Functional analysis II course of the University of Oxford (B4.2)

In a real Hilbert space, weak convergence in a Hilbert space meansBy the Riesz representation theorem, every continuous linear functional on $X$ has the form $u\mapsto ⟨u,z⟩$. Thus this is exactly weak convergence, defined by convergence under every continuous linear functional.

Define the continuous linear functionals $f_n(z)=⟨x_n,z⟩$. For each fixed $z\in X$, weak convergence makes $f_n(z)$ a convergent, hence bounded, scalar sequence. The Uniform boundedness principle states that a pointwise bounded family of bounded linear operators from a Banach space to a normed space has uniformly bounded operator norms. Since a Hilbert space is a Banach space, it applies to $(f_n)$.

The Cauchy-Schwarz inequality gives $\Vert f_n\Vert \le \Vert x_n\Vert$, and evaluation at $z=x_n/\Vert x_n\Vert$ gives equality when $x_n\ne 0$; equality also holds at zero. Consequently

By weakly convergent sequence is bounded, $M={\sup }_n\Vert x_n{\Vert }^2$ is finite. Choose any $n_1$. Once $n_1<\cdots <n_k$ have been chosen, put $s_k={\sum }_{j=1}^k{x}_{n_j}$. Since $s_k$ is fixed, weak convergence to zero gives $⟨s_k,x_n⟩\to 0$. Choose $n_{k+1}>n_k$ with $|⟨s_k,x_{n_{k+1}}⟩|\le 1$. This gives the required one-sided inequality as well.

The useful estimate is for the squared norm. Indeed, $\Vert s_1{\Vert }^2\le M$, and the inner product expansion givesBy mathematical induction, $\Vert s_k{\Vert }^2\le Mk+2(k-1)\le (M+2)k$. HenceThis proves norm convergence of the averages, the weak Banach–Saks theorem in a Hilbert space. The printed inequality omits the square on its left side. Our stronger estimate also proves that printed bound, since $\sqrt{(M+2)k}\le (M+2)k$ for $k\ge 1$, but the printed bound alone only bounds the averages and cannot imply their norm convergence to zero.

First suppose the convex set $K$ is closed in the norm topology, and $x_n\in K$ satisfies $x_n\rightharpoonup x$. Apply the weak Banach–Saks theorem in a Hilbert space to the weakly null sequence $x_n-x$. Along the resulting subsequence,in norm. Each $z_m$ is a convex combination of points in $K$, so $z_m\in K$. Since $K$ is a closed set in the norm topology, $x\in K$.

Conversely, suppose $K$ is a sequentially weakly closed set. If $x\in \overline{K}$ in the norm topology, choose $x_n\in K$ with $\Vert x_n-x\Vert <1/n$. The Cauchy-Schwarz inequality gives $|⟨x_n-x,z⟩|\le \Vert x_n-x\Vert \Vert z\Vert \to 0$ for every $z$, hence weak convergence to $x$. The assumed closure property gives $x\in K$. The empty set satisfies both properties too. Thus

The orthogonal complement $x^{\perp }$ is infinite-dimensional: otherwise $X=\mathrm{span}\{x\}\oplus x^{\perp }$ would be finite-dimensional. Inductively choose an orthonormal sequence $(e_n)$ in $x^{\perp }$, each new vector being orthogonal to the preceding finite linear span. The Bessel inequality gives ${\sum }_n|⟨e_n,z⟩{|}^2\le \Vert z{\Vert }^2$ for every $z\in X$, so $⟨e_n,z⟩\to 0$. Thus $(e_n)$ is a weakly null orthonormal sequence.

SetBy orthogonality, $\Vert x_n{\Vert }^2=\Vert x{\Vert }^2+(1-\Vert x{\Vert }^2)=1$, while $⟨x_n-x,z⟩\to 0$ for every $z$. Hence $x_n\rightharpoonup x$. When $\Vert x\Vert =1$, this formula simply gives the constant sequence $x_n=x$. In particular, the weak closure of the unit sphere contains the whole closed unit ball.

