 Solution

ID: past-exam-of-the-analysis-course-of-the-university-of-oxford/2009/b/6/ii/solution

⟨ f, g ⟩ = \int f \overset{g}{ˉ}

. On the positive half-interval, reversing the triangular integration order gives

\int_{0}^{π} (\int_{0}^{t} f (s) d s) \overline{g (t)} d t = \int_{0}^{π} f (s) \overline{(\int_{s}^{π} g (t) d t)} d s .

On the negative half-interval the corresponding calculation gives

\int_{- π}^{0} (- \int_{t}^{0} f (s) d s) \overline{g (t)} d t = \int_{- π}^{0} f (s) \overline{(- \int_{- π}^{s} g (t) d t)} d s .

These exchanges are justified by the Fubini theorem, since on a finite interval

L^{2}

functions belong to

L^{1}

and the absolute double integral is at most

∥ f ∥_{1} ∥ g ∥_{1}

. Therefore

(T^{*} g) (s) = ⎩ ⎨ ⎧ \int_{s}^{π} g (t) d t, - \int_{- π}^{s} g (t) d t, s > 0, s < 0.

Adding

T g

gives

\int_{0}^{π} g

on the positive half and

- \int_{- π}^{0} g

on the negative half, exactly the prescribed

Ug

. Values at zero do not matter in

L^{2}

. Consequently

T + T^{*} = U .

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