Several authors have solved
the Pythagorean functional equation
![|f(x ⊢ iy)|2 = |f(x)|2 + |f(iy)|2,](a100x.png) | (1) |
where f is an entire function and x and y are real variables.
A simple computation shows that, if f is a solution of (1), then f is also a
solution of
![- -
|f(z1 +z2)|2 + |f(z1 − z2)|2 = |f(z1 + z2)|2 + |f(z1 − z2)|2,](a101x.png) | (2) |
where z1 and z2 are complex variables. (If an entire function vanishes at the origin
and is a solution of (2), then it is a solution of (1), and conversely.) If an entire
function f is a solution of Jensen’s functional equation
![f(z1 + z2)+ f(z1 − z2) = 2f(z1),](a102x.png) | (3) |
where z1 and z2 are complex variables, then it is also a solution of
![|f(z + z )+ f(z − z )| = |f(z + z )+ f(z − z )|.
1 2 1 2 1 2 1 2](a103x.png) | (4) |
In this paper we shall prove that a solution of (4) is always a solution of (2). Then
we shall solve certain functional inequalities derived from (2) and use the solutions to
solve (1), (2), (3), and (4).
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