2. (Direct) Assume without loss of generality that $g$ is bounded in some neighborhood of $0$. (If $g$ is unbounded and $f(0) \neq 0$, then $fg$ is unbounded, hence discontinuous.) If $g$ is discontinuous at $0$, there exist sequences $(x_{k}) \to 0$ and $(y_{k}) \to 0$ such that $$ (g(x_{k})) \to \lim_{\delta \to 0^{+}} \inf_{0 < |x| < \delta} g(x),\qquad (g(y_{k})) \to \lim_{\delta \to 0^{+}} \sup_{0 < |x| < \delta} g(x), $$ and the former is strictly smaller than the latter. Standard results about limits of sequences show that if $f(0) = M \neq 0$, then $(fg(x_{k}))$ and $(fg(y_{k}))$ have different limits, so $fg$ is discontinuous at $0$.

    (If it's too early in your analysis career to use $\liminf$ and $\limsup$, you can instead show there exists a sequence converging to $0$ along which the values of $g$ do not converge, or use some other method to argue that there exist two sequences converging to $0$ along which the limiting values of $g$ differ.)

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