I have to decide whether an operator is closed

Question

So here is my problem,

I have to decide whether the following operator is closed,

$$\frac{\mathrm{d}}{\mathrm{d}x}:C^2([0,1])\subset C^0([0,1])\rightarrow C^0([0,1])$$

with the $||\cdot||_{\infty}$ norm on $C^0([0,1])$

I managed to show that the same operator defined on $C^1([0,1])$ is closed but I am not sure about $C^2$. I tried to compose a similar proof but I realised that the theorem I used for $C^1$ does not work for the upper case. So I am assuming that it is not closed but I am not able to construct a sequence in the graph such that the limit is outside.

Can someone give me a hint?

Thanks a lot

Answer 1

Consider the sequence $(f_n, \partial_t f_n)_{n\in \mathbb{N}} \subset \Gamma (B)$ given by $f_n(t)= \int_{0}^{t} \sqrt{(s+\frac{1}{n})}ds$ and $\partial_t f_n(t) = \sqrt{t+\frac{1}{n}}$.

Since

$$\|\partial_t f_n(t)-\sqrt{t}\| = \sup_{t\in [0,1]}|\sqrt{t+\frac{1}{n}}-\sqrt{t}|=\sqrt{\frac{1}{n}}\rightarrow 0$$

and similarly$ \|f_n(t)-\frac{2}{3}t^{\frac{3}{2}}\|\rightarrow 0$ we can conlude $(\frac{2}{3}t^{\frac{3}{2}},\sqrt{t}) \notin \Gamma (B)$ implying that $B$ is not closed.