Is a function $f$ with $f(X)\perp (I-f)(X)$ necessarily linear?

Question

Let $X$ be a real or complex inner-product space, and let $f : X\rightarrow X$ be a function such that every element of $f(X)$ is orthogonal to every element of $(I-f)(X)$. Prove or give a counterexample to the claim that $f=P$ where $P$ is a linear, orthogonal projection, i.e., $$ P^{2}=P,\;\; (Px,y)=(x,Py),\;\; x,y \in X. $$

Answer 1

Let $Y:={\rm span\,}f(X)$, the subspace generated by the range of $f$. If $Y=\{0\}$, we are ready.

Now, for any $x\in X,\ z\in f(X)$, we have $\langle z,\ x-f(x)\rangle=0$, i.e. $\langle z,x\rangle=\langle z,\,f(x)\rangle$.

It follows that for all $z\in f(X)$, $$ \langle z,\,f(x_1)\rangle+\langle z,f(x_2)\rangle\ =\ \langle z,\,x_1\rangle + \langle z,\,x_2\rangle\ =\ \langle z,\,x_1+x_2\rangle\ =\ \langle z,\,f(x_1+x_2)\rangle\,,$$ so the vectors $f(x_1+x_2)\,-\,f(x_1)-f(x_2)$ are orthogonal to all $f(X)$, hence $\in Y^\perp$.
Similarly for any $x$, $\ f(\lambda x)-\lambda\, f(x)\ \in Y^\perp$.

But, observe also that, by definition of span, all these vectors are also in $Y$.

Consequently, $f$ must be linear.