Terminology/Definition: The set $$ \Gamma(A):=\{(x,Ax):x\in\text{Dom}A\}\subseteq X\times Y $$ is called the graph of $A$. We define the norm in $X\times Y$ by $||(x,y)||:=||x||_X+||y||_Y$. We say that $A$ is a closed graph operator if $\Gamma(A)$ is a closed set in $X\times Y$. This means that whenever $x_n\in\text{Dom}A$, $x_n$ converges to $x$ and $Ax_n$ converges to $y$, then $x\in\text{Dom}A$ and $y=Ax$.
Currently I'm self studying functional analysis, namely closed graph operators. In the text, the other gives the following example:
Example: The operator $Ax=dx/dt$ in $L^2[0,1]$ with $\text{Dom}A=\{x\in C[0,1]: x'\in C[0,1],x(0)=0\}$ is not a closed graph operator.
I'm trying to figure out why it's not a closed graph operator. I believe it boils down to finding a sequence of functions $x_n(t)\in C^1[0,1]$ with $x(0)=0$ such that $x_n(t)\to x(t)$ in $C[0,1]$ and $x'_n(t)\to y(t)$ in $L^2[0,1]$ but $x'(t)\neq y(t)$. However, I've tried many examples and can't find such a function.
Let $y_n$ be continuous with $0\leq y_n(t) \leq 1$, $y_n(t)=0$ for $t \leq \frac 1 2$ and $y_n(t)=1$ for $t \geq \frac 1 2+\frac 1 n$. Let $x_n(t)=\int_0^{t} y_n(s)ds$. Then $(x_n,Ax_n)$ is a sequence in the graph of $A$ converging in $L^{2}$ to $(x(t),y(t))$ where $x(t)=\int_0^{t} \chi_{(\frac 1 2 ,1)}(s)ds$. Also, $x_n'=y_n \to \chi_{(\frac 1 2 ,1)}$ in $L^{2}$. Note that $x_n \to x$ uniformly. Since $x(t)$ is not even differentiable at $\frac 1 2$ (and hence does not belong to the domain of $A$) it follows that that graph of $A$ is not closed.