0
$\begingroup$

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 $x_0$.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ Yeah ok, makes sense. $\endgroup$ Feb 6 '18 at 15:00
2
$\begingroup$

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 $x=x_{0}$.

$\endgroup$
0
1
$\begingroup$

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 $\delta = \epsilon/M.$

$\endgroup$
2
  • $\begingroup$ Thank you, but I think I prefer using sequences. $\endgroup$ Feb 6 '18 at 15:18
  • $\begingroup$ I am sorry, I don't understand you. $\endgroup$ Feb 6 '18 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.