# Continuous function times a bounded function

The following:

Let $f,g : D \rightarrow \mathbb{R}$ be functions with $D \subset \mathbb{R}$ such that $f$ is continuous and $g$ is bounded. Let $x_0 \in D$ and define $h(x) = (f(x) - f(x_0)) \cdot g(x)$. Prove that $h$ is continuous at $x_0$.

I have an idea how to solve it, but it looks wrong:

Let $(x_n)$ be a sequence such that $\lim_{n \rightarrow \infty} x_n = x_0$. Because $f$ is continuous, it follows that

$$\lim_{n \rightarrow \infty} (f(x_n) - f(x_0)) = f(\lim_{x \rightarrow \infty}x_n) - f(x_0) = f(x_0) -f(x_0) = 0$$

So $f(x_n) - f(x_0)$ is a null sequence. Up to this point everything seems fine. But now I don't know if my argumentation is correct:

Because $g$ is bounded, it follows that the sequence $(g(x_n))$ is bounded too ($\color{red}{???}$)

Because $(f(x_n) - f(x_0))$ is a null sequence and because $(g(x_n))$ is a bounded sequence it follows that:

$$\lim_{n \rightarrow \infty} h(x_n) = \lim_{n \rightarrow \infty} (f(x_n) - f(x_0)) \cdot g(x_n) = 0 =h(x_0)$$

So $h$ is continuous at $x_0$.

Yes, $g$ bounded means there is a constant $M$ such that $|g(x)|< M$ for every $x$, so in particular for each $x_n$

• Yeah ok, makes sense. Feb 6 '18 at 15:00

Let $M>0$ be such that $|g(x)|<M$ for all $x\in D$, then \begin{align*} |h(x)|&\leq M|f(x)-f(x_{0})|, \end{align*} and we have $\lim_{x\rightarrow x_{0}}|f(x)-f(x_{0})|=0$ by the continuity of $f$. By Squeeze Theorem we have $\lim_{x\rightarrow x_{0}}h(x)=0$. But $h(x_{0})=0$, so $\lim_{x\rightarrow x_{0}}h(x)=h(x_{0})$, this proves the continuity of $h$ at $x=x_{0}$.

Use $\epsilon$, $\delta$ formalism.

With $M$, real, positive we have:

$|g(x)| \lt M,$ for $x \in D.$

Let $\epsilon \gt 0$ be given.

$f(x)$ is continuos at $x=x_0:$

There is a $\delta >0$ such that:

$|x-x_0| \lt \delta$ implies

$|f(x)-f(x_0)| < \epsilon$.

$|h(x)-h(x_0)| =$

$|f(x)-f(x_0)||g(x)| \lt \delta M.$

Choose $\delta = \epsilon/M.$

• Thank you, but I think I prefer using sequences. Feb 6 '18 at 15:18
• I am sorry, I don't understand you. Feb 6 '18 at 17:55