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Marianna Stone

Marianna Stone

Answered question

2022-05-27

Let ( X , A , μ ) be a finite measure space. Fix 1 < p , q < with 1 p + 1 q = 1. Let { f n } be a sequence in L p ( μ ) such that sup f n p = M < . Suppose for every g L q ( μ ) we have lim n f n g d μ converges.
Prove that f n weakly converges to some f L p ( μ ).
I have shown that ϕ ( g ) = lim n f n g d μ defines a linear functional ϕ ( L q ( μ ) ) . I think f is the element in Lp(μ) corresponding to ϕ ( L q ( μ ) ) by the Riezs' theorem isomorphism, but how do I prove this? How do I find this f?

Answer & Explanation

Hugo Bruce

Hugo Bruce

Beginner2022-05-28Added 10 answers

If ϕ is a bounded linear functional on L q ( μ ) then there exists f L p ( μ ) with ϕ ( g ) = f g d μ . You don't have to find it; Riesz did that for you.
Under your definition of ϕ this means
lim n f n g d μ = f g d μ
for all g L q ( μ ), verifying weak convergence.
Carly Roy

Carly Roy

Beginner2022-05-29Added 4 answers

Clearly, ϕ ( g ) = lim X f n g d μ is well-defined and linear. Moreover,
| ϕ ( g ) | M g q
so ϕ M.
By Riesz Representation, there exists a unique f L P such that ϕ ( g ) = X f g d μ. But this means lim X ( f n f ) g d μ = 0.

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