Lwt $f, g $ be differentiable and $f', g'$ bounded. Show $\int_{\mathbb R}fg' = -\int_{\mathbb R}f'g$

Question

Let $f $ and $g $ be differentiable function on $\mathbb R $. Assume that $f'$ and $g'$ are bounded, and that $f $ and $g $ are integrable. Show that $\int_{\mathbb R}fg' = -\int_{\mathbb R}f'g$ (in particular, check that both sides of the equation are defined.

I've shown that both sides of the equation are well-defined. By chain rule and Fundamental Theorem of Calculus, it suffices to show $fg (x) \to 0$ as $x $ goes to infinity. How would I be able to show this?

Answer 1

If the integral with limits of infinity is convergent then you know that the function fg(x) must also be convergent

Answer 2

Lemma. Let $f$ be uniformly continuous and Lebesgue integrable on $\mathbb{R}$. Then $f(x) \to 0$ as $|x|\to\infty$.

Proof. Assume otherwise. Then we can find $|x_n| \to \infty$ and $\epsilon > 0$ such that $|f(x_n)| \geq \epsilon$. By the uniform continuity of $f$, we can find $\delta > 0$ such that $|f(x)| \geq \epsilon/2$ whenever $|x-x_n| \leq \delta$. Assuming WLOG that $|x_{n+1}| > |x_n| + 2\delta$, this leads to a contradiction

$$ \int_{-\infty}^{\infty} |f(x)| \,dx \geq \sum_{n=1}^{\infty} \int_{x_n-\delta}^{x_n+\delta} |f(x)| \, dx \geq \sum_{n=1}^{\infty} \epsilon\delta = \infty. $$

Therefore the claim follows. ////

Returning to the original problem, we make several observations.

1. Since $f$ is integrable and $g'$ is bounded, we find that $fg'$ is also integrable. (Here, Lebesgue integrability is being considered.)

2. It is tempting to apply Fundamental Theorem of Calculus (FToC). The issue is that the classical FToC assumes continuity of the derivative. In other words, one needs a stronger version of FToC for a full generality. Toward this direction, it is known that if $f$ is everywhere-differentiable on $[a, b]$ then it is also absolutely continuous on $[a, b]$ and hence FToC holds.

Combining altogether, both $fg'$ and $f'g$ are in $L^1(\mathbb{R})$ and we have

$$ \int_{a}^{b} (f'g + fg') \, dx = [fg]_{a}^{b} = f(b)g(b) - f(a)g(a) $$

Letting $a\to-\infty$ and $b\to\infty$ together with Lemma and Dominated Convergence Theorem proves the desired result.