Why is the derivative operator $D:C^\infty([0,1])\to C([0,1])$, $Df\equiv f'$, closed and unbounded?

Question

This is an example I read around closed graph theorem. Let $Y=C[0, 1]$ and $X$ be its subset $C^\infty[0, 1]$. Equip both with uniform norm. Define $D: X \to Y$ by $f \mapsto f'$. Suppose $(f_n, f_n') \to (g, h)$ in $X \times Y$. Thus, $f_n \to g, f_n' \to h$ both uniformly on $[0, 1]$. It is well known that $g' = h$. Thus, the graph of $D$ is closed. But since $D(t^n) = nt^{n-1}$, $D$ is not continuous. I have three questions about this example.

1. Why $f_n \to g, f_n' \to h$ both uniformly on $[0, 1]$?
2. Why $g' = h$ even though it is well known?
3. Why $D$ is not continuous because $D(t^n) = nt^{n-1}$?

Could anyone explain these questions to me, please? Thank you!

Answer 1

3.

$D(t^n) = nt^{n-1}$. But $\|t^n\| = 1$ and $\|D(t^n)\| \to \infty$. So $D$ is not a bounded operator. You know, right, that for linear operator on normed spaces, continuity holds iff the operator is bounded?

Answer 2

1. The topology on $X\times Y$ is given by the norm $\|(x,y)\|=\max\{\|x\|,\|y\|\}$. So if $(f_n,f_n')\to(f,g)$, this means that $\|f_n-f\|\to0$, $\|f_n'-g_n\|\to0$, i.e. $f_n\to f$, $f_n'\to g$ uniformly.

2. Fix $x\in X$. Then, using the Mean Value Theorem for the equality in the middle,
   \begin{align} \left|\frac{f(x+h)-f(x)}h-g(x)\right| &\leq\left|\frac{f(x+h)-f_n(x+h)}h\right|+\left|\frac{f_n(x+h)-f_n(x)}h-g(x)\right|\\[0.2cm] &\qquad +\left|\frac{f_n(x)-f(x)}h\right|\\ \ \\ &\leq\frac{2\|f_n-f\|}h+\left|\frac{f_n(x+h)-f_n(x)}h-g(x)\right|\\ \ \\ &=\frac{2\|f_n-f\|}h+\left|f_n'(\xi_{h,n})-g(x)\right| \\ \ \\ &\leq\frac{2\|f_n-f\|}h+\left|f_n'(\xi_{h,n})-f_n'(x)\right|+\left|f_n'(x)-g(x)\right|\\ \ \\ &\leq\frac{2\|f_n-f\|}h+\left|f_n'(\xi_{h,n})-f_n'(x)\right|+\left\|f_n'-g\right\|\\ \ \\ \end{align} As this holds for all $n$, we get \begin{align} \left|\frac{f(x+h)-f(x)}h-g(x)\right| &\leq\limsup_n\left\|f_n'(\xi_{h,n})-f_n'(x)\right\|. \end{align} Now fix $\varepsilon>0$. For $n$ big enough, $\|f_n'-g\|<\frac\varepsilon3$. Then \begin{align} \left\|f_n'(\xi_{h,n})-f_n'(x)\right\| &\leq \|f_n'(\xi_{h,n})-g(\xi_{h,n})\|+\|g(\xi_{h,n})-g(x)\|+\|g(x)-f_n'(x)\|\\ \ \\ &\leq \frac{2\varepsilon}3+\|g(\xi_{h,n})-g(x)\|. \end{align} As $g$ is uniformly continuous, there exists $\delta>0$ such that $|g(t)-g(s)|<\frac\varepsilon3$ whenever $|t-s|<\delta$. So, if $h<\delta$, $$ \left|\frac{f(x+h)-f(x)}h-g(x)\right| <\varepsilon. $$

3. If $f(t)=t^n$, then $\|f\|=\sup\{|t^n|:\ t\in[0,1]\}=1$. And $$ \|Df\|=\sup\{|nt^{n-1}|:\ t\in [0,1]\}=n. $$ So $D$ is unbounded.