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This is an old homework problem of mine that I was never able to solve. The solution may or may not involve the Baire category theorem, which I am terrible at applying.

Let $C[0, 1]$ denote the space of continuous functions $[0, 1] \to \mathbb{R}$. Suppose $|| \cdot ||$ is a norm on $C[0, 1]$ with respect to which the evaluation functions $f \mapsto f(x), x \in [0, 1]$ are continuous. Show that the topology induced by $|| \cdot ||$ is the same as the usual topology induced by the sup norm.

It is straightforward to show that any sequence converging uniformly converges with respect to $|| \cdot ||$. I am stuck on proving the converse; I cannot seem to figure out how to use the assumption that $|| \cdot ||$ is a norm.

Edit: Zhen Lin informs me that $|| \cdot ||$ is supposed to be complete. That should make the problem statement true now!

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If you can show that $C[0,1]$ is complete with respect to the new norm, then you're done by the open mapping theorem (which does indeed involve Baire). – Chris Eagle Jan 5 '11 at 16:33
@Chris: thanks! I am having some trouble doing this; it doesn't seem to me like the assumption that evaluation is continuous is enough to conclude that a Cauchy sequence has a pointwise limit. – Qiaochu Yuan Jan 5 '11 at 16:49
Why is it clear that any uniformly converging sequence converges with respect to the other norm? It would suffice to show that the various evaluation functionals are uniformly continuous with respect to the other norm, but I don't see why this is immediate. – Akhil Mathew Jan 5 '11 at 17:08
@Akhil: dang. Okay, the argument I thought I had for this is nonsense. – Qiaochu Yuan Jan 5 '11 at 17:17
@William: being a norm is far more restrictive than being a metric. Trevor's claim is correct. It follows easily from the standard theorem that continuous is equivalent to bounded for operators between normed spaces. – Chris Eagle Jan 8 '11 at 10:35
up vote 23 down vote accepted

Here's $\DeclareMathOperator{\ev}{ev}$ the answer if $(C[0,1], \|\cdot\|)$ is assumed to be complete. Consider the family $\mathcal{F} = \{\ev_{x}\}_{x \in [0,1]}$ of continuous linear functionals on $(C[0,1],\|\cdot\|)$. For each $f \in C[0,1]$ we have $\sup_{x \in [0,1]} |\ev_{x}(f)| \leq \|f\|_{\infty}$, so the family $\mathcal{F}$ is pointwise bounded. By the uniform boundedness principle the family is uniformly bounded, that is to say $\sup_{x \in [0,1]} \|\ev_{x}\| \leq M$ for some constant $M$. On the other hand $|f(x)| = |\ev_{x}(f)| \leq \|ev_{x}\|\|f\| \leq M \|f\|$ gives $\|f\|_{\infty} = \sup_{x \in [0,1]} |f(x)| \leq M\|f\|$, so the identity $(C[0,1], \|\cdot\|) \to (C[0,1],\|\cdot\|_{\infty})$ has norm at most $M$. Since both spaces are complete, we may apply the open mapping theorem in order to conclude that its inverse is also continuous. In other words, the norms $\|\cdot\|$ and $\|\cdot\|_{\infty}$ are equivalent.

Edit. Here's an example that shows that completeness of the norm is necessary:

Choose a discontinuous linear functional $\varphi: (C[0,1],\|\cdot\|_{\infty}) \to \mathbb{R}$ and define a norm on $C[0,1]$ by \[ \|f\| = \|f\|_{\infty} + |\varphi(f)|. \] Since $\|f\|_{\infty} \leq \|f\|$, we have that the identity $(C[0,1],\|\cdot\|) \to (C[0,1],\|\cdot\|_{\infty})$ is continuous. Since the evaluation functionals are continuous with respect to the sup-norm, they are also continuous with respect to the norm $\|\cdot\|$. But as $\varphi$ is discontinuous, there is a sequence $f_{n}$ with $\|f_{n}\|_{\infty} = 1$ and $|\varphi(f_{n})| \to \infty$, hence the norms cannot be equivalent. Of course, $\|\cdot\|$ cannot be complete because the last sentence would be in contradiction to the open mapping theorem.

Edit 2.

I forgot to argue why the topologies in the above counterexample are not the same. This is obvious: The functional $\varphi$ is continuous with respect to $\|\cdot\|$ but it isn't continuous with respect to $\|\cdot\|_{\infty}$.

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I've completely forgotten the solution, but I think something like this might work: The set $\mathcal{F}$ of all evaluation maps is a family of bounded linear operators on $C[0, 1]$. It suffices to show that $\mathcal{F}$ is uniformly bounded, and then the uniform bound gives you the constant needed to establish the equivalence with the uniform norm. I don't remember how to establish uniform boundedness, but I imagine the uniform boundedness principle applies.

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You don't know a priori that $C[0,1]$ with the new norm is complete, hence the uniform boundedness principle does not apply. (The uniform boundedness principle applies only in the case $\mathcal{F} \subset \mathcal{L}(X,Y)$ with $X$ complete and $Y$ normed). – t.b. Jan 8 '11 at 11:37
@Theo Buehler: I don't know if the question was changed, but the current version says that $C[0, 1]$ is complete w.r.t. to the unknown norm. – Zhen Lin Jan 8 '11 at 12:16
Where does that link come from? Anyway, Qiaochu's formulation does not specify completeness. – t.b. Jan 8 '11 at 12:26
@Theo Buehler: The question comes from a problem set from a lecture course (Part II Linear Analysis, Cambridge) both Qiaochu and I attended. – Zhen Lin Jan 8 '11 at 12:28
And yes, $\DeclareMathOperator{\ev}{ev}$ if you assume completeness, consider the family $\mathcal{F} = \{\ev_{x}\}_{x \in [0,1]}$ and observe that $|\ev_{x}(f)| \leq \|f\|_{\infty}$, so the family $\ev_{x}$ is pointwise bounded, hence it is uniformly bounded, that is $\sup_{x \in [0,1]}\|\ev_{x}\| \leq M$ for some $M$. But this means $|f(x)| = |\ev_{x}(f)| \leq \|\ev_{x}\| \|f\| \leq M \|f\|$, so $\|f\|_{\infty} \leq M \|f\|$. In other words, the identity $(C[0,1],\|\cdot\|) \to (C[0,1],\|\cdot\|_{\infty})$ has norm $\leq M$. Now apply the open mapping theorem. – t.b. Jan 8 '11 at 12:35

OK, assuming completeness, Theo provided a simple proof below. My previous proof also works if one assumes completeness, but it is too convoluted.

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I haven't read through this argument in detail, but I don't think this argument can work because I don't think the problem statement is true if || \cdot || is not assumed to be complete... – Qiaochu Yuan Jan 8 '11 at 13:56

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