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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?

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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
1  
@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
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1 Answer

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

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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
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