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!
 A: 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 $X$.
