Extending derivative to $L^2$

Question

I have written in my notes that the differential operator $Tf(x) = f'(x)$ defined on $C^1[0,1]$ cannot be extended to a closed operator on all of $L^2[0,1]$, although I'm forgetting why. Is it just that if this were the case, then by the closed graph theorem $T$ would be bounded, but the differential operator is not bounded? The problem with this explanation is that $T$ is unbounded in the supremum norm on $C^1$, while the closed graph theorem would suggest $T$ is bounded in the $L^2$ norm, which I'm not sure is a contradiction.

Answer 1

Too long for a comment.

You can approximate the step function ${\mathbb 1}_{x>\frac 1 2}$ with a sequence of continuous functions by using a piecewise linear approximation $f_n$ at the interpolation points $0$, $\frac 12-\frac 1n$, $\frac 12$ and $1$. Don't worry about those functions having a few points where the derivative is discontinuous. It doesn't change or affect the result. You can convince yourself that you could slightly change those functions to obtain $C^1$ functions.

That sequence is bounded in both $L^2$ and $L^{\infty}$ norms.

You can show that the $L^2$ norm of $Tf_n$ grows as $\sqrt{n}$ (and the $L^{\infty}$ norm grows as $n$.

Edit: if you are bothered by the piecewise linear approximations not being continuously differentiable, you can use $$f_n(x)= \text{arctanh}\left( n\left (x-\frac 12\right ) \right) $$