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Assume that $V$ is a real vector space of smooth non-analytic functions with a compact support $[a, b]$ (i.e vector space of bump functions) and $f \in V$. I'm trying to show that:

$$I=\int_{b}^{a}{f(t)D{f(t)}} \, dt=0$$

where $D$ is the first order derivative operator and $[a, b]$ (which also is bounds of integral as noticed) represents the closed interval of compact support.

I've noticed that all bump functions I've seen are even within their compact interval. Therefore obviously:

$$f(-x)=f(x) \implies D(-f(x))=-D(f(x)) \implies I=0$$

Is this true for all bump functions? Concordantly, is $I \neq 0$ possible?

Example

Consider:

$$f(x)=\begin{cases} \textrm{exp}({\frac{-1}{(x-a)^2(x-b)^2}})\ \ x\in [a,b]\\ 0\end{cases}$$

The function above is perhaps one of the most frequently occurring bump functions, satisfying all of its properties, including: smoothness, non-analytic property, "faster-than-polynomial" growth rate and compact support.

It can be easily verified that $f(-x)=f(x), \forall x \in [a, b]$; from which it can be deduced that $\int_{b}^{a}{f(x)D{f(x)}} \, dx=0$.

You can also see this on Desmos.

Question

Is it true that all bump functions need to be evenly constructed to satisfy all their properties? Can $f(t)Df(t)$ be odd if $f(t)$ is not even? Furthermore if both questions can be falsified by some $g \in V$, why would $I=0$ hold?

Note:

As you noticed in the previous paragraph, I'm assuming that $I=0$. This is due to the fact that the derivative operator $D$ is skew symmetric (i.e $D^T=-D$) with respect to the billinear form $\langle f, g \rangle = \int_{b}^a {f(t)g(t) \, dt}$ which implies that the condition $I=0$ must hold (source).

ShellRox
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  • Of course they don‘t have to be even. In fact, any function that you can „smoothly draw“ and that is 0 outside some compact interval is a bump function. It is just somewhat annoying to construct bump functions – Maximilian Janisch Nov 23 '19 at 14:55
  • Also, why do you impose that the function is not analytic? – Maximilian Janisch Nov 23 '19 at 14:59
  • No, bump functions are obviously not even. For example, if $f$ is a bump function, $c\in\Bbb R$ and $g(t)=f(t+c)$ then $g$ is a bump function... – David C. Ullrich Nov 23 '19 at 15:04
  • @MaximilianJanisch I apologize, I didn't mention that $f \in V$ must be a non-zero bump function and therefore it must be non-analytic by identity theorem. – ShellRox Nov 23 '19 at 15:04
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    The fact that $I=0$ is just the fundamental theorem of calculus. Note that $f(a)=f(b)=0$ and $fDf=\frac12D(f^2)$. – David C. Ullrich Nov 23 '19 at 15:04
  • @DavidC.Ullrich I'm sorry for a ignorant question, but why is it true that $f(a)=f(b)=0$? Considering the fact that compact support also contains $a$ and $b$? – ShellRox Nov 23 '19 at 15:12
  • @MaximilianJanisch Quite interesting, my assumption about non-analytic property came from this source, which I perhaps did not understand correctly. – ShellRox Nov 23 '19 at 15:14
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    Well first, when you said "with a compact support $[a,b]$" of course you meant "with a compact support contained in $[a,b]$". With your version $V$ is not a vector space, since $0\notin V$ because $\emptyset\ne[a,b]$. – David C. Ullrich Nov 23 '19 at 15:23
  • @ShellRox The source was correct and I was wrong: There is a version of the Theorem for real analytic functions that I wasn't aware of – Maximilian Janisch Nov 23 '19 at 15:27
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    Say $f\in V$ and let $K$ be the support of $f$. We're assuming that $K\subset[a,b]$, and by definition $K$ is the closure of the set of $x$ with $f(x)\ne0$. In particular, ${x:f(x)\ne0}\subset K$, so that $f(x)=0$ for every $x\in\Bbb R\setminus K\subset\Bbb R\setminus[a,b]$. Since $f$ is continuous this implies $f(a)=0$ and $f(b)=0$. – David C. Ullrich Nov 23 '19 at 15:27
  • @DavidC.Ullrich Now it makes sense, I didn't consider vector space axioms before considering compact support interval. Thank you very much for elaboration. – ShellRox Nov 23 '19 at 15:33

1 Answers1

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The right way to go is integration by parts: Since $f(a)=f(b)=0$ (see below), we have $$\int_a^b f(x)\cdot f'(x)\,\mathrm dx=f(b)^2-f(a)^2-\int_a^b f'(x)\cdot f(x)\,\mathrm dx=-\int_a^b f'(x)\cdot f(x)\,\mathrm dx.$$

We conclude that your integral is indeed $0$.

Now, "most" bump functions are not at all even. In fact, the set of bump functions is dense in the set of continuous functions with compact support by the Stone Weierstrass Theorem. So for any non-even continuous function with compact support, there are infinitely many non-even bump functions that look very similar to that continuous function.

EDIT: The support of $f$ is defined as the closure of the set of all points where $f(x)\neq 0$. Hence, for any $\varepsilon>0$, we have $f(a-\varepsilon)=f(b+\varepsilon)=0$. Since $f$ is continuous (as it is even infinitely differentiable), we have $f(a)=f(b)=0$.

Maximilian Janisch
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