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

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

• 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. – Damian Sobota Apr 15 '13 at 18:02
• 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 $R^3$ – josh Apr 15 '13 at 18:09
• 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 $\|\cdot\|$. – Damian Sobota Apr 15 '13 at 18:11
• 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 $\mathbb{R}$). – Damian Sobota Apr 15 '13 at 18:13
• not sure what that is – josh Apr 15 '13 at 18:15

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