What is the difference between a Hilbert space and an Inner Product space? They both seem to be defined as simply a vector space equipped with an inner product.
Also can a metric always be defined by an inner product?
[Math] Hilbert vs Inner Product Space
hilbert-spacesinner-productsmetric-spacesvector-spaces
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[Math] Inequivalent norms (given by different inner products) on infinite dimensional Hilbert space.
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 $X$.
- Yes, if $H$ is infinite-dimensional, it will have infinite-dimensional subspaces.
- Asserting that a subset $S$ of $H$ is dense means that $\overline S=H$. So, the only subset of $H$ which is both closed and dense if $H$ itself.
- Yes, $\mathscr S$ inherits an inner product from $H$, but nothing prevents you from defining a new one.
Best Answer
If you have a vector space $X$ with an inner product $\langle \cdot, \cdot \rangle$, this defines a norm $\|\cdot\|$ by $\|x\|=\sqrt{\langle x, x\rangle}$ (it is a good exercise to prove that this is in fact a norm). Similarly, this defines a metric, $d(x,y)=\|x-y\|$ (it is again a good exercise to prove that this is in fact a metric). This is the case for any inner product space, so yes, an inner product always defines a metric. However, not every metric is defined by an inner product!
A sequence of elements $\{x_n\}$ in $X$ is called a Cauchy sequence if $\|x_n-x_m\|\to0$ as $n,m\to\infty$. An inner product space $X$ is called a Hilbert space if it is a complete metric space, i.e. if $\{x_n\}$ is a Cauchy sequence in $X$, then there exists $x\in X$ with $\|x-x_n\|\to0$ as $n\to\infty$.