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Let $f:[0,1]^2 \to \mathbb R^3$ be continuous. Prove there exists r=max {$\|f(x)\|: x \in [0,1]^2$}

Am I suppose to use intermediate value theorem or extreme value theorem?

so since $[0,1]^2$ is a compact set, and f continuous, by extreme value theorem there exists a max?But does extreme value thm hold for functions from Rn to R? This function goes to

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    $\begingroup$ Hints: 1) Image of a compact set by a continuous mapping is compact. 2) Every continuous mapping from a compact set into $\mathbb{R}$ attains a maximum and a minimum. 3) Norm is continuous. $\endgroup$ – Damian Sobota Apr 15 '13 at 18:02
  • $\begingroup$ so since $[0,1]^2$ is a compact set, and f continuous, by extreme value theorem there exists a max? But does extreme value thm hold for functions from $R^n$ to R? This function goes to $R^3$ $\endgroup$ – josh Apr 15 '13 at 18:09
  • $\begingroup$ Almost done, but in the last step you have to consider function $\|f(\cdot)\|$ which is also continuous as a composition of two continuous mappings: $f$ and $\|\cdot\|$. $\endgroup$ – Damian Sobota Apr 15 '13 at 18:11
  • $\begingroup$ And you have to know the topological version of the extreme value theorem (ie. one for domains being compact spaces, not only close intervals in $\mathbb{R}$). $\endgroup$ – Damian Sobota Apr 15 '13 at 18:13
  • $\begingroup$ not sure what that is $\endgroup$ – josh Apr 15 '13 at 18:15
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Assume there is a sequence $x_n\in[0,1]^2$ such that $\|f(x_n)\|$ is an increasing unbounded sequence. Since $[0,1]^2$ is compact, there exists a subsequence $x_{n_k}$ of $x_n$ convergent to $x_0\in[0,1]^2$. The sequence $\|f(x_{n_k})\|$ is also unbounded and increasing. As $\|f(\cdot)\|$ is continuous, $\lim_{k\to\infty}\|f(x_{n_k})\|=\|f(x_0)\|<\infty$ which is a contradiction. This proves that $A:=\{\|f(x)\|:\ x\in[0,1]^2\}$ is bounded. Let $s=\sup A$. The same argument with sequences shows that $s$ must be the maximum of $A$, that is $s\in A$.

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  • $\begingroup$ since $[0,1]^2$ is compact and f continuous, then this implies that f($[0,1]^2) is compact. Isnt this enough to conclude that there is a max, since compact means bounded(sup is finite) and closed(contains its sup). $\endgroup$ – josh Apr 15 '13 at 21:25
  • $\begingroup$ It is enough (actually, you have to talk about $\|f([0,1]^2)\|$, not just $f([0,1]^2)$). But the proof of boundedness of a compact space just goes as I presented. It is still the same idea. $\endgroup$ – Damian Sobota Apr 16 '13 at 0:03
  • $\begingroup$ Can you show why s $\in$ A with sequences? Do you create a subsequence convergent to x such that f(x)=s? thanks $\endgroup$ – josh Apr 16 '13 at 18:10
  • $\begingroup$ Yes. You have to use compactness (completeness) to have the limit in $[0,1]^2$ of a sequence giving $s$. $\endgroup$ – Damian Sobota Apr 17 '13 at 17:52
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The function $\Lambda:\Bbb{R}^3\to\Bbb{R}$ with $\Lambda(x)=||x\|$ is continuous, and $f:[0,1]^2\to\Bbb{R}^3$ is continuous too. So the function $\psi=\Lambda\circ f:[0,1]^2\to\Bbb{R}$ is continuous. The set $[0,1]^2$ is compact. So $\sup\{\Lambda(x):\ x\in[0,1]^2\}$ exists and equals to it's maximum i.e. there exists a $x_0\in [0,1]^2$ such that $\Lambda(x_0)=\sup\{\Lambda(x):\ x\in[0,1]^2\}$ which means $$\sup\{\Lambda(x):\ x\in[0,1]^2\}=\max\{\Lambda(x):\ x\in[0,1]^2\}.$$ But obviously $\Lambda(x)=\|f(x)\|$ for all $x\in [0,1]^2$.

In fact with your notation $r=\Lambda(x_0)$.

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