Functions that satisfy $\int_{-a}^a f(x) \ dx = a*\int_{-1}^1 f(x) \ dx$

Question

I want to find all elementary functions that satisfy this equation for arbitrary $a$. The equation is automatically true when $a=0$, but when $a\neq0$ I can think of two types of functions that solve the equation. If $f(x)=c$ then $$\int_{-a}^a f(x) \ dx = a*\int_{-1}^1 f(x) \ dx$$ $$\int_{-a}^a c \ dx = a*\int_{-1}^1 c \ dx$$ $$ c(a - (-a)) = a*c(1- (-1)) $$ which gives $2ac=2ac$. If $f(x)$ is an odd function, then the integrals on both sides will be zero, satisfying the equation. In an attempt to find more solutions, I replaced $a$ with the variable $t$ and considered both sides of the equation to be functions of $t$, so $$\int_{-t}^t f(x) \ dx = t * \int_{-1}^1 f(x) \ dx$$ then differentiated to get $$f(-t) + f(t) = \int_{-1}^1 f(x) \ dx$$ and differentiated again to get $$-f'(-t) + f'(t) = 0$$ $$f'(t) = f'(-t)$$ which implies $f'(x)$ is an even function. This leaves me with multiple questions.

1. Am I correct in my conclusion that $f'(x)$ is even?
2. Are there solutions for $f(x)$ besides what I found?
3. If the answer to (2) is no, does that mean that the derivative of an odd function is an even function?

Answer 1

Judging from your work, I will assume that we are looking for all functions $f$ such that $$\int_{-a}^a f(x)\,dx = a\int_{-1}^1f(x)\, dx,\quad \forall a\in\mathbb R.\tag{1}$$ Also, I will assume $f$ is continuous.

As you observe, this is satisfied by any integrable odd function since both integrals vanish.

Assume that $f$ is even and define $F(a) = \int_0^a f(x)\, dx$. Since $f$ is continuous, $F$ is differentiable with $F' = f$. From $(1)$ and $f$ being even we have $$ F(a) = \int_0^a f(x)\, dx = \frac 12 \int_{-a}^a f(x)\, dx = \frac a2 \int_{-1}^1 f(x)\, dx = a F(1),$$ so differentiating gives us $f(a) = F'(a) = F(1)$, i.e. $f$ is a constant function.

Finally, since any function is a sum of an even and an odd function: $$f(x) = \frac{f(x)+f(-x)}2 + \frac{f(x)-f(-x)}2,$$ we conclude that all $f$ continuous on $\mathbb R$ that satisfy $(1)$ are of the form $f(x) = g(x) + c$, where $g$ is odd and $c\in\mathbb R$.

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To answer your specific questions:

1. Yes, under interpretation that $(1)$ holds for all $a$, not just a fixed constant.
2. Answered above under the assumption that $f$ is continuous and $(1)$ holds for all $a$.
3. Yes, derivative of an even function is odd, and vice versa.

Answer 2

1. If you assume $f$ is differentiable, then yes, that assertion should hold.
2. There are solutions for $f$, in fact, infinitely many. I leave it to you to verify that $f(x) = 1 + g(x)$, where $g$ is any odd integrable function, satisfies your functional equation.
3. The derivative of an odd function is an even function. For if $f(x) = -f(-x)$, then $f'(x) = f'(-x)$, which means $f'$ is even.

Answer 3

Point 1: If you are looking for the differentiable functions which satisfy the equation for every positive $a,$ then $f'$ is even indeed.

Point 3: The derivative of an odd (differentiable) function is always an even function. (similarly, the derivative of an even differentiable function is odd). But that is not "meant" by the following negative answer to 2. As we shall see, we can solve this equation without assuming that $f$ is differentiable.

Point 2: You already found that all constant or odd functions are solutions. Since the equation is linear with respect to $f,$ sums of such solutions (i.e. functions of the form $g(x)+c$ with $g$ odd) are also solutions. Let us prove that they are the only (continuous) solutions. Let $f$ be a solution, $c=f(0),$ and $g=f-c.$ Then (by linearity) $g$ is again a solution hence (differentiating like you did) $g(t)+g(-t)$ does not depend on $t$ i.e. it is equal to $2g(0)=0,$ q.e.d.