Caracterizing a subspace of L^2

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Homework Help Overview

The discussion revolves around characterizing a subspace of L², specifically focusing on the subspace M consisting of functions with a vanishing mean over the interval [0,1]. The original poster is investigating the dimension of the orthogonal complement M^{\perp} and is questioning whether it consists solely of constant functions.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to prove that M^{\perp} contains only constant functions, raising questions about the nature of functions in this orthogonal complement.

Discussion Status

Some participants express curiosity about the conclusion regarding M^{\perp}, with one participant suggesting that it must consist of constant functions based on the definition of vanishing mean. There is an acknowledgment of the original poster's struggle with the proof, but no consensus is reached.

Contextual Notes

Participants mention the constraints of the problem, including the focus on functions defined almost everywhere and the implications of the L² norm. There is a reference to the original poster's busy schedule with finals, which may affect participation.

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[SOLVED] Caracterizing a subspace of L^2

Homework Statement



Call M the subspace of L²([0,1]) consisting of all functions of vanishing mean. I.e., [itex]u\in M \Rightarrow \int_0^1u(s)ds=0[/itex].

I am trying to find the dimension of the orthogonal of M,

[tex]M^{\perp}=\{x\in L^2([0,1]):\int_0^1x(s)u(s)ds=0 \ \forall u\in M\}[/tex]

I would be surprised if [itex]M^{\perp}[/itex] was anything other than the constant functions, but my attempts at a proof have been unsuccessful.

Any idea how to go at this?
 
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Last edited:
Was it the (a.e.) constant functions in the end?

Sorry I couldn't help -- been busy with studying for my finals!
 
morphism said:
Was it the (a.e.) constant functions in the end?

Sorry I couldn't help -- been busy with studying for my finals!

How could it be anything else? The set of vanishing mean functions is basically defined as the set of all u(x) such that <u(x),1>=0. L^2 is a norm up to a.e.
 
Well, yeah... :-p
 
morphism said:
Well, yeah... :-p

Nice to talk to an erudite and sophisticated gentleman for a change. Instead of a bonehead. :)
 

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