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

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.