Proving $\{x \cdot f(x)\} \rightarrow 0$ given $\{x\} \rightarrow 0$

Question

One of my proofs that I am doing is dependent on me trying to piece this last part out. How can I prove that this is true (or is it even true)?

If $\{x\} \rightarrow 0$, then $\{x \cdot f(x)\} \rightarrow 0$, where $f: \mathbb{R} \rightarrow \mathbb{R}$ is bounded.

Answer 1

Hint: If $f=0$ then it's easy. Suppose not and fix $\epsilon>0$ and find a big enough $N$ with $|x_n|<\frac{\epsilon}{\sup_{n}|f(x_n)|}$ whenever $n\geq N$.

Answer 2

Since $f$ is bounded (say, $|f(x)| \leq M$) we have the following estimate for all $x$: $$ |xf(x)| \leq |x|M. $$ Now since $M$ is a constant, what happens as $x \to 0$?

Answer 3

Let $(x_{n})$ be a be a sequence that converges to zero, and let $f:\mathbb{R} \rightarrow \mathbb{R}$ a bounded function. Let $(p_{n}):= nf(n)$. Since $f$ is bounded, there exists $M_{f} \geq 0 $ such that $M_{f} = sup_{x \in \mathbb{R}}(f(x))$. $|p_{n}| = |x_{n}f(n)| \leq x_{n}M_{f}$. Since $x_{n}$ converges to zero, for every $\epsilon >0$ there exists $N_{\epsilon} \in \mathbb{Z}_{\geq 0}$ such that for all integers $k \geq N_{\epsilon}$, $|x_{k}| < \epsilon$. Since this is true for every $\epsilon >0$, there exists $N_{\frac{\epsilon}{M_{f}}}$ such that for all $k > N_{\frac{\epsilon}{M_{f}}}$, $|x_{n}| < \frac{\epsilon}{M_{f}}$. This implies that for all $k > N_{\frac{\epsilon}{M_{f}}}$, $|p_{k}| \leq |x_{k}M_{f}| < \frac{\epsilon}{M_{f}}M_{f} = \epsilon$. Since $\epsilon$ was arbitrary, it follows that $\lim_{n \rightarrow \infty}p_{n} = 0$. $\square$