problem: Consider the space $Y = C([0,1], \mathbb{R})$ of continuous functions from $[0,1]$ to $\mathbb{R}$. Let $$ X := \left\{ f \in Y \mid f^{'} \ \text{exists and is continuous on} \ [0,1] \right\} $$ Equip both $Y$ and $X$ with the norm $|| . ||_{\infty}$. Consider the operator $D: X \rightarrow Y$ given by $Df = f^{'}$. Clearly $D$ is linear. Suppose that $(f_n)_{n \in \mathbb{N}}$ is a sequence in $X$ such that $(f_n, f_n^{'}) \rightarrow (f,g) $ in $X \oplus_1 Y$. Then $f_n^{'} \rightarrow g$ uniformly on $[0,1]$.
(i) Show that $f^{'} = g$ and hence deduce that the graph of $D$ is bounded.
(ii) Show that $D$ is not a bounded operator (hint: consider $f_n(t) = t^n$).
attempt: (i) I wish to show that $|| f^{'} - g ||_{\infty} = 0$. What I did was $$ || f^{'} - g ||_{\infty} \leq || f^{'} - f_n^{'}||_{\infty} + || f_n^{'} - g ||_{\infty}. $$ Since it is given that $f_n^{'} \rightarrow g$ uniformly, we have that $||f_n^{'} - g||_{\infty} \to 0$ (this is a standard result for the sup norm). But I was stuck at figuring out how to get $|| f^{'} - f_n^{'}||_{\infty} $ small. I can say that $$|| f^{'} - f_n^{'}||_{\infty} \leq || D || \ || f - f_n ||_{\infty}. $$ Then since it is given that $(f_n, f_n^{'}) \rightarrow (f,g)$, I think that $||f - f_n||_{\infty} \to 0$. But I'm worried that since $D$ is not bounded, I cannot get $|| D|| $ small to get this argument to work...
(ii) For this part, for all $n \in \mathbb{N}$ and all $t \in [0,1]$ with $t \neq 0$ I think we have $$ \frac{ | n t^{n-1}|}{ | t^n|} \geq n. $$ This would show that $$ \frac{ || D f_n||}{ || f_n||} \geq n$$ and so $D$ is unbounded. Is this reasoning correct?
