$f : I → R^2$, why does the supremum exist?

Question

set $f : I → R^2$, why does the supremum $|f(x)|$ s.t. $x\in I$ exist? $f$ is continuous and $I=[0,1]$

I can show that this is the case when using $R$.

Since $f$ is continuous and closed intervals are compact -> that the supremum exists by the Extreme Value Theorem.

Answer 1

You've got two functions in play:

• $f:I\to R^2$
• $g : R^2 \to R$, given by $g(x) = \left| x \right|$

So what you're actually looking to do is show that $g \circ f$ assumes a maximum value on $I$. But $g \circ f$ is a function from $I$ to $R$, and you already know this is the case for such functions so long as $g \circ f$ is continuous. You know $f$ is continuous, hence you need only prove that $g$ is also continuous (it is).

Answer 2

If you can already show the analogous property of functions $I\rightarrow\mathbb R$, then the only step remaining is to note that the function $g:\mathbb R^2\rightarrow\mathbb R$ taking vectors $v$ to their norm $|v|$ is continuous. This is easily proven by the triangle inequality. Then, we see that the map taking $x$ to $|f(x)|$ is simply the composition $g\circ f$ and hence must be continuous. Moreover, $g\circ f$ is a function $I\rightarrow\mathbb R$ and so has a maximum value, as you are already familiar with.