For each real $a$, define the sublevel set $K_a=\{u\in K:F(u)\le a\}$. Continuity of $F$ and closure of $K$ make $K_a$ a closed set in $X$; the convex function inequality makes it a convex set. The preceding result therefore makes every $K_a$ a sequentially weakly closed set. Equivalently, $F$ has weak lower semicontinuity along sequences in $K$.

Every $K_a$ is bounded. This is immediate if $K$ is bounded. Otherwise, an unbounded $K_a$ would contain $u_n$ with $\Vert u_n\Vert \ge n$, contradicting the assumed coercive function condition $F(u_n)\to \infty$.

Choose $z\in K$ and put $\ell ={\inf }_{K}F\in [-\infty ,F(z)]$. Choose a minimizing sequence $(u_n)$ entirely in $K_{F(z)}$, with $F(u_n)\to \ell$; when $\ell =-\infty$, require $F(u_n)\le -n$ for sufficiently large $n$. It is bounded by the previous paragraph.

The weak subsequence of a bounded Hilbert-space sequence theorem gives $u_{n_j}\rightharpoonup u_0$. This applies even if $X$ is not separable: the closed linear span of the sequence is a separable Hilbert space; successive extraction makes every coefficient in a countable Hilbertian basis converge, the common norm bound gives square-summable limiting coefficients by the Bessel inequality, and approximation by finite linear combinations proves weak convergence. Orthogonal decomposition extends that conclusion to all test vectors in $X$. Since $K$ is a sequentially weakly closed set, $u_0\in K$.

For any real $a>\ell$, a tail of $(u_{n_j})$ lies in $K_a$, so $u_0\in K_a$. If $\ell =-\infty$, this would give $F(u_0)\le a$ for every real $a$, impossible because $F$ is real-valued. Thus $\ell$ is finite; letting $a\downarrow \ell$ gives $F(u_0)\le \ell$, while the definition of infimum gives the reverse inequality. ThereforeThis proves minimum of a coercive convex functional on a Hilbert space by the direct method in the calculus of variations, with bounded sublevel sets supplying the required weak sequential compactness in a Hilbert space.

The Banach-Steinhaus convergence theorem can be stated as follows. Let $X$ be a normed space, $Y$ a Banach space, and $A_n\in \mathcal{B}(X,Y)$. Ifand $A_{n}d$ converges in $Y$ for every $d$ in a dense subset $D\subseteq X$, then $A_{n}x$ converges for every $x\in X$. The pointwise limit $A$ is a bounded linear operator andIf $X$ is also a Banach space, the uniform operator norm bound is necessary: pointwise convergence implies pointwise boundedness, so the Banach-Steinhaus theorem, namely the Uniform boundedness principle, supplies it. The limit is convergence in the strong operator topology; it need not be convergence in operator norm.

To see the sufficient direction, write $C={\sup }_n\Vert A_n\Vert$. For $d\in D$ the triangle inequality givesFirst approximate $x$ by $d$, then let $m,n\to \infty$. Completeness of $Y$ gives the limit. Linearity passes to limits, and $\Vert Ax\Vert ={\lim }_n\Vert A_{n}x\Vert \le ({\mathrm{liminf}}_n\Vert A_n\Vert )\Vert x\Vert$ gives the displayed bound. If a candidate bounded linear operator $A:X\to Y$ is already given and $A_{n}d\to Ad$ on $D$, the same approximation proves convergence to $Ax$ even when $Y$ is merely a normed space.

Composition with $t\mapsto t^{1+{\alpha }_n}$ preserves continuity, so $A_n$ maps the space of continuous functions on a compact space $C([0,1])$ into itself and is linear. Since $t\mapsto t^{1+{\alpha }_n}$ maps $[0,1]$ onto itself,For $0<t\le 1$, the mean value theorem applied to $r\mapsto t^{1+r}$ givesThe same bound holds at $t=0$. Each $x\in C([0,1])$ has uniform continuity by the Heine-Cantor theorem. With its modulus of continuity ${\omega }_x(\delta )={\sup }_{|s-t|\le \delta }|x(s)-x(t)|$, we obtainAlternatively, the bound proves convergence on every polynomial; the Weierstrass approximation theorem makes these a dense subset of $C([0,1])$, and the Banach-Steinhaus convergence theorem with $\Vert A_n\Vert =1$ gives the same conclusion.

