Q: Let $f: \mathbb{R}\rightarrow \mathbb{R}$ be a bounded function (that is, there exists some $M\geq 0$ so that $|f(x)|\leq M$ for all $x\in\mathbb{R}$). Define a new function $g:\mathbb{R}\rightarrow \mathbb{R}$ by $g(x) = xf(x)$.

a) Prove that $g$ is continuous (cts) at $0$.

b) If $a \neq 0$, prove that $f$ is cts at $a$ iff $g$ is continuous at $a$.

My attempt:

a) Let $N$ and $M$ be lower and upper bounds of $f$. Then $xN\leq xf(x)\leq xM$, and $\lim_{x\rightarrow 0}xN$ = $\lim_{x\rightarrow 0}xM$.

So by the squeeze theorem for functions, $\lim_{x\rightarrow 0}xf(x) = 0$, so $\lim_{x\rightarrow 0}g(x) = 0$.

But we also have $g(0) = 0f(0) = 0$, so $g$ is cts at $0$.

b) Suppose $f$ is cts at $a$. Then $\lim_{x\rightarrow a}g(x) = \lim_{x\rightarrow a}xf(x) = af(a) = g(a)$, so $g$ is cts at a.

Now suppose $g$ is cts at a. Then $\lim_{x\rightarrow a}f(x) = \lim_{x\rightarrow a}\frac{g(x)}{x}= \frac{g(a)}{a} = f(a)$.

So $f$ is cts at $a$.

So, are my proofs convincing enough to you guys?


The proof of part a) needs to be modified a bit. You have used the logic that if $N \leq f(x) \leq M$ then $xN \leq xf(x) \leq xM$. This holds only when $x \geq 0$. It is better to change the argument in a minor way.

We have $|f(x)|\leq M$ and this means that $|xf(x)| \leq |x|M$ and then we get $$-|x|M \leq xf(x) \leq |x|M$$ You can now take limits and use squeeze theorem to get $\lim_{x\to 0} xf(x) = 0$


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