Does $\langle f+h,g\rangle=\langle f,g\rangle+\langle h,g\rangle$ hold for all elements $f, g, h$ of an inner product space?

Are there any exceptions?

I was thinking proof by contradiction i.e. define $\langle f,g\rangle\ \neq0$ for two orthogonal elements of the product space, but positive definiteness would require one of the two elements to be zero.

EDIT: I was told that additivity in the second coordinate of any inner product space $\langle f,g\rangle$ logically follows from and can be proven using only these four properties of inner product spaces:

1. $\langle f,g\rangle=\langle g,f\rangle$ (symmetry)
2. $\langle f+h,g\rangle=\langle f,g\rangle+\langle h,g\rangle$ (additivity)
3. $\langle \text{c}f,g\rangle=\text{c}\langle f,g\rangle$
4. $\langle f,f\rangle>0\,\,\forall\,\left\{f\neq0\right\}\in V$, where $V$ is some linear space (posititive definiteness).

I'm thinking that 1. and 2. imply $\langle f,g+h\rangle=\langle f,g\rangle+\langle f,h\rangle$; how do I use these axioms to explicitly verify this?

• It has to be additive bilinear in both coordinates otherwise it is not an inner product. – Timbuc Mar 25 '15 at 17:39
• I get that - but how do I show that being additive bilinear in one coordinate implies the same for the other coordinate, for any inner product? – Benjamin Loya May 2 '15 at 23:33

$\langle f,g+h\rangle = \langle g+h, f\rangle$ by symmetry. Then, $\langle g+h,f\rangle = \langle g,f\rangle + \langle h,f\rangle = \langle f,g\rangle + \langle f,h\rangle$ by additive in first coordinate and then symmetry.