Past exam of the functional analysis I course of the University of Oxford / 2023 / 2 / d / ii

Solution

When

p = \infty

p^{'} = 1

and

T_{h} f = h^{- 1} \int_{- h}^{h} f (x) d x

f (x) = {sin (lo g ∣ x ∣), 0, x \neq = 0, x = 0.

It represents an element of the Lp space

L^{\infty} ((- 1, 1))

, with L-infinity norm at most one. Since

\int_{0}^{h} sin (lo g x) d x = \frac{h}{2} (sin (lo g h) - cos (lo g h)),

where the lower endpoint contributes zero, its symmetric Lebesgue integral gives

T_{h} f = sin (lo g h) - cos (lo g h) .

For

h_{n} = e^{- 2 πn}

and

h_{n} = e^{- (2 πn + π)}

, both sequences tend to zero through

(0, 1)

, but

T_{h_{n}} f = - 1

and

T_{h_{n}} f = 1

. Therefore

f \in / Y, Y ⊊ L^{\infty} ((- 1, 1)) .

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