Show that Identity is closed but not continuous

Question

I'm trying to show the following

The operator Identity defined as $$I:(C[0, 1], ||\cdot||_1) \rightarrow (C[0, 1], ||\cdot||_\infty)$$ is closed but not continuous. So $(C[0, 1], ||\cdot||_1)$ it's not complete.

I get stuck to prove that is closed. Is there any result which would help me to prove that?

To prove that is not continuous

Take $(f_n) \subset (C[0, 1], ||\cdot||_1)$ defined as $$f_n(t) = \dfrac{1}{(1+t)^n}$$

We have that $||f_n||_1 \rightarrow 0$ however $||f_n||\infty \rightarrow 1$

Thank you for any help

Answer 1

Let $f_n,f,g\in C[0,1]$ and assume $\|f_n-f\|_1\to 0$ and $\|If_n-g\|_\infty = \|f_n-g\|_\infty\to 0$, just as in the assumptions for the definitions of closedness (Note that $f,g$ are already in $C[0,1]$ by assumption!).

We have to show that $g=f$.

There are probably multiple ways to do this, here is one suggestion:

Because of $\|\cdot\|_1 \leq \|\cdot\|_\infty$ we have $$ \|f-g\|_1 \leq \|f_n-f\|_1+\|f_n-g\|_1\leq \|f_n-f\|_1+\|f_n-g\|_\infty \to 0. $$ This implies that $\|f-g\|_1=0$ and hence $f=g$.