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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

$$ ||D(e^x)|| \le M ||e^x|| $$

  1. Does this count as a proof?

  2. 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 $\infty$?

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    $\begingroup$ That makes no sense, you should consider the space of continuous functions for which $\Vert f \Vert $ is finite. $\endgroup$
    – Martin R
    Aug 28, 2021 at 4:13
  • $\begingroup$ The norm you defined is NOT a norm on the space you gave. So your question is flawed as you cannot even define unbounded (or bounded) operators. $\endgroup$ Aug 28, 2021 at 4:22

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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 $n$.

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  • $\begingroup$ A very pertinent answer. I just want to point out that the scaling should be $\sqrt{n} e^{-n^2 x^2}$, and then the norm ration is proportional to $n$. $\endgroup$
    – orangeskid
    Aug 28, 2021 at 5:33
  • $\begingroup$ @orangeskid That's right, I was freezing the $L^1$ norm and not the $L^2$ norm. $\endgroup$
    – Ian
    Aug 28, 2021 at 13:22
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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.

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  • $\begingroup$ Not every class in $L^2$ has a continuous representative. Actually, taking the limit does not really matter for this problem. All that really matters is that you have a sequence of functions whose norms are bounded, while the norms of their derivatives are not bounded. $\endgroup$
    – Ian
    Aug 28, 2021 at 4:50
  • $\begingroup$ That's true I overlook the problem, we do not need to check that the limit is in the space (and I agree that in general there may not exist continuous representatives). However, in this case there exists one continuous representative since there is only one discontinuity jump at 0, and we can take the limit to be 0. $\endgroup$
    – Rundasice
    Aug 28, 2021 at 4:55
  • $\begingroup$ Jumps can't be mollified away by a change on a set of measure zero. They can be mollified by a change on a set of arbitrarily small positive measure, of course (Lusin's theorem) but that's something else. $\endgroup$
    – Ian
    Aug 28, 2021 at 4:56
  • $\begingroup$ But the limit is identically 0 except at $x = 0$. $\endgroup$
    – Rundasice
    Aug 28, 2021 at 5:00
  • $\begingroup$ Alright, I see your point, but it is sweeping some details under the rug in a way that feels a bit misleading if you're not familiar with this kind of analysis. (In particular why can you make the part outside $[0,1]$ have $L^2$ norm no larger than the $e^{-nx}$ part? You can, but the OP likely doesn't know why.) $\endgroup$
    – Ian
    Aug 28, 2021 at 13:24
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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 $n\rightarrow\infty$.
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  • $\begingroup$ Don't the $f_n$ fail to be continuous, thus failing to be in our space to begin with? $\endgroup$ Aug 28, 2021 at 16:50
  • $\begingroup$ @user1770201 : The graph of $f_n$ is piecewise linear and continuous. The derivative is not continuous. $\endgroup$ Aug 28, 2021 at 19:54

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