Hi, I am trying to solve the following functional equation
f(x)-f(2a-x)=b(x-a)
considering "a" and "b" are positive constants.
there are two parameters so let's try a function of the form
$f(x) = c_1 x + c_0$
$c_1 x + c_0 - (c_1(2a-x)+c_0) = b(x-a)$
$2c_1 x - 2 a c_1 = b x - a b$
$2c_1 = b$
$c_1 = \dfrac b 2$
$-2ac_1 = -a b$
$a$ is specified positive so
$2c_1 = b$
$c_1 = \dfrac b 2$
and we have agreement in the value of $c_1$
so we have
$f(x) = \dfrac b 2 x + c_0$
There are of course higher order polynomials that satisfy this functional relationship but this is the lowest order one.
Well, the first order polynomial is the special case of the second order polynomial, but to solve the functional equation one should give a complete solution, maybe there is a more general function that becomes, for example, second order polynomial in the special case but still satisfies the equaion.
for example, let's solve following functional equation:
f(x)-2f(1/x)=2^x
we can replace x by 1/x
f(1/x)-2f(x)=2^(1/x)
from these two equations, one can determine that f(x) =-(1/3)(2^x+2^(1-1/x))
If we take $x = a$ in the given equation we have $f(a)-f(a) = 0$ so $f(a)$ is arbitrary. Let's call $f(a)=C$. Now let's assume $f$ is differentiable and differentiate the equation giving
$f'(x) + f'(2a-x) = b$ and put $x=a$ in that. That gives $f'(a) + f'(a) = b$ so $f'(a) = \frac b 2$. Let's differentiate again:
$f''(x) - f''(2a-x) = 0$. Putting $x=a$ in that gives $f''(a)-f''(a) = 0$, so $f''(a)$ is arbitrary. Lets call $f''(a) = D$. Let's differentiate again:
$f'''(x) + f'''(a-x) = 0$. Putting $x=a$ in that gives $f'''(a)+f'''(a) = 0$, so $f'''(a)$ is $0$.
Now a pattern becomes clear. From now on all the odd derivatives will be $0$ and the even ones will be arbitrary when $x=a$. So let's look at the Taylor series
for $f(x)$ about $x=a$.
$$f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!}(x-a)^k = C + \frac b 2 (x-a) + \frac D 2 (x-a)^2 +\text{more terms with even powers of }(x-a)$$
Notice that if we take the arbitrary constant $D=0$ we get $f(x) = C + \frac b 2 (x-a) = C-\frac{ab} 2 + \frac b 2 x$ which agrees with the earlier first degree
solution. And it satisfies the functional equation. So what about all those arbitrary terms of the form $K(x-a)^{2k}$. Notice that such expressions satisfy $f(x)-f(2a-x)=0$ so they don't disturb the functional relationship that is already satisfied by the first two terms. So the functional equation can be satisfied by adding any convergent sum of the form$$
\sum_{k=1}^\infty a_k (x-a)^{2k}$$ to the first two terms.
In fact, there is no real reason not to include the constant term as one of the even powers and write the answer as
$$f(x) = \frac b 2 (x-a) + \sum_{k=0}^\infty a_k (x-a)^{2k}$$
So, for example, a non polynomial example is $f(x) = \frac b 2(x-a) + \cos(x-a)$.