Mean value theorem for Improper Integrals

Question

If $f$ is continuous on $[a,\infty)$ and $\phi(x)\geq0$ and be integrable in $[a,\infty) $ then there exists some $c\in(a,\infty)$ such that

$$\int_{a}^{\infty} f(x)\phi(x) dx = f(c)\int_{a}^{\infty}\phi(x) dx$$

I tried $\int_{a}^{\infty} f(x)\phi(x) dx = \lim_{n\to\infty} \int_{a}^{n} f(x)\phi(x)dx$. Can we apply mean value theorem for proper integrals to the latter integral and then take limit?

Answer 1

Caution: The statement is wrong, even not well-defined, if the integrals are allowed to be non-finite! However, if the integrals are finite, the statement is true.

Following assumptions are crucial:

Assume $\int_I \phi$ and $\int_I f\phi$ to be finite where $I = [a, \infty)$ (as stated in the comment). For convenience, assume $\int_I \phi = 1$ without loss of generality, as for $\int_I \phi = 0$ we have $\phi = 0$ almost everywhere and the statement is trivially true.

Proof:

Let $m = \inf_I f$ and $M = \sup_I f$, which may be non-finite. As $m \le f \le M$ and the integral is monotone, we have $$ m \le \int_I f \phi \le M. $$ (That is obviously still valid if $m = -\infty$ or $M = \infty$.)

1. Case $M = \int_I f\phi$. As we assume $\int_I f\phi$ to be finite, we have $M < \infty$. Assume the contrary that $f(x) < M$ for every $x\in I$. Then, for every non-empty compact subset $K\subseteq I$ we have $\min_{K} (M - f) > 0$ by continuity of $f$. Now, as $$ 1 = \int_I \phi = \lim_{b\to \infty} \int_a^{b} \phi,$$ there exists some $b_0$ with $\int_{a}^{b_0} \phi > 0$. In particular, we have $$ \int_I \underbrace{(M-f)\phi}_{\ge 0} \ge \int_a^{b_0} (M-f)\phi \ge \min_{x\in [a, b_0]}(M-f(x)) \int_a^{b_0}\phi > 0, $$ a contradiction. Thus, $f < M$ is false, and there exists some $c_M\in I$ with $f(c_M) = M$.

2. Case $m = \int_I f\phi$: Likewise, we have $m > -\infty$ and there exists some $c_m\in I$ with $f(c_m) = m$.

3. Case $m < \int_I f\phi < M$: We have $\int_I f\phi \in f(I)$, as $f$ is continuous and, thus $f(I)$ is an interval. In particular, there exists $c\in I$ with $$ \int_I f \phi = f(c). $$

The proof (with minor modification) also applies to $f$ restricted on $(a,\infty)$. As the integral remains the same, thus there is also a $c\in (a, \infty)$ with $\int_I f\phi = f(c)$.

Alternatively, if $\int_I f\phi = f(a)$ then there exists some $c\in (a,\infty)$ with $f(a) = f(c)$. To prove this, assume the contrary: For every $x\in(a,\infty)$ we have $f(x) \ne f(a)$. Then, as consequence of the intermediate value theorem, we have either $f(x) > f(a)$ for all $x\in(a,\infty)$ or $f(x) < f(a)$ for all $x\in(a,\infty)$. Thus we obtain $\int_I f\phi > f(a)$ or $\int_I f\phi < f(a)$ respectively, which is a contradiction to $\int_I f\phi = f(a)$.