# Hilbert space image of basis under bicontinuous map

Let $X$ and $Y$ be separable Hilbert spaces and $T:X \to Y$ be linear continuous with linear continuous inverse $T^{-1}:Y \to X$.

If $x_n$ is a countable orthnormal basis of $X$, then can I say that $Tx_n$ is a basis for $Y$? Because for $y \in Y$, there is some $x \in X$ such that $$y = Tx = T\sum_{n}(x,x_n)x_n = \sum_{n}T(x,x_n)Tx_n$$ Am I allowed to do that with pulling the operator $T$ into the sum?

Does it follow that $Tx_n$ is orthonormal too? Thanks.

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$(Tx_n)$ would be a (Schauder) basis of $Y$. It need not be orthonormal. To see why the latter is true, one may use the fact: if $X=Y$, and $(z_i)$ are such that $\sum\Vert x_i-z_i\Vert$ is sufficiently small, then $(z_i)$ is a basis of $X$ and the map defined by $x_i\mapsto z_i$ is an isomorphism. Of course, one may "peturb" $(x_i)$ so that the resulting sequence $(z_i)$ is no longer orthogonal. – David Mitra Jan 20 '13 at 12:41
Thanks @DavidMitra. Is my "proof" correct though? The wiki says that the order matter and that the sum does not converge unconditionally, so I am unsure. – george.s Jan 20 '13 at 13:47
You don't want "$T(x,x_n)$" ($(x,x_n)$ is a scalar), just $(x,x_n)$. Since $T$ is continuous, you do have $y=\sum(x,x_n) Tx_n$. You also need to show that this representation is unique. To do that, use the fact that $T^{-1}$ is bounded. What Wiki are you looking at? If $T$ is an isomorphism, then it will preserve unconditionality of sums. In particular, $(T(x_i))$ will be an unconditional basis of $Y$. – David Mitra Jan 20 '13 at 14:17
Thanks. I was looking at the wiki for Schauder basis. I really have no idea about this unconditionality of sums thing, I guess I have to read up on it. – george.s Jan 20 '13 at 14:20

The right hand side of your displayed equation should be $\sum(x,x_n)T(x_n)$. With this correction, your argument shows that any $y\in Y$ has a representation $y=\sum\alpha_i Tx_i$. To show that $(Tx_i)$ is, in fact, a (Schauder) basis of $Y$, you need to show that this representation is unique. So here, you have to prove that if $y=\sum\alpha_iTx_i$, then necessarily, $\alpha_i=(x,x_i)$ for every $i$. This is easily done using the fact that $T^{-1}$ is continuous and the fact that $(x_i)$ is a basis of $X$.
In fact, you can show considerably more: not only is $(Tx_i)$ a basis of $Y$; but also that the basis $(Tx_i)$ is equivalent to the basis $(x_i)$. That is, $\sum\alpha_i x_i$ converges if and only if $\sum\alpha_iTx_i$ converges. It follows then that $(Tx_i)$ is an unconditional basis of $Y$ and that $\sum\alpha_iTx_i$ converges if and only if $(\alpha_i)\in\ell_2$. A normalized basis satisfying the latter condition is called a Riez basis. (Incidentally, one can show, using the parallelogram law, that a normalized basis of a separable Hilbert space is unconditional if and only if it is a Riez basis.)
So, $(Tx_i)$ shares one important property that orthonormal bases possess. But, $(Tx_i)$ is not necessarily orthonormal or even orthogonal. This follows from the fact that if one "peturbs" a basis $(x_n)$ by a sufficiently small amount, it remains a basis equivalent to $(x_n)$. More precisely: if $(x_n)$ is a normalized basis of $X$ with basis constant $K$ and if $(y_n)$ is a sequence from $X$ with $\sum\Vert x_n-y_n\Vert\le 1/2K$, then $(y_n)$ is a basis of $X$ which is equivalent to $(x_n)$ (c.f. Lindenstrauss and Tzafriri, Classical Banach Spaces I, proposition 1.a.9.).