# Inequivalent norms (given by different inner products) on infinite dimensional Hilbert space.

I have this question in reviewing for my exam.

Let $H$ be an infinite dimensional Hilbert space. Write down an inner product on $H$ that gives a norm inequivalent with the original norm. Is $H$ complete under the norm determined by the new inner product?

In my understanding, as the norm of a Hilbert space is induced by the inner product it equipped, there are different inner products, all satisfy linearity, conjugate symmetry, positive-definiteness, hence different norms. My confusion is, why the question is asking "original norm"? Is a Hilbert space uniquely determined by the inner product it uses? For example, the collection of square integrable functions on $R^d$ is a Hilbert space and equipped with a inner product of integral form. Finite-dimensional complex Euclidean space with dot product. Or, there is a space can be equipped with different inner product (to define its norm) to form the same Hilbert space? For example, the Euclidean space?

BTW, how is this question related to the fact that.

All infinite-dimensional (separable) Hilbert spaces are $l^2$$(Z) in disguise Due to my poor understanding of functional analysis, I am not sure how is this question trying to motivate my thinking. Any help would be appreciated! Thanks in advance! • Regarding the "original norm": If someone writes "let H be a Hilbert space", they are usually assuming that some fixed scalar product on H is given. Hence you are supposed to find a new inner product on H such that the induced norm is not equivalent (in the sense of \Vert \cdot \Vert_1 \leq C_1 \Vert \cdot \Vert_2 \leq C_2 \Vert \cdot \Vert_1) to the norm induced by the original inner product. The hint tells you (more or less) to first consider the case H =\ell^2 (with the usual inner product). After that, try to deduce the general case from that. – PhoemueX Nov 16 '14 at 8:02 ## 1 Answer Even in finite dimensions you can easily change the inner product. Let \{ e_{n} \} be an orthonormal basis of \mathbb{C}^{N} and define the new inner product$$ (x,y)_{\mbox{new}}=\sum_{n=1}^{N}\lambda_{n}(x,e_{n})(e_{n},y). $$where \lambda_{n} > 0 for all n. All the norms are equivalent on \mathbb{C}^{N}. This can be written as$$ (x,y)_{\mbox{new}}=(Ax,y)_{\mbox{old}} $$where A is a positive definite selfadjoint matrix. In finite dimensions, this describes every possible inner-product. The inner products are in one-to-one correspondence with positive definite matrices. Because selfadjoint A can be diagonalized by an orthonormal basis of eigenvectors, (Ax,y) = \sum_{n}\lambda_{n}x_{n}y_{n}^{\star} always looks like a weighted inner product when viewed with respect to a correctly chosen orthonormal basis. If X is an infinite dimensional linear space on which two topologically equivalent Hilbert inner products are defined, say (\cdot,\cdot) and (\cdot,\cdot)_{1}, then the same thing happens. There exists a unique positive bounded selfadjoint A such that$$ (x,y)_{1}=(Ax,y),\;\;\; x,y\in X.$$But it also goes the other way:$(x,y)=(Bx,y)_{1}$where$B$is positive. You end up with$(x,y)=(Bx,y)_{1}=(ABx,y)$which gives$AB=I$. Similarly$BA=I$. The existence of such$A$and$B$comes from the Lax-Milgram Theorem, which is proved using the Riesz Representation Theorem for bounded linear functionals on a Hilbert Space. But there are bounded positive linear operators$A$on a Hilbert space$X$which are not positive definite. Such an$A$gives rise to$(x,y)_{1}=(Ax,y)$with$\|x\|_{1} \le C\|x\|$for a constant$C$, but the reverse inequality need not hold, which can lead to an incomplete$X$under$\|\cdot\|_{1}$. For example, let$X=L^{2}[0,1]$with the usual inner product. Define a new inner product by$(f,g)_{1}=\int_{0}^{1}xf(x)g(x)\,dx$. This is achieved as$(Af,g)$where$Af=xf(x)$. This space is not complete because$1/\sqrt{x}$is in the completion of$L^{2}$under the norm$\|\cdot\|_{1}$. The completion of$(X,\|\cdot\|_{1})$consists of$\frac{1}{\sqrt{x}}L^{2}[0,1]$. However, if you instead define$\|f\|_{1}^{2}=\int_{0}^{1}(x+\epsilon)|f(x)|^{2}\,dx$for some$\epsilon > 0$, then you end up with an equivalent norm on$X$. • Thanks for answering. While$l^2$(Z) is complete and separable, how to show all infinite-dimensional (separable) Hilbert spaces are$l^2$(Z) in disguise? – Bob Nov 18 '14 at 23:18 • @Bob : They're not. You can let$S$be any set, and define$l^{2}(S)$as the set of complex functions$f$on$S$for which$f$is zero except for a finite or countable number of points, and such that$\sum_{s\in S}|f(s)|^{2} < \infty$. This is a general Hilbert space. Such a Hilbert is$l^{2}(\mathbb{Z})$iff$S$is countable. The correspondence comes by finding a complete orthonormal basis$\{ e_{s}\}_{s\in S}$for the Hilbert space$\mathcal{H}$. Then$\mathcal{H}\simeq l^{2}(S)$comes from the correspondence$x\in\mathcal{H} \sim f_{x} \in l^{2}(S)$given by$f_{x}(s)=(x,e_{s})$– DisintegratingByParts Nov 18 '14 at 23:40 • thanks. I do mean some countable$Z$, I was looking at p163 of S&S real analysis. Thank you. – Bob Nov 18 '14 at 23:45 • @Bob : The cardinality of an orthonormal basis for a Hilbert space$\mathcal{H}$is unique. That is, if you have two different orthonormal bases, then the bases have the same cardinality. Then, through the$l^{2}(S)$correspondence, you find that Hilbert spaces with the same cardinality orthonormal bases are equivalent. So all Hilbert spaces are isometrically isomorphic to$l^{2}(S)$where the cardinality of$S\$ is uniquely determined to be the cardinality of a complete orthonormal basis. So dimension becomes cardinality and the spaces are isomorphic iff they have the same dimension. – DisintegratingByParts Nov 18 '14 at 23:51
• yes, I came across this before, thank you for the reminders and further remarks. – Bob Nov 18 '14 at 23:55