- #1
futurebird
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My current understanding of the Riesz representation theorem is that it is useful since it tells you what all bounded linear functionals on Lp look like. They look like the integral of fg where g is some function in Lq. So, I was trying to think of an example of a bounded linear functional on an Lp space (1 <= p < infinity) that didn't look like the integral of fg where g is some function in Lq, and then find it's "riesz representation" --
I had a hard time thinking of a bounded linear functional, mostly because I'm still unsteady about the concept of "bounded" in this context. But I'm pretty sure that for f(x) in L2, T(f(x)) = f(1) would be a bounded linear functional on L2. T(af(x)+bh(x)) = af(1)+bh(1)= T(af(x)) +T(bh(x)), so it is linear. Now is it bounded? That is where I'm stuck.
||T|| = sup |f(1)|/||f||2
I don't know what to make of the numerator. It's smaller than the denominator? So ||T|| <= 1 and it is bounded (?) That means there must be a function g in L2 (I choose L2 since conjugate indicies are confusing otherwise) with T(f(x)) = integral [fg]. OK. What is g? That is where I'm stuck again.
I had a hard time thinking of a bounded linear functional, mostly because I'm still unsteady about the concept of "bounded" in this context. But I'm pretty sure that for f(x) in L2, T(f(x)) = f(1) would be a bounded linear functional on L2. T(af(x)+bh(x)) = af(1)+bh(1)= T(af(x)) +T(bh(x)), so it is linear. Now is it bounded? That is where I'm stuck.
||T|| = sup |f(1)|/||f||2
I don't know what to make of the numerator. It's smaller than the denominator? So ||T|| <= 1 and it is bounded (?) That means there must be a function g in L2 (I choose L2 since conjugate indicies are confusing otherwise) with T(f(x)) = integral [fg]. OK. What is g? That is where I'm stuck again.