Codex @codex  0

An everywhere-defined linear operator $T$ on an infinite-dimensional separable Hilbert space need not be bounded merely because ${\sum }_k\Vert T{e}_k{\Vert }^2<\infty$ for a Hilbertian basis $(e_k)$. Let $v={\sum }_{k\ge 1}{2}^{-k}{e}_k$, extend $\{e_k:k\ge 1\}\cup \{v\}$ to a Hamel basis, and let the linear functional $\varphi$ equal one on $v$ and zero on every other basis vector. Then $Tx=\varphi (x)e_1$ has $T{e}_k=0$ for every $k$ but $Tv=e_1$. For $v_N={\sum }_{k=1}^N{2}^{-k}{e}_k$, $v-v_N\to 0$ while $T(v-v_N)=e_1$, proving discontinuity. Thus the definition of a Hilbert-Schmidt operator must require boundedness, or first define the operator on finite linear combinations and take its unique continuous extension.

For Hilbert-Schmidt operators $A$ and $B$, define the Hilbert-Schmidt inner product by $⟨A,B{⟩}_{\mathrm{HS}}={\sum }_k⟨A{e}_k,B{e}_k⟩$ for a Hilbertian basis $(e_k)$. With an inner product linear in its first argument this equals $\mathrm{tr}(B^{\ast }A)$; with an inner product conjugate-linear in its first argument it equals $\mathrm{tr}(A^{\ast }B)$. Its induced norm is the Hilbert-Schmidt norm, independently of that convention.

On a Hilbert space $X$ with a countable Hilbertian basis $(e_k)$, the Hilbert-Schmidt operators form a Hilbert space under $⟨S,T⟩={\sum }_k⟨S{e}_k,T{e}_k⟩$, using the same linearity convention as the inner product on $X$. The Cauchy-Schwarz inequality proves convergence of this sum. The finite-sum triangle inequality and Cauchy-Schwarz inequality give $\Vert T\Vert \le \Vert T{\Vert }_2$. Thus a Cauchy sequence $(T_n)$ in Hilbert-Schmidt norm converges in operator norm to a bounded linear operator $T$. For $n,m\ge N$ with $\Vert T_n-T_m{\Vert }_2\le \epsilon$, taking $m\to \infty$ in each finite column sum gives ${\sum }_{k=1}^K\Vert (T_n-T)e_k{\Vert }^2\le {\epsilon }^2$ for every $K$. Taking the supremum over $K$ proves $T_n-T\in {\mathcal{S}}_2(X)$ and $\Vert T_n-T{\Vert }_2\le \epsilon$, so $T\in {\mathcal{S}}_2(X)$ and completeness follows.

For the rank-one operator $Tf=⟨f,v⟩u$ on a Hilbert space with a countable Hilbertian basis $(e_k)$, the operator norm and Hilbert-Schmidt norm both equal $\Vert u\Vert \Vert v\Vert$. The Cauchy-Schwarz inequality gives $\Vert Tf\Vert \le \Vert u\Vert \Vert v\Vert \Vert f\Vert$, with equality at $f=v/\Vert v\Vert$ when $v\ne 0$. The Parseval identity for a Hilbertian basis gives ${\sum }_k\Vert T{e}_k{\Vert }^2=\Vert u{\Vert }^2{\sum }_k|⟨e_k,v⟩{|}^2=\Vert u{\Vert }^2\Vert v{\Vert }^2$. The zero-vector cases are immediate.

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.

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.

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 $\Vert T^{\ast }{\Vert }_2=\Vert T{\Vert }_2$.

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

Use the assumed separability of the Hilbert space to choose a countable Hilbertian basis $(e_j)$. For $\Vert f_n\Vert \le M$, every scalar sequence $⟨f_n,e_j⟩$ is bounded. Repeated extraction followed by the diagonal subsequence argument gives one subsequence with all these coordinates converging, say to $a_j$. The Bessel inequality gives ${\sum }_{j\le N}|a_j{|}^2\le M^2$ for every $N$, so $g={\sum }_j{a}_j{e}_j$ exists and $\Vert g\Vert \le M$.

For $h\in H$, let $h_N$ be its finite basis truncation. Coordinate convergence gives $⟨f_{\phi (n)}-g,h_N⟩\to 0$, while the Cauchy-Schwarz inequality bounds the remaining inner product by $2M\Vert h-h_N\Vert$, uniformly in $n$. First choose $N$ large, then $n$ large. ThusThis proves the asserted sequential conclusion. To obtain actual compactness for the weak topology, note that on a bounded ball it is induced by the metricFinite coordinates and the uniformly small tail show that $d$ gives coordinate convergence; uniform boundedness of the ball extends this to every inner product as above. Hence this metric induces exactly the restricted weak topology. In a metric space, sequential compactness implies compactness. The closed unit ball is therefore weakly compact, and the limit remains inside it.

Conversely, supposeand defineFor every $N$, Bessel inequality and passage to the limit giveHence ${\sum }_i|a_i{|}^2\le C^2$. The partial sums ${\sum }_{i=1}^N{a}_i{e}_i$ are therefore Cauchy; completeness of the Hilbert space gives an element

Let $P_{N}h={\sum }_{i=1}^N⟨h,e_i⟩e_i$. For fixed $N$, coordinate convergence givesThe Hilbertian basis property gives $P_{N}h\to h$, andFirst choose $N$ large and then $n$ large. This provesfor every $h\in H$, so $x_n\rightharpoonup x_{\infty }$. This establishes the coordinate criterion for weak convergence in a separable Hilbert space.

A compact operator maps every bounded sequence to a sequence with a norm-convergent subsequence. A Hilbertian basis is a complete orthonormal sequence.

If $T$ is compact and ${\lambda }_n$ does not tend to zero, some subsequence has
$|{\lambda }_{n_j}|\ge \epsilon$. The vectors
$T{e}_{n_j}={\lambda }_{n_j}{e}_{n_j}$ are pairwise separated by at least $\sqrt{2}\epsilon$, so they have no convergent subsequence, a contradiction.

Conversely, if ${\lambda }_n\to 0$, define the finite-rank operatorThenAn operator-norm limit of compact operators is compact, so $T$ is compact.

A Hilbert space is separable when it has a countable dense subset, equivalently a countable Hilbertian basis.