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Let $A$ be a Hilbert Matrix, ie $a_{ij}=(1+i+j)^{-1}$. Would you help me to prove that $\Vert A\Vert\leq \pi$ where the norm is standard euclidean norm.

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Could you tell us with which norm you are working with? Thanks! –  Davide Giraudo Oct 1 '12 at 15:30
    
usual Euclidean norm –  beginner Oct 1 '12 at 15:31

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For euclidean norm the norm of matrix $A$ may be defined as follows $$ \Vert A\Vert=\sup\{\langle Ax,y\rangle:\quad x,y\in\mathbb{R}^n,\quad\Vert x\Vert_2\leq 1,\quad\Vert y\Vert_2\leq 1\} $$ Hence (why?) it is enough to prove that for all positive reals $\{x_i\}_{i=1}^n$, $\{y_j\}_{j=1}^n$ holds $$ \sum\limits_{i=1}^n\sum\limits_{j=1}^n\frac{x_i y_j}{i+j+1}\leq \pi\left(\sum\limits_{i=1}^n x_i^2\right)^{1/2}\left(\sum\limits_{j=1}^ny_j^2\right)^{1/2} $$ This is so called Hilbert's inequality. You can find its proof on page 155 of the book The Cauchy-Schwarz Master Class by J. Michael Steele.

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