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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.

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  • $\begingroup$ $f(x)=\boldsymbol 1_{\mathbb Q}(x)-\boldsymbol 1_{\mathbb R\setminus \mathbb Q}(x)$ is discontinuous everywhere but $|f|$ is continuous. $\endgroup$ – Surb Feb 22 at 16:28
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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.

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  • $\begingroup$ I don't understand why $|(||x||-||y||)|\leq ||x-y||$ implies that the norm is continuous. $\endgroup$ – davidllerenav Feb 22 at 16:41
  • $\begingroup$ Let $\epsilon>0$. Then you can take $\delta=\epsilon$ and you get that if $||x-y||<\delta$ then $|(||x||-||y||)|\leq ||x-y||<\delta=\epsilon$. So the norm is even uniformly continuous. $\endgroup$ – Mark Feb 22 at 16:45
  • $\begingroup$ And how do I use that to prove that the norm is continuous? $\endgroup$ – davidllerenav Feb 22 at 16:55
  • $\begingroup$ If a function is uniformly continuous then of course it is continuous. There is nothing left to prove. $\endgroup$ – Mark Feb 22 at 17:06
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    $\begingroup$ Maybe it will help to write $f(x)=\|x\|.$ Then, $f$ is continuous at $x_0$ iff for all $\epsilon>0$ there is a $\delta>0$ such that if $\|x-x_0\|<\delta$, then $|f(x)-f(x_0)|<\epsilon.$ Now unpack this and you'll see the reverse triangle inequality is exactly the condition you need for this to hold. i.e. take $\delta=\epsilon. $ It shows in fact, that the continuity is uniform. $\endgroup$ – Matematleta Feb 22 at 18:15
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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.

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  • $\begingroup$ Why $\|x-y\|\ge |\|x\|-\|y\||$ implies thaat the norm $\|\cdot \|:\mathbb R^n\to \mathbb R$ is continuous? $\endgroup$ – davidllerenav Feb 22 at 16:40

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