Proving the Bounded Linearity of A in l^{p} Space

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Homework Help Overview

The discussion revolves around proving that a matrix operator \( A \) defines a bounded linear functional on \( l^{p} \) space for \( 1 \leq p \leq \infty \). The original poster questions the continuity of the operator and its implications for boundedness.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants explore the definitions of continuity and boundedness in the context of linear transformations. There is a discussion about whether the continuity of \( A \) is obvious and what it means for \( A \) to be a functional versus a transformation.

Discussion Status

The conversation is ongoing, with participants questioning the clarity of definitions and the implications of continuity. Some suggest using specific characterizations to approach the problem, while others express uncertainty about the continuity of \( A \) and seek further clarification.

Contextual Notes

There is a mention of needing to find a relationship between \( \epsilon \) and \( \delta \) to establish continuity, as well as considerations for different cases based on the value of \( p \). The discussion also hints at potential complexities in proving continuity when mapping between different \( l^{p} \) spaces.

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Homework Statement


Let 1[tex]\leq[/tex]p[tex]\leq[/tex][tex]\infty[/tex] and suppose ([tex]\alpha_{ij}[/tex] is a matrix such that (Af)(i)=[tex]\sum^{\infty}_{j=1}[/tex][tex]\alpha[/tex][tex]_{ij}[/tex]f(j) defines an element Af of [tex]l^{p}[/tex] for every f in [tex]l^{p}[/tex]. Show that A is a bounded linear functional on [tex]l^{p}[/tex]


Homework Equations


Isn't this obvious if we apply theorem that says following are equivalent for A:X-->X a linear transformation on normed space?
(a)A is bounded linear functional
(b)A is continuous at some point
(c)There is a positive constant c such that ||Ax||[tex]\leq[/tex]c||x|| for all x in X.


The Attempt at a Solution


Isn't the contiunuity of f obvious? So by the theorem, I think A is bounded linear functional on [tex]l^{p}[/tex]. Could you guys correct me if I am wrong? Or if I am right could you just say it is right?

Thanks
 
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Er, the continuity of f? f is an element of l^p, not a function. Also, how is A a functional on l^p? To me a functional on X is a mapping into the scalar field, and I believe this is standard terminology; A looks like a mapping from l^p to l^p (namely f [itex]\mapsto[/itex] Af).

Edit:
Looking at what you said again, I think you might've meant to say "continuity at f". If so, what f?
 
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Maybe then I should say that A is bounded linear transformation?
But still isn't continuity of A obvious by construction?
 
Why is it obvious? Can you post your proof?
 
Sorry. It's not obvious. It seems continuous though.
I know for given epsilon > 0, I need to find delta>0 such that ||f||<delta implies
||Af||<epsilon. Hmm.. How can I find such delta? Or use cauchy sequence?
 
It's not that it's a difficult problem, but I just wouldn't say that the continuity of A is obvious. Anyway, I would use characterization (c) in the first post. Namely, try to compute ||Ax||_p given an arbitrary sequence x=(x_1, x_2, ...) in l^p. It will help to split this into three cases, depending on whether p=1, 1<p<[itex]\infty[/itex], or p=[itex]\infty[/itex]. Holder's inequality (and Cauchy-Schwarz) will be helpful in the first two cases.

If you need any more hints, post back.

Also, a much more interesting (and more difficult!) problem is to prove that the same map, x [itex]\mapsto[/itex] Ax, is continuous if it maps sequence in l^p to sequences in l^p', where 1 < p, p' < [itex]\infty[/itex]. You might want to try to tackle this one if you're looking for a challenge.
 
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