Functional logarithmic equation: $\ln\left(\int_1^2f(x)\,\mathrm{d}x\right)=\int_1^2 \ln f(x)\,\mathrm{d}x$

Question

Today, during a timecut from work, I casually came up with an interesting question.

I was trying to find examples of specific functions that are commutative towards composition, that is $f \circ g=g \circ f$. I tried to find a possible answer, but all my attempts lead me nowhere, with zero progress.

What are all $C^1(\mathbb{R})$ functions $f:\mathbb{R}\to\mathbb{R}^+$, such that: $$ \ln\left(\int_1^2f(x)\,\mathrm{d}x\right)=\int_1^2 \ln f(x)\,\mathrm{d}x $$ In this sense, it "looks" that the natural logarithm is interchangeable with the integral, which of course is not valid in the general sense. Are there functions $f(x)$ that allow such interchange?

EDIT:

After slightly more thinking, I have determined that:

• Thanks to the comments, $f(x)=1$ is a solution.

• If $f(x)$ is a solution, then all functions $cf(x)$, $c>0$ are also a solution. Therefore, $f(x)=c$, $c>0$ are all solutions.

• If $f(x)$ and $g(x)$ are solutions, then: $$ \int_1^2 f(x)g(x)\,\mathrm{d}x=\left(\int_1^2 f(x)\,\mathrm{d}x\right)\left(\int_1^2g(x)\,\mathrm{d}x\right) $$

• Linear functions $f(x)=cx$ are not solutions.

Answer 1

No nontrivial solutions exist.

Note that since $\ln$ is a concave function, we have for any $c \in \mathbb{R}^+$ \begin{align} \ln(x) \leq \frac{1}{c}(x-c) + \ln(c) = \frac{x}{c} - 1 + \ln(c) \end{align} Therefore, we get \begin{align} \int_1^2 \ln(f(x))dx \leq& \int_1^2 \frac{f(x)}{c}-1+\ln(c) dx \\ =& \frac{1}{c}\int_1^2 f(x)dx -1 + \ln(c) \end{align} Substituting $c = \int_1^2 f(x)dx$, we find that \begin{align} \int_1^2 \ln(f(x))dx \leq \ln \left( \int_1^2 f(x)dx \right) \end{align} Hence, if $f$ solves the equation, we must have for any $h: \mathbb{R} \to \mathbb{R}$ that \begin{align} \left. \frac{d}{dt} \right|_{t=0} \left( \ln \left( \int_1^2 f(x)+t \cdot h(x) dx \right) - \int_1^2 \ln(f(x) + t \cdot h(x))dx \right) = 0 \end{align} Simplifying this, we get that if $f$ is a solution, we must have for any $h: \mathbb{R} \to \mathbb{R}$ \begin{align} \frac{\int_1^2 h(x)dx}{\int_1^2 f(x)dx} = \int_1^2 \frac{h(x)}{f(x)}dx \end{align} This is clearly a quite strong condition on $f$. It is equivalent to saying that \begin{align} \int_1^2 h(x) \left(1 - \frac{\int_1^2 f(t)dt}{f(x)} \right)dx = 0 \end{align} for all $h$. In particular, it must hold for $h = \left(1 - \frac{\int_1^2 f(t)dt}{f(x)} \right)$. From this we find \begin{align} \int_1^2 \left(1 - \frac{\int_1^2 f(t)dt}{f(x)} \right)^2 dx =& 0 \\ \left(1 - \frac{\int_1^2 f(t)dt}{f(x)} \right) =& 0 \\ f(x) =& \int_1^2 f(t)dt \end{align} These last two equations must hold almost everywhere on $[1,2]$. Hence, $f$ must be constant almost everywhere on $[1,2]$. The condition of differentiability of $f$ was not necessary.

Answer 2

Alright, so let's vary the upper limits in the integrals, and write $$\log\int_1^xf(t)\mathrm dt=\int_1^x\log f(t)\mathrm dt.$$ Differentiating with respect to $x$ then gives $$\frac{f(x)}{\int_1^xf(t)\mathrm dt}=\log f(x),$$ or $$\frac{f(x)}{\log f(x)}=\int_1^xf(t)\mathrm dt,$$ and writing $y=f(x)$ and differentiating again gives $$y\frac{-1/y}{\log^2y}+\frac{y'}{\log y}=y,$$ which simplifies to $$-1+y'\log y=y\log^2y.$$ Now separating variables and integrating gives $$\int\frac{\log y\,\mathrm dy}{1+y\log^2y}=x+C,$$ from where I got my claim.