Must all inner-product preserving maps be linear?

Question

Must all inner-product preserving maps be linear?

Answer 1

For all $x,y,z,\mu,\nu$, you want $\langle f(x),f(\lambda y+\mu z)\rangle=\langle x,\lambda y\rangle+\langle x,\mu z\rangle$ and, so, for all $x'\in\operatorname{im} f$ the map $\langle x',f(\bullet)\rangle$ is linear.

Assume as a contradiction that there are $x,y,\mu,\nu$ such that $w=f(\lambda x+\mu y)-\lambda f(x)-\mu f(y)\ne 0$. Then, \begin{align}0<&\langle w,w\rangle=\langle w,f(\lambda x+\mu y)\rangle-\lambda\langle w,f( x)\rangle-\mu\langle w,f( y)\rangle=\text{(linearity)}\\=&\langle w, f(\lambda x+\mu y-\lambda x-\mu y)\rangle=\langle w,f(0)\rangle=0\end{align}

Absurd.