Why is $D =d/dx$ an unbounded linear operator? Context. Suppose we're in the context of the space of all continuous complex functions over $\mathbb{R}$.  Let our norm be defined as
$$
||f|| = \sqrt{\int f(x) \overline{f(x)}dx}
$$
Definition of unboundedness. The condition for linear operator $L$ to be unbounded is that there does not exist some $M$ such that for all vectors $x$
$$
\|Lx\| \leq M \|x\|,\,
$$
Question: Why does the linear operator $D$, defined as the derivative function $d/dx$, qualify as unbounded?
Attempt: Consider $e^x$, which is in our space.  Then $D(e^x) = \frac{d}{dx}e^x = e^x$.  Then since
$$
||e^x|| = \sqrt{\int e^x \overline{e^x}}dx = \int e^{2x}dx = \infty
$$
then there can't exist some $M \in \mathbb{R}$ such that
$$
||D(e^x)|| \le M ||e^x||
$$

*

*Does this count as a proof?


*If so, the example feels "cheap" to me since it deals with an integral of a function evaluating to $\infty$.  Is there a way to show that $D$ must be an unbounded linear operator without appealing to functions that integrate to $\infty$?
 A: One sensible way to approach this is to define this norm on $L^2(\mathbb{R}) \cap C^1(\mathbb{R})$, the space of functions which are both square integrable and continuously differentiable. Then look at $f_n(x)=\sqrt{n} e^{-n^2 x^2}$. These functions all have the same $L^2$ norm, but the derivatives have norm scaling like $n$.
A: This does not count as a proof since the function $e^x$ is not square integrable. However, the idea can be modified to give the desired proof. Consider the sequence of function $f_n(x) = e^{-nx}$ between $0$ and $1$ and you extend it with fast decay so that it stays square integrable and the sequence is uniformly bounded in $(L^2(\mathbb{R}) \cap C^0(\mathbb{R}), \lVert \cdot \rVert_{L^2})$. Then we compute $$\int_0^1\left(\frac{d}{dx}f_n(x)\right)^2dx = \int_0^1 n^2e^{-2nx} dx = e^{-n}n\sinh(n).$$
This diverges as $n \rightarrow \infty$. Note that the limit is not continuous per say (there is a discontinuity at $0$). Nevertheless, in $(L^2(\mathbb{R}) \cap C^0(\mathbb{R}), \lVert \cdot \rVert_{L^2})$ functions are defined a.e., and hence there exists a continuous representative.
A: For $n=1,2,3,\cdots$, let $f_n$ be a function whose graph is piecewise linear, continuous, and connects the following points in order:
$$
          (-\infty,0),(-1/n,0),(0,1),(1/n,0),(\infty,0).
$$
The derivative of this function is
$$
                  f_n'(x)=n\chi_{[-1/n,0]}-n\chi_{[0,1/n]}.
$$
Therefore, $\|f_n'\|^2=2n^2\frac{1}{n}=2n$, and
$$
       \|f_n\|^2 = 2\int_{0}^{1/n}(1-nx)^2dx 
         = \left.-2(1-nx)^3\frac{1}{3n}\right|_{0}^{1/n}=\frac{2}{3n}
$$
So $\|f_n'\|/\|f_n\|$ is unbounded as $n\rightarrow\infty$.
