Solve this functional equation: $h(-s)=a-h(s)$

Question

Let $h$ be an analytic function. My question is : Solve this functional equation: $$h(-s)=a-h(s)$$ holds true for all $s∈ℂ$. Here, $a∈ℂ$, $a≠0$.

Answer 1

Since you know that $h$ is analytic, you can take derivatives. So your equation becomes $$\tag{1} -h'(-s)=-h'(s), $$ or $h'(s)=h'(-s)$. Thinking of the series expansion of $h'$ around $0$, we would have $h'(z)=\sum_{k=0}^\infty c_kz^k$. The property of $h'$ now reads $$ \sum_{k=1}^\infty c_kz^k=\sum_{k=1}^\infty c_k(-z)^k. $$ We can see here that all terms of odd degree should vanish, and any $$ h'(z)=\sum_{k=0}^\infty c_{2k}z^{2k} $$ satisfies $(1)$. We see from the original equation that $h(0)=a/2$. So $$ h(z)=\frac{a}2+\sum_{k=1}^\infty d_{k}z^{2k-1} $$ for any choice of $\{d_k\}$ that makes $h$ entire.

Answer 2

Inserting $s = 0$ into the functional equation, we obtain

$$h(0) = a - h(0),$$

hence $h(0) = a/2$. That suggests we look at $f(s) = h(s)-a/2$, and for that we find

$$f(-s) = h(-s) - a/2 = (a - h(s)) - a/2 = a/2 - h(s) = - f(s),$$

so $f$ is an odd function.

Conversely, for every odd function $f$, the function $h(s) = f(s) + a/2$ satisfies the functional equation

$$h(-s) = f(-s) + a/2 = a/2 - f(s) = a - (a/2+f(s)) = a - h(s).$$

Answer 3

We have $$h(-s)-\frac{a}{2} = -\left( h(s)-\frac{a}{2}\right).$$ So $$h(s)=\frac{a}{2} + g(s)$$ is a solution for any odd analytic $g(s)$.