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This question was on an exam and I am not sure how to answer it. I mostly tried writing zero in different ways and tried lots of algebra to get something out. I also tried to use the fact that $H$ is a hilbert space but couldn't get anywhere. Any suggestions?

If $H$ is a hilbert space and $f:H\to H$ is an isometry ($\|f(x)-f(y)\|=\|x-y\|$) such that $f(0)=0$, then $f$ is linear.

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Are you working over a complex or real Hilbert? –  Davide Giraudo Mar 16 '12 at 19:35
    
First thing to notice : $f$ conserves the inner product ... –  Selim Ghazouani Mar 16 '12 at 19:41
    
Ah yes I should have said that. I believe the problem said complex. –  john w. Mar 16 '12 at 19:43
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This is not true for complex Hilbert spaces ($f$ may be antilinear, e.g. consider complex conjugation acting on $\mathbb{C}$) but it is only barely false; I think $f$ must be linear or antilinear. –  Qiaochu Yuan Mar 16 '12 at 19:43
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An outline is given here –  t.b. Mar 16 '12 at 20:34

3 Answers 3

up vote 5 down vote accepted

In fact, in any strictly convex real Banach space $B$ an isometry $f$ with $f(0)=0$ is linear. Note that for any $a, b \in B$, $(a+b)/2$ is the only $v \in B$ with $\|a-v\| = \|v-b\|=\|a-b\|/2$. So $f((a+b)/2)$ is the only $v$ with $\|f(a) - v\| = \|v - f(b)\| = |f(a) - f(b)|/2$, and thus $f((a+b)/2) = (f(a) + f(b))/2$.
From this it's not hard to show that $f(ra) = r f(a)$ for any rational $r$, and by continuity this is true for any scalar, and then $$f(sa + tb) = f((2sa + 2tb)/2) = (f(2sa)+f(2sb))/2 = s f(a) + t f(b)$$

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There is similar general result by Mazur and Ulam. In fact every bijective isometry $f:E\to F$ between normed spaces $E$ and $F$ is the composition of isometric linear operator $T:E\to F$ and shift on vector $f(0)$.

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While this is certainly relevant to the question, it doesn't really answer it because there are non-surjective isometries, of course. But +1 for linking to the beautiful paper by Väisälä - one of my favorite texts of the past few years :) –  t.b. Mar 16 '12 at 22:52
    
@t.b. I was amazed too –  no identity Mar 16 '12 at 22:56
    
thanks for the article link. –  john w. Mar 17 '12 at 3:08

You can prove that $\langle f(x),f(y) \rangle = \langle x,y \rangle$ for any real Hilbert Space by deriving the following expressions for $\|f(x) - f(y)\|^2$ and $\|x-y\|^2$: \begin{align*} \|f(x)-f(y)\|^2 &= \langle f(x) - f(y), f(x) - f(y)\rangle \\ &= \|f(x)\|^2 - 2\langle f(x),f(y)\rangle + \|f(y)\|^2\\ \end{align*} and \begin{align*} \langle x,y\rangle &= \|x\|^2 - 2\langle x,y\rangle + \|y\|^2 \end{align*} Since $\|f(x)-f(0)\| = \|x - 0\|$, and $f(0) = 0$, you know $\|f(x)\| = \|x\|$. Therefore, after setting the two expressions equal to each other, the correct terms match up to give preservation of the inner product, as claimed.

Then you can use this to show that $$\|f(x+y) - f(x) - f(y)\|^2 = \langle f(x+y) - f(x)-f(y), f(x+y)-f(x)-f(y)\rangle = 0,$$ therefore proving that $f$ is additive. The same should work for $f(\lambda x)$.

Showing the preservation of the inner product for a complex Hilbert space won't work using the same trick; the farthest I've gotten with it is showing that the real part of $\langle f(x),f(y)\rangle$ is equal to the real part of $\langle x,y\rangle$. All of the proofs I've seen that show equality in the complex case assume that $f$ is already linear and just use the polarization identity $$\langle x,y\rangle = \frac{1}{4}(\|x+y\|^2 - \|x-y\|^2 +i\|x+iy\|^2 -i\|x-iy\|^2)$$ on $\langle f(x),f(y)\rangle$. I suspect it is true in the complex case, I just haven't found out why.

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