Prove that if $f$ is continuous then $\Vert f\Vert$ is continuous. Is the converse true?

Question

I need help with this problem:

Given a function $f:D\subset\mathbb{R}\rightarrow\mathbb{R}^n$, consider the function $\Vert f\Vert :D\subset\mathbb{R}\rightarrow\mathbb{R}$ where $\Vert f\Vert (t) = \Vert f(t)\Vert, t\in\mathbb{R}$. Prove that if g is continuous the $\Vert f\Vert$ is continuous. Is the converse true?

I don't understand this problem. I think that the function $\Vert f\Vert :D\subset\mathbb{R}\rightarrow\mathbb{R}$ takes the norm of every number plugged in, and we get the same result that the function $\Vert f(t)\Vert$ would give us, which is taking the norm of the function. How do I prove this problem? I tried to use limits from both sides, but I don't have any function to apply them to.

Answer 1

Hint: By the reverse triangle inequality we know that $|(||x||-||y||)|\leq ||x-y||$ for all $x,y\in\mathbb{R^n}$. Using this you can prove that a norm is continuous, and then $||f||$ will be just a composition of continuous functions.

The converse is obviously false even if $n=1$. Just take the function $D:\mathbb{R}\to\mathbb{R}$ to be $D(x)=1$ if $x\in\mathbb{Q}$, otherwise $D(x)=-1$. This function is nowhere continuous, but $|D|$ is just a constant.

Answer 2

For the first part, note that $\|x-y\|\ge |\|x\|-\|y\||$ so the norm $\|\cdot \|:\mathbb R^n\to \mathbb R$ is continuous. And now, since $g$ is also continuous, so is $\|g\|$, being a composition of continuous functions.

For the second part, take $ g(x) = \begin{cases} -1 & x< 0 \\ 1 & x\ge 0 \\ \end{cases}$

Then, $|g|$ is continuous but $g$ is not.