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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a $C^2(\mathbb{R})$ function which is also even (ie, $f(-x) = f(x)$). Prove that the function $F: \mathbb{R^n} \rightarrow \mathbb{R}$ defined by $F(x) = f(|x|)$ (where $|x|$ denotes the Euclidean norm of the vector $x$) is also $C^2(\mathbb{R^n})$.

It feels so simple, but I'm just not getting it, so any help would be appreciated.

Thanks!

rakhil11
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1 Answers1

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Hint $F(x)=f(x)$ for $x\ge 0$, and for $x<0$ we have $F(x)=f(|x|)=f(-x)=f(x)$, i.e. $F(x)=f(x)$ for all $x\in \mathbb{R}$

HorizonsMaths
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    Does $f$ being 1D versus $F$ being n-D have no effect on the proof, then? – rakhil11 May 12 '15 at 04:41
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    This answer assumes $n=1.$ The case $n>1$ is more interesting. See http://math.stackexchange.com/questions/1277242/proving-function-is-ck/1277505#1277505 – zhw. May 12 '15 at 04:50
  • Indeed, this question appears to be basically a duplicate so I'll mark it as such when the other question has an answer accepted. – rakhil11 May 12 '15 at 05:03