- #1

futurebird

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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.