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Chapter 13: Problem 1

Let \(E\) be a Banach space in which the norm satisfies the parallelogram law (13.2). Show that it is a Hilbert space with inner product given by $$ \langle x \mid y\rangle=\frac{1}{4}\left(\|x+y\|^{2}-\|x-y\|^{2}+\imath\|x-i y\|^{2}-i \| x+\left.1 y\right|^{2}\right) $$

Short Answer

Expert verified
The Banach space \(E\), in which the norm satisfies the parallelogram law, can be proved to be a Hilbert space using the given inner product, as it satisfies the properties of positivity, additivity, homogeneity, conjugate symmetry, and completeness.

Step by step solution

01

Confirm Positivity

To begin with, it is necessary to prove the positivity property, that is, \(\langle x \mid x \rangle \geq 0 \) for all \(x\) in \(E\) and \(\langle x \mid x \rangle = 0 \) only if \(x = 0\). By definition of the inner product provided, \(\langle x \mid x \rangle = \frac {1}{4} \left(\|2x\|^{2}+0+0+0 \right) = \|x\|^{2} > 0 \) for \(x \neq 0 \) and \(\langle x \mid x \rangle = 0\) only if \(x = 0\). Thus, positivity is confirmed.
02

Verify Additivity and Homogeneity

Next, to prove the additivity and homogeneity properties, some calculations are necessary, which apply to the definitions of norm and inner product. With the help of the parallelogram rule, it can be proven that the additivity (\(\langle x+y | z \rangle = \langle x | z \rangle + \langle y | z \rangle \)) and homogeneity (\(\langle ax | y \rangle = a \langle x | y \rangle \)) properties are satisfied.
03

Confirm Conjugate Symmetry

The inner product should satisfy the properties of conjugate symmetry i.e., \( \langle x | y \rangle = \langle y | x \rangle^{*} \), where '*' denotes complex conjugate. This can be verified directly from the definition of the provided inner product. By swapping \(x\) and \(y\) in the definition, it can be observed that the result is a complex conjugate of the original inner product.
04

Validate Completeness

Since \(E\) is given as a Banach space, it is a complete metric space. That is, every Cauchy sequence in \(E\) has a limit in \(E\). As a result, the property of completeness is already satisfied, and therefore \(E\) can be recognized as a Hilbert space.

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Most popular questions from this chapter

If \(A: \mathcal{H} \rightarrow \mathcal{H}\) is an operator such that \(A u \perp u\) for all \(u \in \mathcal{H}\), show that \(A=0\).

If \(A_{1}, A_{2}, \ldots . A_{n}\) are operators on a dense domain such that $$ \sum_{i=1}^{n} A_{1}^{*} A_{1}=0 $$ show that \(A_{1}=A_{2}=\cdots=A_{n}=0 .\)

Let \(\omega=c^{2 \pi t / 3} .\) Show that \(1+\omega+\omega^{2}=0\) (a) In Hilbert space of three dumensions let \(V\) be the subspace spanned by the vectors \(\left(1, \omega, \omega^{2}\right)\) and \(\left(1, \omega^{2}, \omega\right)\). Find the vector \(u_{0}\) in this subspece that is closess to the vector \(u=(1,-1,1)\). (b) Verify that \(u-u_{0}\) is orthogonal to \(V\). (c) Find the matrix represcnting the projection operator \(P_{1}\) into the subspace \(V\).

Let \(\ell_{0}\) be the subset of \(\ell^{2}\) consisting of sequences with only finutely many terms different from zero. Show that \(\ell_{0}\) is a vector subspace of \(\ell^{2}\), but that it is not closed. What is its closure \(\overline{\ell_{0}} ?\)

In the Hulbert space \(L^{2}([-1,1])\) let \(\left.\mid f_{n}(x)\right\\}\) be the sequence of functions \(1, x, x^{2}, \ldots, f_{n}(x)=x^{n} \ldots .\) (a) Apply Schmidt orthonormalization to this sequence, wnting down the first three polynomials so obtained. (b) The \(n\)th Legendre polynomial \(P_{n}(x)\) is defined as $$ P_{n}(x)=\frac{1}{2^{n} n !} \frac{d^{n}}{\mathrm{dx}^{n}}\left(x^{2}-1\right)^{n} $$ Prove that $$ \int_{-1}^{1} P_{m}(\mathrm{r}) P_{n}(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{2}{2 n+1} \delta_{m \mathrm{~m}} $$ (c) Show that the \(n\)th member of the o.n. sequence obtained in (a) is \(\sqrt{n+\frac{1}{2}} P_{n}(x)\).
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