Prove that in $C([0,1])$, $||x||= \text{sup}_{t \in [0,1]}|tx(t)|$ the norms $||.||$ and $||.||_{\infty}$ are not equivalent

banach-spacesfunctional-analysisnormed-spaces

In $C([0,1]),||.||)$ $with ||x||= \text{sup}_{t \in [0,1]}|tx(t)|$ prove that $||.||$ and $||.||_{\infty}$ are not equivalent norms

Try:

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 a $c>0$ 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 that 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

What you did is fine and it proves indeed that there is no constant $c$ so that we always have$$\|f\|_\infty\leqslant c\|f\|.\tag1$$Note that $(1)$ is equivalent to $\frac{\|f\|_\infty}{\|f\|}\leqslant c$. So, proving that $(1)$ doesn't hold for some $c$ and for every $f\in C\bigl([0,1]\bigr)$ is equivalent to asserting that for every $c$ there is some $f_c\in C\bigl([0,1]\bigr)$ such that $\frac{\|f_c\|_\infty}{\|f_c\|}>c$. And, in order no prove that, it is enough to prove that, for every $n\in\Bbb N$ there is some $f_n\in C\bigl([0,1]\bigr)$ such that $\frac{\|f_n\|_\infty}{\|f_n\|}>c$. So, it is natural to take a sequence $(f_n)_{n\in\Bbb N}$ such that $\lim_{n\to\infty}\frac{\|f_n\|_\infty}{\|f_n\|}=\infty$.

And, no, it doesn't have to be a Cauchy sequence.

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