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?
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$\begingroup$ any finite dimensional space is naturally isomorphic to $\mathbb R^d$, so just check that what you want is preserved by isomorphism $\endgroup$– AndrewCommented Jan 25 at 2:05
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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".
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$\begingroup$ So the sup norm and $||•||_1$ are equivalent on $V$ as they will give the same topology right? $\endgroup$ Commented Jan 25 at 3:11
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$\begingroup$ Well, how do you understand my 2nd paragraph in connection with those two norms? $\endgroup$– KCdCommented Jan 25 at 3:13
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$\begingroup$ 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 $V$ $\endgroup$ Commented Jan 25 at 3:16
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$\begingroup$ 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). $\endgroup$– KCdCommented Jan 25 at 3:20