Past exam of the analysis course of the University of Oxford / 2009 / B / 6

The Hilbert projection theorem states that every vector $x$ has a unique decomposition $x=m+n$ relative to a closed linear subspace $M$, with $m\in M$ and $n\in M^{\perp }$. The vector $m$ is the unique nearest point in $M$.

The Riesz representation theorem states that every bounded complex-linear functional $F$ on a Hilbert space has a unique representation $F(x)=⟨x,y⟩$, with $\Vert F\Vert =\Vert y\Vert$, using the convention of linearity in the first argument. If $F=0$, use $y=0$. Otherwise $M=\ker F$ is closed. Project a vector outside $M$ to obtain $z\in M^{\perp }$ with $F(z)\ne 0$, and rescale it so $F(z)=1$. Then $x-F(x)z\in M$, soThis gives the representation. Uniqueness follows because a vector orthogonal to every $x$ is zero. The Cauchy-Schwarz inequality gives $\Vert F\Vert \le \Vert y\Vert$, and evaluation at $y/\Vert y\Vert$ gives equality when $y\ne 0$.

For the adjoint, fix $v$. The functional $u\mapsto ⟨Tu,v⟩$ is bounded with norm at most $\Vert T\Vert \Vert v\Vert$. By the Riesz representation theorem, there is a unique vector $T^{\ast }v$ satisfyingFor scalars $a,b$, conjugate-linearity in the second argument shows that $T^{\ast }(av+bw)=a{T}^{\ast }v+b{T}^{\ast }w$, by uniqueness of the representing vector. The same norm estimate gives $\Vert T^{\ast }v\Vert \le \Vert T\Vert \Vert v\Vert$. Thus $T^{\ast }$ is a unique bounded linear operator, the adjoint operator.

For the adjoint of an integration operator anchored at zero, use the inner product $⟨f,g⟩=\int f\bar{g}$. On the positive half-interval, reversing the triangular integration order givesOn the negative half-interval the corresponding calculation givesThese 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 $\Vert f{\Vert }_1\Vert g{\Vert }_1$. ThereforeAdding $Tg$ gives ${\int }_0^{\pi }g$ on the positive half and $-{\int }_{-\pi }^{0}g$ on the negative half, exactly the prescribed $Ug$. Values at zero do not matter in $L^2$. Consequently

Since $f(0)=0$, the fundamental theorem of calculus gives $f=T{f}^{\prime }$, with the same oriented integral on the negative half-interval. Set $h(t)=t^2$. The adjoint calculation gives the real-valued functionChoose any value at zero. This function is bounded and hence belongs to $L^2(-\pi ,\pi ]$. Since it is real, the adjoint operator identity yieldsOne can also check this by integration by parts separately on each half-interval: $g^{\prime }=-t^2$, $g(\pm \pi )=0$, and the two possible jump terms at zero vanish because $f(0)=0$.