Maximizing a function containing an integral Problem. Let $\rho\colon[-1,\infty)\to\mathbb{R}$ be a function such that $$\int_{-1}^\infty\rho(x)\,dx=1.$$ Let $G\colon[0,1]\to\mathbb{R}$ be a function that is defined with $$G(f) := \int_{-1}^\infty \log(1+fx)\,\rho(x)\,dx.$$ Show that $G$ is either monotonically increasing, or has one local maximum on the interval $(0,1)$ (depending on how “fat” the “tail” of the function $\rho$ is).
Real life motivation. Suppose that we start with capital $V_0$, and we can repeatedly invest it in some venture with a percentage return given by a fixed probability distribution. This distribution is continuous, but it never takes values below $-1$ (we can never lose more than our investment). Its density function is the function $\rho$ as shown above.
Now, let $X_1,X_2,\ldots$ be independent random variables having this distribution. Furthermore, assume that we always invest a fixed fraction $f$ of our current capital. After $n$ investments, then, our capital will be $$V_n = V_0(1+fX_1)\cdots(1+fX_n).$$
We wish to measure how fast our capital is growing, and to that end we define $$G := \lim_{n\to\infty} \frac{1}{n}\log\frac{V_n}{V_0}.$$ Clearly, the faster $V_n$ grows, the bigger $G$ is. And by the weak law of large numbers, we see that $G$ is indeed the integral shown above. Our goal is to say something useful about the fraction $f$ of our current capital that we invest each time.
 A: In your motivation, you say that $\rho$ is a density, so I'm assuming that $\rho\geq0$.  Then one can differentiate under the integral, provided $\rho$ decays sufficiently rapidly ($\rho \in L^p$ for $1\leq p \leq 2$ is sufficient by dominated convergence theorem) to obtain
$\displaystyle G'(f)=\int_{-1}^\infty \frac{f\rho(x)}{1+fx} dx \geq 0$
for $f\in (0,1)$, hence $G$ is monotone.  Since $\rho$ is a density, it is in particular an $L^1$ function, and so this works for all $\rho$.  
Doesn't seem like you can get a local maximum on $(0,1)$.  If $\rho$ is indeed a density, then $G$ is actually stritly monotone for $f\in(0,1)$.  But someone please correct me if I'm wrong.
A: I found out that I’m simply wrong.
What I want to show is only true if the expected value corresponding to the density is positive. To see that, note that the first derivative of $G(f)$ at $f=0$ is the expected value. Note also (as @user4143 has mentioned) that the second derivative of $G(f)$ is always negative.
If the expected value is negative, however, $G(f)$ is—because of the properties of the derivatives mentioned above—a decreasing function.
