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.

What goes up must come down

4 thoughts on “What goes up must come down

  1. 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.

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  2. 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.

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