How to Estimate the Operator Norm ||A||_2 for a Difference Operator?

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SUMMARY

The discussion focuses on estimating the operator norm ||A||_2 for a difference operator A. The user proposes calculating a non-trivial upper bound using the maximum of the L1, L2, and L∞ norms, arriving at ||A||_1 = 4 and ||A||_∞ = 2. The user estimates ||A||_2 as 4√2 mn based on the absolute row and column sums. The conclusion suggests that the maximum bound b is equal to ||A||_2 = 4√2 mn, while also emphasizing the need to investigate the application of A on any vector x to ensure the operator's boundedness.

PREREQUISITES
  • Understanding of operator norms, specifically L1, L2, and L∞ norms.
  • Familiarity with difference operators in linear algebra.
  • Knowledge of inner product spaces and their properties.
  • Basic concepts of bounded operators in functional analysis.
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  • Research the properties of difference operators in linear algebra.
  • Learn about the definitions and applications of operator norms in functional analysis.
  • Study the implications of bounded vs. unbounded operators.
  • Explore the use of the operator norm in estimating solutions to linear equations.
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Mathematicians, students of linear algebra, and researchers working with operator theory and functional analysis will benefit from this discussion.

Max Fleiss
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Greetings everyone!

I have a set of tasks I need to solve using using operator norms, inner product... and have some problems with the task in the attachment. I would really appreciate your help and advice.

This is what I have been thinking about so far:
I have to calculate a non trivial upper bound, so maybe it could be done by:
b=max( ||A||_1,||A||_2,||A||_\infty )

Since A is a difference operator I estimated the following:
||A||_1= 4
||A||_\infty= 2
But how can I estimate ||A||_2=?
If I know that abs row sum is 2 (besides 0 there appears only one 1 and one -1 in the rows) and abs column sum is 4 (it is two times the size of row length dim(A)=2mn x mn). Can I estimate ||A||_2 by:
||A||_2=\sqrt{rows^2+columns^2 }=\sqrt{(2 \cdot 2mn)^2+(4 \cdot mn)^2}=4 \sqrt{(mn)^2+(mn)^2}=4 \sqrt{2} \sqrt{m^2n^2} since mn are positive ||A||_2=4 \sqrt{2} mn
So I would say b=max(L_1,L_2,L_\infty)=L_2=4 \sqrt{2} mn

Is my conclusion, approximation of a non trivial upper bound b right?

Thank you in advance for your help!
 

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It is not clear why you use different norms here. In any case, you need to investigate ##||Ax||## for any ##x##. Application of ##A## on ##x## should give you an idea for un upper bound, i.e. you need to use the definition of ##A## somewhere, since not all operators are bounded.
 
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