Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $x\in H$ $$ A||x||^2\leq\sum_{k}\langle x,x_k \rangle^2\leq B||x||^2? $$

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.