# Equivalent norms on a finite dimensional subspaces of $C([0,1])$

Consider the space $$C([0,1])$$. Then the norm on this space is the sup norm. Now let $$V$$ be a finite dimensional subspace of $$C^1([0,1])$$. Then we can consider $$V$$ as a subspace of $$C([0,1])$$. Then we have two norms on $$V$$ namely the sup norm and $$||•||_1$$ norm. My question is are these two norms equivalent on $$V$$. I am aware of the fact that all norms on finite dimensional spaces are equivalent. Can we give that argument here also?

• any finite dimensional space is naturally isomorphic to $\mathbb R^d$, so just check that what you want is preserved by isomorphism Commented Jan 25 at 2:05

You say "Then the norm on this space is the sup-norm". That is not a logical deduction, but simply a choice. It's not the only one possible (e.g., there are $$L^p$$-norms where $$p \geq 1$$).

Since $$V$$ is finite-dimensional, we can apply the theorem that all norms on a finite-dimensional real (or complex) vector space are equivalent. More precisely, when $$V$$ has a norm on it, that gives $$V$$ a topology, and all norms on a finite-dimensional real (or complex) vector space give $$V$$ the same topology. Giving $$V$$ the same topology is one way to define two norms as being equivalent; there is another description of equivalent norms on a real vector space, where each norm is bounded above by a constant multiple of the other, but that turns out to be the same property as the two norms giving the vector space the same topology.

Note: the "trivial norm" where $$|\!|v|\!| = 1$$ when $$v \not= 0$$ and $$|\!|v|\!| = 0$$ when $$v = 0$$ in $$V$$ makes $$V$$ a discrete space, but that doesn't contradict the theorem I mentioned because the "trivial norm" on $$V$$ is not a vector space norm since the rule $$|\!|cv|\!| = |c||\!|v|\!|$$, where $$c$$ is real and $$|\cdot|$$ is the usual absolute value on the real numbers, is not valid on $$V$$ with the "trivial norm".

• So the sup norm and $||•||_1$ are equivalent on $V$ as they will give the same topology right? Commented Jan 25 at 3:11
• Well, how do you understand my 2nd paragraph in connection with those two norms?
– KCd
Commented Jan 25 at 3:13
• As I understand from the first line that for finite dimensional space all norms are equivalent, I think sup norm and $||•||_1$ norm are equivalent on $V$ Commented Jan 25 at 3:16
• Sure. When you have two vector space norms, they are equivalent when $V$ is finite dimensional. You just need to be working with actual vector space norms. That's why I pointed out in my Note that the "trivial norm" is not actually a vector space norm (when the real numbers have their usual absolute value).
– KCd
Commented Jan 25 at 3:20