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

Let H be a Hilbert space .Is there always a non orthogonal Riesz basis $D$ on it such that following holds?

$$\sup_{g\in D }\sum_{g'\in D,g'\not=g}|\langle g,g'\rangle|<1/3 $$

And is there Riesz such that the inequality does not hold?

share|cite|improve this question
There is only one Riesz, and he satisfies $1880\le\text{Riesz}\le1956$. – joriki Oct 28 '12 at 15:08
Thanks, could you give me a link to proof or sketch of proof? – StudentMath Oct 28 '12 at 15:10
I wanted to construct it using orthonormal basis of H – StudentMath Oct 28 '12 at 15:12
But for an orthonormal basis you even have $\sup_{g\in D }\sum_{g'\in D,g'\not=g}|\langle g,g'\rangle|=0$? – joriki Oct 28 '12 at 15:14
@joriki: there's at least a second Riesz (younger brother) and he satisfies $1886 \leq \text{M. Riesz} \leq 1969$. :-) – commenter Oct 28 '12 at 16:09
up vote 1 down vote accepted

Fix an orthonormal basis $\{e_j\}_{j\in J}$. For each $j\in J$ we denote by $j+1$ some fixed element different from $j$ (it can actually be $j+1$ if $J=\mathbb N$).

Define $$ f_j=e_j+\frac18\,e_{j+1},\ \ j\in J. $$ The set $\{f_j\}$ clearly spans $H$. Also, $$ \left\|\sum_jc_jf_j\right\|^2=\sum_{j,k}\langle c_je_j+\frac{c_j}8e_{j+1},c_ke_k+\frac{c_k}8e_{k+1}\rangle=\sum_j|c_j|^2+\frac18\,\sum_j|c_j|^2+2\text{Re}\,\frac18\,\sum_jc_j\overline{c_{j+1}}. $$ Note that, by Cauchy-Schwarz, $|\sum_jc_j\overline{c_{j+1}}|\leq\sum_j|c_j|^2$. Then $$ \left\|\sum_jc_jf_j\right\|^2\leq\frac98\sum_j|c_j|^2+\frac14\,\sum_j|c_j|^2=\frac{11}8\sum_j|c_j|^2. $$ Also, $$ \left\|\sum_jc_jf_j\right\|^2\geq\frac98\sum_j|c_j|^2-\frac14\,\sum_j|c_j|^2=\frac{7}8\sum_j|c_j|^2. $$ All this shows that $\{f_j\}$ is a Riesz basis.

Finally, for any $j\in J$, $$ \sum_{k\ne j}|\langle f_k,f_j\rangle|=\frac18|\langle f_{j+1},f_j\rangle|+\frac18|\langle f_j,f_{j+1}\rangle|=\frac18+\frac18<\frac13. $$

share|cite|improve this answer
Thanks, I just did not understand the last inequality, where did the sum go after equality sign? – StudentMath Nov 1 '12 at 9:29
There were a few typos that I corrected. But there is no sum, only two of the inner products are nonzero. – Martin Argerami Nov 1 '12 at 13:09
ok, I got it, thank you very much – StudentMath Nov 1 '12 at 18:59

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.