Past exam of the functional analysis II course of the University of Oxford / 2023 / 3

Use the convention that the inner product is linear in its first argument. By the defining identity for the adjoint operator,For the last implication choose $y=Tx$. This proves the first image-kernel orthogonality for an adjoint identity. Taking orthogonal complements and using the double orthogonal complement theorem for the vector subspace $\mathrm{Im}{T}^{\ast }$ gives

Put $A=\lambda I-T$. Since $T$ is a self-adjoint operator, $A^{\ast }=\overline{\lambda }I-T$. Applying the preceding image-kernel orthogonality for an adjoint identities to $A^{\ast }$ givesIf $\lambda$ is real, $A^{\ast }=A$ and $\ker A=\{0\}$ because $\lambda$ is not an eigenvalue. If $\lambda$ is nonreal and $A^{\ast }x=0$, then $Tx=\overline{\lambda }x$. The scalar $⟨Tx,x⟩$ is real by self-adjointness, while it equals $\overline{\lambda }\Vert x{\Vert }^2$. Its imaginary part forces $x=0$. Thus $\ker A^{\ast }=\{0\}$ in both cases, andThe conjugation on $\lambda$ in the adjoint operator is essential when $\lambda$ is nonreal.

Again put $A=\lambda I-T$. Surjectivity gives, by image-kernel orthogonality for an adjoint,If $\lambda$ is real, $A=A^{\ast }$, so $\ker A=\{0\}$. If $\lambda$ is nonreal and $Ax=0$, then $Tx=\lambda x$; reality of $⟨Tx,x⟩=\lambda \Vert x{\Vert }^2$ again implies $x=0$. Thus $A$ is injective as well as surjective, without needing to assume separately that $\lambda$ is not an eigenvalue.

Since $A$ is a bounded linear operator on the Banach space $X$, the bounded inverse theorem makes $A^{-1}$ bounded. ThereforeThis is exactly membership in the resolvent set of an operator.

Here the intended space must consist of bounded linear operators, namely Hilbert-Schmidt operators. The printed restriction to “all linear mappings” is insufficient. To see this, choose $v\in X$ outside the algebraic linear span of $(e_k)$, for example $v={\sum }_{k\ge 1}{2}^{-k}{e}_k$. Extend the linearly independent family $\{e_k:k\ge 1\}\cup \{v\}$ to a Hamel basis. Define a linear functional $\varphi$ on this Hamel basis by $\varphi (v)=1$ and zero on every other basis vector, and extend it linearly. Then the everywhere-defined linear operator $Rx=\varphi (x)e_1$ satisfiesIts printed column sum is zero although $R\ne 0$. More explicitly, $v_N={\sum }_{k=1}^N{2}^{-k}{e}_k\to v$ but $R(v-v_N)=e_1$ for every $N$. Thus $R$ is discontinuous; it has no everywhere-defined adjoint operator of the bounded kind used in the question. This proves finite column sum does not imply boundedness. We use $T\in \mathcal{B}(X)$ in the intended argument below.

By the Parseval identity for a Hilbertian basis for $(e_j)$, followed by the adjoint operator identity and then the Parseval identity for a Hilbertian basis for $(h_k)$,The interchange is valid by Tonelli's theorem for nonnegative double series, even before finiteness is known. Repeating with $h_k=e_k$ shows that the right side also equals ${\sum }_k\Vert T{e}_k{\Vert }^2$. HenceIn particular the Hilbert-Schmidt norm is independent of the Hilbertian basis, and $\Vert T^{\ast }{\Vert }_2=\Vert T{\Vert }_2$.

As in the preceding part, take $T$ to be a bounded linear operator. For $x\in X$, the Hilbertian basis expansion and continuity of $T$ giveThe triangle inequality, the Cauchy-Schwarz inequality for finite sums, and the Parseval identity for a Hilbertian basis givePassing to the limit and taking the supremum over the unit ball provesThus the Hilbert-Schmidt norm controls the operator norm. Without the boundedness interpretation, the nonzero $R$ constructed in the preceding part has zero printed column sum and makes the requested inequality false.

Use the intended space of Hilbert-Schmidt operators, with the given Hilbert-Schmidt inner product linear in its first argument. It remains to prove completeness in its Hilbert-Schmidt norm. Let $(T_n)$ be a Cauchy sequence in that norm. The bound $\Vert T_n-T_m\Vert \le \Vert T_n-T_m{\Vert }_2$ makes it a Cauchy sequence in operator norm. The space $\mathcal{B}(X)$ is a Banach space, so $T_n\to T$ in operator norm for some bounded linear operator $T$.

For completeness, that last standard fact follows by taking $Tx={\lim }_n{T}_{n}x$ for each $x$: the Cauchy sequence $(T_n)$ is uniformly bounded in operator norm, so its limit is linear and bounded, and the uniform Cauchy estimate on the unit ball gives operator norm convergence.

Given $\epsilon >0$, choose $N$ so that $\Vert T_n-T_m{\Vert }_2<\epsilon$ for $n,m\ge N$. Fix $n\ge N$ and any positive integer $K$. Since $T_m{e}_k\to T{e}_k$ for each $k$, passage to the limit in a finite sum givesTaking the supremum over $K$ gives ${\sum }_k\Vert (T_n-T)e_k{\Vert }^2\le {\epsilon }^2$. Also,Thus $T\in S_2(X)$, and $\Vert T_n-T{\Vert }_2\le \epsilon$ for $n\ge N$. This proves Hilbert-Schmidt operators form a Hilbert space:Under the literal unrestricted “linear mappings” definition, the nonzero $R$ from part (i) satisfies $⟨R,R{⟩}_{S_2(X)}=0$. The form would fail positive definiteness and would not be an inner product; the boundedness correction is needed here too.

Write $u(t)=e^t$ and $v(t)=e^{-t}$. Both lie in the Hilbert space $L^2(-\pi ,\pi )$, and the integral operator factors as the rank-one operatorThe Cauchy-Schwarz inequality gives $\Vert Tf\Vert \le \Vert u\Vert \Vert v\Vert \Vert f\Vert$, and equality is attained at $f=v/\Vert v\Vert$. Hence the operator norm is $\Vert T\Vert =\Vert u\Vert \Vert v\Vert$. Direct integration givesFor any Hilbertian basis $(h_k)$, the Parseval identity for a Hilbertian basis givesThis is the general formula for the norms of a rank-one operator. In particular $T$ is a Hilbert-Schmidt operator, andThis computation needs no explicit Fourier basis. The printed optional basis formula omits the variable $t$: the intended Hilbertian basis is $e_n(t)=(2\pi {)}^{-1/2}{e}^{int}$, $n\in \mathbb{Z}$. Taken literally, the printed expressions are constant functions and cannot form an orthonormal basis of $L^2(-\pi ,\pi )$.