functional-analysis « Bedtime mathematics

A bound for matrix norm

November 16, 2009 mymbl Leave a comment

We wish to better the bounds for the norm of a matrix ${M = (k(i,j))}$ with scalar entries when it defines a linear map with from ${l^p}$ to ${l^p}$ , that is the series ,

$\displaystyle Mx(i) = \sum_{j=1}^\infty k_{i,j} x(j)$

is convergent for each ${i}$ and ${Mx \in l^p}$ . The standard bounds for matrix norm given in the text for ${p \geq 2}$ , are ${\alpha_1^{1/p} \alpha_\infty^{1/q}}$ and ${\beta_p}$ , for ${l^p}$ spaces. We give yet another bound using the power mean inequality:

1. Power Mean Inequality :

Consider positive weights ${p_k}$ , ${k = 1, 2, \ldots, }$ which have total mass ${p_1 + p_2 + \ldots= 1}$ , then for nonnegative real numbers ${x_k}$ , ${k = 1, 2,\ldots }$ and for all ${-\infty < s < t < \infty}$ one has

$\displaystyle (\sum_{k=1}^\infty p_k x_k^s)^{1/s} \leq (\sum_{k=1}^\infty p_k x_k^t)^{1/t} \ \ \ \ \ (1)$

2. Bound for matrix norm

Consider,

$\displaystyle \begin{array}{rcl} \left | \left | Mx \right | \right |_p^p &= \sum_{i=1}^\infty \left|\sum_{j = 1}^\infty k_{i,j} x(j) \right |^p \\ &\leq \sum_{i=1}^\infty \left( \sum_{j=1}^\infty \left |k_{i,j} x(j) \right | \right )^p \\ &= \sum_{i=1,r_i \neq 0}^\infty r_i^p \left ( \sum_{j=1}^\infty \left |\frac{k_{i,j}}{r_i} x(j) \right | \right )^p \end{array}$

where for some fixed ${s > 0}$ and ${s <= 1}$ , ${r_i = \sum_{j=1}^\infty |k_{i,j}^{s}|}$ .

$\displaystyle \begin{array}{rcl} \left | \left | Mx \right | \right |_p^p & \leq \sum_{i=1, r_i \neq 0}^\infty r_i^p \left ( \sum_{j=1}^\infty \frac{\left |k_{i,j}^{s} \right |}{r_i} \left|k_{i,j}\right|^{1-s}\left| x(j) \right | \right )^p \end{array}$

Now we use power mean inequality with ${\alpha=1}$ { and } ${\beta = p}$ , ${ p_k = \frac{|k_{i,j}^{s}|}{r_i}}$ { and } ${x_k = | k_{i,j}^{1-s} x(j)|}$ . So,

$\displaystyle \begin{array}{rcl} \left | \left | Mx \right | \right |_p^p &\leq \sum_{i=1,r_i\neq 0}^\infty r_i^p \sum_{j=1}^\infty \left | \frac{|k_{i,j}^{s}|}{r_i} k_{i,j}^{p(1-s)}x(j)^p \right | \\ &= \sum_{i=1}^\infty\sum_{j=1}^\infty r_i^{p-1}\left | { k_{i,j}^{s+p(1-s)}} \right | \left|x(j)\right|^p \\ & = \sum_{j=1}^\infty \left|x(j)\right|^p \sum_{i=1}^\infty r_i^{p-1}\left | { k_{i,j}^{s+p(1-s)}} \right | \\ \end{array}$

Now if we take,

$\displaystyle \gamma_s = \sup_j \left \{ \sum_{i=1}^\infty r_i^{p-1} \left| k_{i,j}^{s + p(1-s)}\right| \right \}^{1/p}$

Then we have ${Mx \leq \gamma_s \left| \left | x \right |\right |_p}$ . So we have infinite number of bounds for the norm as we vary ${s}$ among positive reals (of course, not all values of ${s}$ might be useful )

Example : Consider the matrix

$\displaystyle A = \begin{pmatrix} 1 & 1/2^{3} & 1/3^3 & \ldots \\ 1/2 & 1 & 0 & \ldots \\ 1/3 & 0 & 1 & \ldots \\ \vdots & & & \vdots \\ \end{pmatrix}$

For this matrix ${\alpha_1 = \infty}$ and ${\beta_2}$ is also infinity, however ${\gamma}$ is finite with ${s =1/2}$ . This is because ${\sup_i r_i = \sum_n (1/n^3)^{1/2} < \infty }$ and ${\sup_{i} \sum_{j=1}^n k_{i,j}^{(2+1)/2} =\sum_{j=1}^n 1/n^{3/2} < \infty}$ . Thus ${\gamma_{2} \leq \sup_i r_i \sup_i \sum_{j=1}^n k_{i,j}^{3/2} < \infty}$ .

You can further see that ${\gamma_s}$ is a better bound than ${\alpha_\infty^{1/q}\alpha_1^{1/p}}$ always, since if we take ${s = 1}$ and ${r = \infty}$ , we have

$\displaystyle \sup_j \sum_{i=1}^\infty r_i^{p-1} \left|k_{i,j}^{s + p(1-s)}\right| = \sup_i {r_i^{p/q}} \sup_j \sum_{i=1}^\infty \left |k_{i,j}\right| \leq \alpha_\infty^{p/q} \alpha_1,$

where we have used the fact that ${p-1 = p/q}$ .

Categories: Uncategorized Tags: functional-analysis, matrix, norm

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