Let $(H, <\cdot ,\cdot>) $ be a real pre-Hilbert space and let $T: H \to H$ be an isometry so that $||T(x)-T(y)|| = ||x - y||$ for all $x, y \in H$. Prove $T$ is affine.
How do I prove this?
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What is your definition of "affine" here? I haven't seen it used in this context before. – Math1000 Oct 26 '17 at 16:13
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1$T $ is affine if and only if there exists a linear map $L: H \to H $ and $x_0 \in H $ such that $T(x) = Lx + x_0$ – Chiray Oct 26 '17 at 16:16
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Ah, we aren't assuming that $T$ itself is linear. The question makes sense now:) – Math1000 Oct 26 '17 at 16:35
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Exaclty, that would be trivial then, but we don't know that $T(0) = 0$ so $T$ is not necessarily linear. – Chiray Oct 26 '17 at 16:41
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You can't prove that $T$ is surjective; that doesn't follow. – David C. Ullrich Oct 26 '17 at 17:01
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You are right, that should be true if $T(0) = 0$, shouldn't it? – Chiray Oct 26 '17 at 17:34
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1No, $T(0)$ does not imply it's surjective. Consider the right shift on $\ell^2(\mathbb N)$. That's a non-surjective linear isometry. – David C. Ullrich Oct 26 '17 at 18:17
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@NateEldredge but we can't prove that $T$ is surejective – Chiray Oct 27 '17 at 08:16
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Replace $T$ by $S(x) = T(x) - T(0)$. $S$ is an isometry with $S(0) = 0$, and the task is to prove that $S$ is linear.
Now $S$ preserves both distances and norms. Writing out $$ ||S(a)-S(b)||^2 = d(S(a), S(b))^2 = d(a, b)^2 = ||a-b||^2, $$ and expanding gives $$\langle S(a), S(b)\rangle = \langle a, b \rangle.$$ Using this, compute $$ \langle S(a + \lambda b) - S(a) - \lambda S(b), S(a + \lambda b) - S(a) - \lambda S(b)\rangle = 0. $$ Hence $S(a + \lambda b) - S(a) - \lambda S(b) = 0$, and $S$ is linear.
Addendum: Maybe the last step is more palatable in the following form: For all elements $S(z)$ in the range of $S$, one has $$ \langle S(a + \lambda b) - S(a) - \lambda S(b), S(z) \rangle = 0. $$ Hence for all $y$ in the linear span of the range (which we have to consider as we don't yet know the range is a linear space), $$ \langle S(a + \lambda b) - S(a) - \lambda S(b), y \rangle = 0. $$ Hence, in particular, $$ \langle S(a + \lambda b) - S(a) - \lambda S(b), S(a + \lambda b) - S(a) - \lambda S(b)\rangle = 0. $$
Comment: Every complex inner product spaces is in particular a real inner product space under $(a, b) \mapsto \operatorname{Re}\langle a, b\rangle $, and the real inner product gives, of course, the same norm. So an isometry is necessarily real affine.
fred goodman
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