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We know that a collection of vectors $\{x_{k}\}$ in a Hilbert space called Riesz basis if it is an image of orthonormal for H under invertible linear transformation. How to prove that there is constants $A,B$ such that for all $x\in H$ $$ A||x||^2\leq\sum_{k}\langle x,x_k \rangle^2\leq B||x||^2? $$

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I think what you exact mean is that there exist a bounded linear operator $L$ on $H$ with bounded inverse and an orthonormal basis $\{e_k\}$ of $H$, such that $Le_k=x_k$ for every $k$. If so, then $ \langle x,x_k \rangle = \langle x,Le_k \rangle = \langle L^*x,e_k \rangle $, where $L^*$ is the adjoint operator of $L$. Therefore, $\sum_k| \langle x,x_k \rangle |^2=\sum_k| \langle L^*x,e_k \rangle |^2=\|L^*x\|^2$. Since $L$ is bounded with a bounded inverse, $L^*$ is also bounded with a bounded inverse. The conclusion follows.

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Thank you, could you also answer-is the converse true? That is, if inequality holds for basis $x_{k}$ then there is bounded linear operator and orthonormal basis such that $x_{k}$ is an image of this basis under this operator? –  StudentMath Oct 28 '12 at 12:26
    
@StudentMath, I think it is false. Let $\{e_k\}$ be an orthonormal basis of $H$. For every $k≥0$ and $i≥1$, let $x_{2^k}=e_{2^k}$ and $x_{2^k(2i+1)}=e_{2^k(2i-1)}+\frac{e_{2^k(2i+1)}}{k+2}$. It is easy to check that $\{x_k\}$ is a basis and it satisfies your inequality. However, as $k\to\infty$, $x_{3\cdot 2^k}−x_{2^k}\to 0$, which implies that $\{x_k\}$ cannot be the image of an orthonormal basis under any invertible bounded linear operator. –  23rd Oct 28 '12 at 13:31

We assume the Hilbert space separable (otherwise, take $x$ orthogonal to all the $x_n$ and different from $0$). Let $T\colon H\to H$ linear invertible and $\{e_n\}$ an orthonormal basis such that $Te_n=x_n$ for all $n$. We have by Bessel-Parseval that $$\sum_k\langle x,x_k\rangle^2=\sum_k\langle x,Te_k\rangle^2=\sum_k\langle T^*x,e_k\rangle^2= \lVert T^*x\rVert^2\leq \lVert T^*\rVert^2\lVert x\rVert^2.$$ As $T$ is invertible, so is $T^*$, so $\lVert T^*x\rVert$ is below bounded by an universal constant (which doesn't depend on $x$) when $x$ is in the unit ball.

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But how to show that such linear invertible $T$ and orthonormal $\{e_n\}$ exists? –  StudentMath Oct 28 '12 at 13:54
    
I don't understand: isn't it assumed in the OP that $T$ and $\{e_n\}$ exists (then we have to find the constants $A$ and $B$). –  Davide Giraudo Oct 28 '12 at 13:55
    
Oh sorry, I was confused, I thought you were answering to my second question that I gave to richard -whether converse it true? –  StudentMath Oct 28 '12 at 13:59

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