A Counter Example about Closed Graph Theorem

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

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 \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| +\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| \ \ \ \ \ \ \textit{(Mean Value Theorem)}\\ \ \\ &\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} Now fix $\varepsilon>0$. As $f_n'$ is uniformly continuous and $\xi_{h,n}$ lies between $x$ and $x+h$, there exists $\delta>0$ such that if we take $|h|<\delta$ we get $|f_n'(\xi_{h,n})-f_n'(x)|<\varepsilon/3$. So for $|h|<\delta$, we have \begin{align} \left|\frac{f(x+h)-f(x)}h-g(x)\right| &\leq\frac{2\|f_n-f\|}h+\frac\varepsilon3+\left\|f_n'-g\right\|, \end{align} for every $n$. Now we can take $n$ big enough so that $\|f_n-f\|<h/6\varepsilon$ and $\|f_n'-g\|<\varepsilon/3$, so \begin{align} \left|\frac{f(x+h)-f(x)}h-g(x)\right| &\leq\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon. \end{align}

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.

Answer 2

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?