Prove that $C([0,1]),||.||)$ with $||x||= \text{sup}_{t \in [0,1]}|tx(t)|$ is not complete using equivalence of norms

banach-spacesfunctional-analysisnormed-spaces

Prove that $C([0,1]),||.||)$ with $||x||= \text{sup}_{t \in [0,1]}|tx(t)|$ is not complete

In https://math.stackexchange.com/posts/3908897/edit I presented this exercise solved as in my lecture notes: by taking a Cauchy sequence and proving that it does not converges to something that is not in $C([0,1]),||.||)$. I 've made another solution inspired in other proofs I've seen and I'd like to know if it is correct and ask you some doubts about which where the same I had in the inspiration of the proof

Try: If I can proof this norm is not equivalent to the ||.||∞ norm, that means that the space C([0,1]),||.||) is not complete.

In $0 \leq t \leq 1 $ $|tf(t)| \leq |f(t)|$ so $||f(t)|| \leq ||f(t)||_{\infty}$

Now the question is if there exists an $c>0$ such that $||f(t)||_{\infty} \leq c||f(t)||$

We take a sequence defined as follows:

$$ f_n(t)= \begin{cases} \frac{1}{ t} && ,\frac{1}{n} \leq t \leq 1 \\ n &&, 0 \leq t \leq \frac{1}{n} \end{cases} $$

$||f_n||_{\infty}= \text{sup}_{t \in [0,1]} |f_n(t)|=n$
$||f_n||=\text{sup}_{t \in [0,1]} |t f_n(t)|=\text{max} \{\text{sup}_{0 \leq t \leq \frac{1}{n}} |t f_n(t)|,\text{sup}_{\frac{1}{n} \leq t \leq 1} |t f_n(t)| \}=\text{max} \{\text{sup}_{0 \leq t \leq \frac{1}{n}} nt,\text{sup}_{\frac{1}{n} \leq t \leq 1} \frac{1}{t}t \}= \text{max} \{1,1\}=1 $

Then $\frac{||f_n||_{\infty}}{||f_n||} \to \infty$, when $n \to \infty$. So the norms are not equivalent

My questions are, in addition to knowing if this is correct :

  1. In this approach I don't need the sequence to be Cauchy, right?. in order to prove tht no such c exists , why do we consider a sequence in the first place? what's the idea?
  2. Why do they take the limit of $\frac{||f_n||_{\infty}}{||f_n||} \to \infty$ at the end, how does this proves that no constant c exists that makes the norms equivalent ?

Best Answer

You can prove that $(C([0,1]),||.||)$ is not complete by comparing the norm $||f||=\sup_{0\leq t\leq 1}|tf(t)|$ with the usual norm $||f||_{\infty}$ but first let me answer to your questions:

  1. No, you dont need to show that $(f_n)$ is Cauchy in neither of two norms in order to show that there is no constant $c>0$ such that $$\tag{1}||f||_{\infty}\leq c||f||$$

We consider the sequence $(f_n)$ because it has the property that $||f_n||=1$ whereas $||f_n||_{\infty}=n$. In other words, we are trying to find some elements $f_n$ that belong to both spaces such that the norm $||f_n||_{\infty}$ is big whereas the norm $||f_n||$ is small so that we make $(1)$ fail.

  1. Now, as you said, we have $\lim_{n\to \infty}\frac{||f_n||_{\infty}}{||f_n||}\to \infty$ but we dont have necessarily have to take the limit in order to prove that $(1)$ is false. To see this, if there was some $c>0$ with $$||f||_{\infty}\leq c||f||$$ for every $f$, then in particular, for the $f_n's$ we would have that $n\leq c$ for every $n$, which leads us to a contradiction.

Now why the non-equivalance of the two above norms implies that $(C([0,1]),||.||)$ is not complete its because of the open mapping theorem.

To see this, take the identity operator $I:(C([0,1]),||.||_{\infty})\to (C([0,1]),||.||)$ with $I(f)(t)=f(t)$ for $f\in C([0,1])$ and $0\leq t\leq 1.$

Since $||f||\leq ||f||_{\infty}$ it follows that $I$ is continuous, one to one and onto. Now, if $C([0,1])$ was complete with the norm $||f||=\sup_{0\leq t\leq 1}|tf(t)|$, then, by the open mapping theorem $$I^{-1}:(C([0,1]),||.||)\to (C([0,1],||.||_{\infty})$$ would also be continuous. But this is equivalent to the existence of some $c>0$ that satisfies $(1)$ which you have shown there cant be such $c>0$.

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