# What goes up must come down

Let $f \in C^1(\mathbb{R})$ such that $f,f' \in L^1(\mathbb{R})$. Prove that $\int_\mathbb{R}f'=0$.

## 4 thoughts on “What goes up must come down”

1. This may be total nonsense since I’ve forgotten all my real analysis, but doesn’t this follow from the density of $C_c^\infty(\Bbb{R})$ in the Sobolev space $W^{1,1}(\Bbb{R})$?

Like

1. Actually I don’t remember the definition of the Sobolev spaces, but everything is possible!

Like

2. Some details: the Sobolev space $W^{1,1}$ is the space of $L^1$ functions with weak derivative also in $L^1$. In particular, any $f$ as in the problem statement is in $W^{1,1}$. This space is a Banach space with norm given by $\Vert f\Vert_{W^{1,1}} = \Vert f\Vert_1 + \Vert f'\Vert_1$.

Since $C_c^\infty(\Bbb{R})$ is dense in $W^{1,1}(\Bbb{R})$, we have a sequence $\{f_n\}$ of smooth, compactly supported functions converging in $W^{1,1}(\Bbb{R})$ to $f$. Notice that this convergence implies $L^1$ convergence of both $f_n$ and $f_n'$. Since the integral of $f_n'$ vanishes by Barrow’s rule, the result follows passing to the limit.

Liked by 1 person

3. Another proof is the following: We have $f \in L^1(\mathbb{R})$ so there exist two sequences of real numbers $(a_n)_{n\in\mathbb{N}}$, $(b_n)_{n\in\mathbb{N}}$ such that $\max\{|f(a_n)|,|f(b_n)|\} <1/n$ for every $n \in \mathbb{N}$ and $\lim_{n\to\infty}a_n=-\infty$, $\lim_{n\to\infty}b_n=+\infty$.
By the Lebesgue's Dominated Convergence Theorem and Barrow's rule it follows that $\int_\mathbb{R}f' = \lim_{n\to\infty}\int_\mathbb{R}f'\chi_{[a_n,b_n]} = \lim_{n\to\infty}(f(b_n)-f(a_n)) = 0$.

Corollary: Such function $f$ tends to $0$ at infinity. Counterintuitively, the hypothesis $f'\in \L^1(\mathbb{R})$ is necessary as you can check by constructing a counterexample.

Liked by 1 person