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?



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

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.


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?


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).

  • $\begingroup$ 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 $\endgroup$ Commented Nov 23, 2019 at 14:55
  • $\begingroup$ Also, why do you impose that the function is not analytic? $\endgroup$ Commented Nov 23, 2019 at 14:59
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    $\begingroup$ The fact that $I=0$ is just the fundamental theorem of calculus. Note that $f(a)=f(b)=0$ and $fDf=\frac12D(f^2)$. $\endgroup$ Commented Nov 23, 2019 at 15:04
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    $\begingroup$ 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 $\emptyset\ne[a,b]$. $\endgroup$ Commented Nov 23, 2019 at 15:23
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    $\begingroup$ 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 $f(b)=0$. $\endgroup$ Commented Nov 23, 2019 at 15:27

1 Answer 1


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 $f(a)=f(b)=0$.

  • $\begingroup$ Thank you for the answer. As stated in the previous comment (I do not have rigorous analysis background), why is it true that $f(a)=f(b)=0$? I'm inquiring this due to the fact that $a$ and $b$ belong in the compact support, and concordantly I notice that for most bump functions that $f(a)=f(b)=\textrm{lim}_{h \rightarrow 0}{h}$. $\endgroup$
    – ShellRox
    Commented Nov 23, 2019 at 15:22
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    $\begingroup$ I will add an explanation in a minute @ShellRox $\endgroup$ Commented Nov 23, 2019 at 15:25
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    $\begingroup$ @ShellRox See my edit $\endgroup$ Commented Nov 23, 2019 at 15:30
  • $\begingroup$ Thank you, it makes sense now! $\endgroup$
    – ShellRox
    Commented Nov 23, 2019 at 15:38

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