Distance preserving function on a Hilbert space Let $\Bbb F = \Bbb R$. Show that every preserving function $f$ on Hilbert space $H$ has the form $f(x) = f(0) + Tx$ for some isometry $T$ in $B(H)$.
If $f$ is linear then $f$ is an isometry. Suppose $f$ is not linear, then I define $T:H\to H$ such that $Tx:= f(x) - f(0)$ but I can not show that it's linear. Please give me a hint. Thanks 
 A: By subtracting $f(0)$ we can assume that $f(0)=0$.
Let $x$ and $y$ be orthogonal. Then $\|x-y\|^2=\|x-0\|^2+\|y-0\|^2$. But this equation is equivalent to $\|f(x)-f(y)\|^2=\|f(x)-f(0)\|^2+\|f(y)-f(0)\|^2$. Therefore, $f(x)$ and $f(y)$ are orthogonal if and only if $x$ and $y$ are orthogonal.
For every $z$ that is orthogonal to $x$ and $y$ (and therefore to $x+y$) we have $$0=(f(x+y)-f(x)-f(y),f(z)).$$ 
Therefore $f(x+y)-f(x)-f(y)=0$. From this it follows that $f(rx+sy)=rf(x)+sf(y)$ for all rational $r,s$. From continuity it follows that it is linear.
This proof is not complete. There are arguments missing, but these are the ideas to use: That it preserves orthogonality, and the idea of using continuity to pass from additivity to linearity.
A: Given that $\|f(x)-f(y)\|=\|x-y\|$, define $g(x)=f(x)-f(0)$. Then $\|g(x)\|=\|x\|$. By the Parallelogram law,
$$
         \|g(x)+g(-x)\|^{2}+\|g(x)-g(-x)\|^{2}=2\|g(x)\|^{2}+2\|g(-x)\|^{2}
$$
Hence,
$$
              \|g(x)+g(-x)\|^{2}+\|x-(-x)\|^{2}=2\|x\|^{2}+2\|-x\|^{2} \\
                     \|g(x)+g(-x)\|^{2} = 0 \\
                       \implies g(-x)=-g(x).
$$
Using this with the Parallelogram Law again,
$$
          4(g(x),g(y))= \|g(x)+g(y)\|^{2}-\|g(x)-g(y)\|^{2} \\
                = \|g(x)-g(-y)\|^{2}-\|g(x)-g(y)\|^{2} \\
                = \|x-(-y)\|^{2}-\|x-y\|^{2}=4(x,y).
$$
If you expand $\|g(x+y)-g(x)-g(y)\|^{2}$, you'll end up with inner product terms of the form $(g(a),g(b))$, all of which can be replaced with $(a,b)$ and, hence,
$$
        \|g(x+y)-g(x)-g(y)\|^{2} = \|x+y-x-y\|=0.
$$
Similarly,
$$
\begin{align}
       \|\alpha g(x)-g(\alpha x)\|^{2}& =\alpha^{2}\|g(x)\|^{2}-2\alpha(g(x),g(\alpha x))+(g(\alpha x),g(\alpha x)) \\
       & = \alpha^{2}\|x\|^{2}-2\alpha(x,\alpha x)+(\alpha x,\alpha x) = 0.
\end{align}
$$