The open mapping theorem states that a surjective bounded linear operator $T:X\to Y$ between Banach spaces is an open map: $T(U)$ is an open set in $Y$ whenever $U$ is an open set in $X$. Equivalently, there is $\delta >0$ such thatwhere both displayed balls are open. In particular, a bijective bounded linear operator between Banach spaces has a bounded inverse, the bounded inverse theorem.

Use the closed unit ball ${\overline{B}}_X$ first, and put $C=\overline{T({\overline{B}}_X)}$. This is a convex set symmetric about zero. If $B_Y(y_0,r)\subseteq C$, then $-y_0\in C$, so taking midpoints givesWrite $\delta =r/2$. By scaling,Fix $y$ with $\Vert y\Vert <\delta /2$. Starting with $r_0=y$, recursively choose $x_j\in X$ such thatAt each step the previous residual lies in the open ball on the left of the scaled inclusion, so the definition of closure permits this approximation. The series ${\sum }_j{x}_j$ converges to some $x$ because $X$ is a Banach space, and the triangle inequality gives $\Vert x\Vert \le {\sum }_j{2}^{-j}=1$. Continuity of $T$ gives $Tx={\lim }_N{\sum }_{j=1}^{N}T{x}_j=y$, since $r_N\to 0$. ThusScaling by $1/2$ now gives the slightly smaller open ballThis proves the assertion whether the original unit ball is taken open or closed. Only completeness of $X$ was used; the normed space $Y$ need not be complete. The argument is geometric correction for approximate surjectivity.

For $n\ge 1$ put $C_n=\overline{T(n{\overline{B}}_X)}$, with closure taken in $Y$. ThenIf every $C_n$ has empty interior, then each $C_n$ is a nowhere dense set, because it is closed. Each $T(n{\overline{B}}_X)$ is then a nowhere dense set too. Consequently $T(X)$ is a meagre set, that is, a set of the first category in $Y$.

Otherwise some $C_n$ has nonempty interior. Scaling gives nonempty interior of $\overline{T({\overline{B}}_X)}$. The geometric correction for approximate surjectivity proved in the preceding part gives an open ball about zero inside $T({\overline{B}}_X)$. Since $T(X)$ is a vector subspace, it then equals $Y$: scale any $y\in Y$ into this ball and scale its preimage back. Hence $T(X)$ is a closed set.

Thus, in fact, the stronger operator range category dichotomy holds:In particular $T(X)$ is closed or of the first category, as required. The only substantive theorem used here is the preceding geometric correction for approximate surjectivity: for a bounded linear operator from a Banach space to a normed space, nonempty interior of the closure of the image of the unit ball implies nonempty interior of the image itself. Its series proof was given there, so this proof does not require the Baire category theorem or completeness of $Y$.

Let $J:C([0,1])\to L^1(0,1)$ send a continuous function to its almost everywhere equivalence class. The domain with the supremum norm and the target with the L1 norm are Banach spaces, andThus $J$ is a bounded linear operator, and the requested vector subspace is $J(C([0,1]))$.

To show this image of a linear map is not closed, consider the continuous functionsand $g=1_{(1/2,1)}$. Their L1 norm difference is the area of one triangle:But $g$ has no continuous function representative. Indeed, a continuous function equal to zero almost everywhere on $(0,1/2)$ must be identically zero there: any nonzero value would persist on an open interval of positive Lebesgue measure. Likewise any continuous representative would be identically one on $(1/2,1)$. Continuity at $1/2$ makes these requirements incompatible. Therefore $g\notin J(C([0,1]))$, although it lies in its closure.

The preceding operator range category dichotomy proves continuous functions are meagre in L1:

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 $L^2(-\pi ,\pi )$.