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.

Let $(f_n)$ be a singular Weyl sequence for $L$ at $\lambda \in {\Sigma }_{\mathrm{e}}(L)$. Since $K$ is compact and $f_n\rightharpoonup 0$, the preceding result gives $K{f}_n\to 0$. ThereforeThe norms remain one and the weak limit remains zero. The singular Weyl sequence criterion yields $\lambda \in {\Sigma }_{\mathrm{e}}(L+K)$. Apply the same argument to $L+K$ and the compact self-adjoint operator $-K$ for the reverse inclusion. Both operators are bounded and self-adjoint. Thus the Weyl theorem for compact self-adjoint perturbations isThe corrected shifted-range definition is necessary. For the diagonal example in the preceding solutions, a rank-one perturbation changing the entry $2$ to $0$ removes $2$ from the spectrum. The unshifted printed definition had classified $2$ as essential merely because the original range was not closed, so it would make this invariance false.

Suppose $K$ is a compact operator and $f_n\rightharpoonup f$. The Uniform boundedness principle makes $(f_n)$ bounded. If $K{f}_n$ did not converge in norm to $Kf$, some subsequence would stay a fixed positive distance away. Compactness provides a further norm-convergent subsequence, say to $g$. On the other hand, for every $h$, $⟨K{f}_n,h⟩=⟨f_n,K^{\ast }h⟩\to ⟨Kf,h⟩$, so its norm limit must be $Kf$, a contradiction.

Conversely, if $K$ sends every weakly convergent sequence to a norm-convergent sequence, take any sequence in the closed unit ball. Weak compactness of that ball supplies a weakly convergent subsequence, whose images converge in norm by hypothesis. Thus every sequence in the image has a convergent subsequence in $H$. Its closure is also sequentially compact: approximate its $n$th member by an image point within $1/n$. Since $H$ is a metric space, that closure is compact. We concludeThis is the principle that compact operators send weak convergence to norm convergence.

Use the corrected essential spectrum of a bounded self-adjoint operator and put $A=L-\lambda I$. Essential spectral points are real. If $\ker A$ is infinite-dimensional, choose an orthonormal sequence in the kernel. It converges weakly to zero by the Bessel inequality, and its residuals vanish.

If $\ker A$ is finite-dimensional, membership in the essential spectrum means the range is not closed. Choose unit $v_n\in (\ker A{)}^{\perp }$ with $A{v}_n\to 0$, using the closed-range bound on the kernel complement. A bounded Hilbert space sequence has a weakly convergent subsequence. Its weak limit $v$ satisfies $Av=0$ because bounded operators preserve weak convergence, and $v\in (\ker A{)}^{\perp }$; hence $v=0$. This subsequence is a singular Weyl sequence.

Conversely a singular Weyl sequence first places $\lambda$ in the spectrum of a bounded operator. If it were not essential, the sequential properness for a self-adjoint operator equivalence for $A$ would yield a norm-convergent subsequence. Its weak limit is zero, whereas norm convergence of unit vectors gives a unit norm limit, a contradiction. ThereforeFor nonreal $\lambda$, the resolvent lower bound excludes such a sequence, so the equivalence covers all $\lambda \in \mathbb{C}$.

Use the positive variable $Y$ from part (a). Since $0\le (S-B{)}^{+}\le S$ and $(S-B{)}^{+}\ge S-B$,Combining these inequalities gives the one-period call price boundsFor the strict assertion, both $\{S>B\}$ and $\{S<B\}$ have positive probability. Positivity of $Y$ givesIf $s-b\ge 0$, the second inequality is the required strict bound; if $s-b<0$, the first is. Hence $\boxed{c>(s-b{)}^{+}}$. The weighted expectations are finite because $P$ and $Y$ are bounded.

Applying death-only rate correction under exponential censoring to the death fraction and combined-rate estimate, the corrected plug-in death rate isThe corresponding censoring-rate estimate is $0.1{\text{year}}^{-1}$. This correction is consistent under the independent exponential competition model: the death-only conditional rate estimates $s$ and the observed fraction estimates $\lambda /s$.

There is an important distinction from the full-data maximum-likelihood estimate. If $T_D$ and $T_C$ are total observed follow-up for the 100 deaths and 50 censored subjects, the full likelihood for both rates isThe reported death-only rate determines $T_D=100/0.3$, but supplies no $T_C$, so the exact full-data estimate cannot be reconstructed from the numerical summaries alone. The value $0.2$ is the requested corrected rate using those summaries; it need not equal the realised full-data maximum-likelihood estimate. The full analysis should use the actual censored follow-up whenever available.