# Real Analysis, Folland problem 5.5.55 Hilbert Spaces

problem 3.5.55 - Let $$\mathcal{H}$$ be a Hilbert space.

a.) (The polarization identity) For any $$x,y\in\mathcal{H}$$ $$ = \frac{1}{4}(\lVert x + y \rVert^2 - \lVert x - y\rVert^2 + i\lVert x + iy\rVert^2 - i\lVert x - iy\rVert^2)$$ b.) If $$\mathcal{H}'$$ is another Hilbert space, a linear map from $$\mathcal{H}$$ to $$\mathcal{H}'$$ is unitary if and only if it is isometric and surjective.

Attempted proof a.)$$\lVert x + y \rVert^2 = \lVert x \rVert^2 + \lVert y \rVert^2 + + $$ $$-\lVert x - y \rVert^2 = -\lVert x \rVert^2 - \lVert y \rVert^2 + + $$ $$i\lVert x + iy\rVert^2 = i\lVert x \rVert^2 + i\lVert y \rVert^2 + - $$ $$-i\lVert x - iy\rVert^2 = -i\lVert x \rVert^2 - i\lVert y \rVert^2 + - $$ Adding the above gives the result

Attempted proof b.): Suppose, $$f$$ is a unitary linear map from $$\mathcal{H}$$ to $$\mathcal{H}'$$. We need to show that $$f$$ is isometric and surjective. Since $$f$$ is a unitary linear map we know from Folland that $$f$$ is also an invertible linear map that preserves inner products: $$\langle fx,fy\rangle_{2} = \langle x,y\rangle _{1} \ \ \forall \ x,y\in\mathcal{H}$$ Set $$y = x$$, then we have $$\lVert fx\rVert_{2} = \lVert x\rVert_{1}$$ thus an isometry.

I am not sure how to proceed further or if this approach is incorrect somehow. Any suggestions is greatly appreciated, I will update more as I keep thinking about it.

• You use the polarization identity to show that preserving the norm is the same as preserving the inner product Commented Mar 27, 2016 at 18:45
• For part a, your 4th line should start with: $-i\lVert x-iy\rVert^2$, not $-i\lVert x+iy\rVert^2$. Commented Mar 28, 2016 at 1:56

You have proved that it is an isometry. It must be surjective for else it wouldn't be invertible.

Now assume that we have a surjective isometric linear mapping. We have to prove that it preserves inner products.

$<fx,fy>=\frac{1}{4}(||fx+fy||^2+||fx-fy||^2+i||fx+ify||^2-i||fx-ify||^2)$

$=\frac{1}{4}(||f(x+y)||^2+||f(x-y)||^2+i||f(x+iy)||^2-i||f(x-iy)||^2)$

$=\frac{1}{4}(||x+y||^2+||x-y||^2+i||x+iy||^2+||x-iy||^2)$

$=<x,y>$

QED.

• You put QED at the bottom. But isn't the converse of the argument for b.) complete? Commented Mar 28, 2016 at 17:51
• @Wolfgang What do you mean with that? Commented Mar 28, 2016 at 18:08
• Since you have shows that it preserves inner products. Then have we proved the converse of the proof for b.)? Commented Mar 28, 2016 at 18:10

Unitary means invertible and inner-product-preserving: $$\langle fx,fy\rangle_2=\langle x,y\rangle_1.\tag1$$ On the other hand, surjectivity is logically "half" of bijectivity, and isometric means $$\lVert fx\rVert_2 = \lVert x\rVert_1.\tag2$$ However, a linear map $f:\mathcal{H}\to\mathcal{H}'$ is invertible if it's bijective and its inverse is bounded: $$\lVert fx\rVert_2\ge c\lVert x\rVert_1 \qquad\text{for some}\quad c>0.\tag3$$

As you noted (and Folland does right after the definition of a unitary map, on page 176), setting $y=x$ shows that $(1)\implies(2)$, inner-product-preserving linear (unitary) maps are isometries. Since unitary maps are also invertible, they must be surjective. This proves one direction: an invertible inner-product-preserving linear map is a surjective isometry.

However (to prove that a surjective isometric linear map is invertible and inner-product-preserving), note that an isometry also has bounded inverse, $(2)\implies(3)$, as seen from their definitions on page 154, taking $c=1$. Isometries are clearly injective since $x\ne y\implies$ $0\ne\lVert{x-y}\rVert=$ $\lVert{f(x-y)}\rVert=$ $\lVert{f(x)-f(y)}\rVert\implies$ $f(x)\ne f(y)$, but if we are given that $f$ is also surjective, then it is therefore bijective and hence also invertible, by definition. We thus only need to show that a surjective isometry is also inner-product-preserving. For this, we'll use the polarization identity from part a to show that preserving the norm entails preserving the inner product.

To verify part a, first note what Folland uses in the proof of the Parallelogram Law, proposition 5.22: $$\lVert x\pm y\rVert^2=\lVert x\rVert^2\pm2\, \text{Re}\langle x,y\rangle+\lVert y\rVert^2,$$ where $2\text{Re}\langle{x,y}\rangle= \langle{x,y}\rangle+\langle{y,x}\rangle$ since the summed terms are complex conjugates. Noting that $$\lVert x\pm iy\rVert^2=\lVert x\rVert^2\pm2\, \text{Re}\langle x,iy\rangle+\lVert iy\rVert^2,$$ and $\langle x,iy\rangle=-i\langle x,y\rangle$ so that $$\text{Re}\langle x,iy\rangle=\text{Im}\langle x,y\rangle,$$ we conclude that \begin{align} \langle x,y\rangle &=\text{Re}\langle x,y\rangle+i\,\text{Im}\langle x,y\rangle\\ &=\text{Re}\langle x,y\rangle+i\,\text{Re}\langle x,iy\rangle\\ &=\tfrac14\left[ \left(\lVert x+y\rVert^2-\lVert x-y\rVert^2\right)+i \left(\lVert x+iy\rVert^2-\lVert x-iy\rVert^2\right)\right]\\ &=\tfrac14\left( \lVert x+y\rVert^2-\lVert x-y\rVert^2+i \lVert x+iy\rVert^2-i\lVert x-iy\rVert^2 \right). \end{align} Note the negative sign by the second term, correcting a typo in the book for part a. Finally, the conclusion of part b follows from $(2)$ since $f$ is an isometry: \begin{align} \langle fx,fy\rangle &=\tfrac14\left( \lVert fx+fy\rVert^2-\lVert fx-fy\rVert^2+i \lVert fx+ify\rVert^2-i\lVert fx-ify\rVert^2 \right)\\ &=\tfrac14\left( \lVert f(x+y)\rVert^2-\lVert f(x-y)\rVert^2+i \lVert f(x+iy)\rVert^2-i\lVert f(x-iy)\rVert^2 \right)\\ &=\tfrac14\left( \lVert x+y\rVert^2-\lVert x-y\rVert^2+i \lVert x+iy\rVert^2-i\lVert x-iy\rVert^2 \right)\\ &=\langle x,y\rangle. \end{align}

• To understand you correctly, are you saying that I have proved the forward direction? Now, I need to prove the converse? Commented Mar 28, 2016 at 0:12
• Yes, @Wolffang. The meat of one direction is showing (1) implies (2) and we've done that. The meat of the converse is showing (2) implies (1) and Hasan Saad has done that, as Adam Hughes suggested. The first paragraphs of my post elaborate the conditions of each side, drawing the plan for both directions of the argument. Commented Mar 28, 2016 at 2:03