Write \(T=D+L\) , where \((Dx)_k=x_k/k\) and \(L\) is the left shift. Both are bounded with norm \(1\) (Example 5.2 and Example 2.4), so \(T\) is bounded and \(\lVert T\rVert\le 2\) . For a lower bound, \(Te_2=e_1+\tfrac12e_2\) has norm \(\sqrt{5}/2\approx 1.118\) , so \(1.118\le\lVert T\rVert\le 2\) .

\(T\) is not compact. The operator \(D\) is compact, because its diagonal entries tend to \(0\) (Example 9.1). If \(T\) were compact, then \(L=T-D\) would be compact, since the compact operators form a subspace. But \(L\) maps the bounded sequence \((e_{k+1})_{k\ge 1}\) to \((e_k)_{k\ge 1}\) , which has no convergent subsequence because its terms are at mutual distance \(\sqrt2\) . So \(L\) is not compact, and neither is \(T\) . The decomposition shows how to analyse such an operator: the compact part \(D\) can be ignored for questions of compactness, and what remains is a shift.

Example. In \(\ell^2\) take \(x_n=y_n=e_n\) . Then \(x_n\rightharpoonup 0\) and \(y_n\rightharpoonup 0\) , but \(\langle x_n,y_n\rangle=1\) for every \(n\) , which does not converge to \(\langle 0,0\rangle=0\) .

Proof with one strong factor. Suppose \(x_n\rightharpoonup x\) and \(y_n\to y\) in norm. Weakly convergent sequences are bounded (Proposition 6.2), say \(\lVert x_n\rVert\le C\) . Then

The first term tends to \(0\) by norm convergence of \((y_n)\) , and the second by weak convergence of \((x_n)\) tested against the fixed vector \(y\) (Corollary 8.2). So \(\langle x_n,y_n\rangle\to\langle x,y\rangle\) . The inner product is continuous for the norm in both variables jointly, and for weak convergence in each variable separately, but not jointly; this is the Hilbert-space form of Exercise 5 of Chapter 6Chapter 6.

On \(C[0,1]\) . \(\lvert\phi_n(f)\rvert\le n\cdot\tfrac1n\lVert f\rVert_\infty=\lVert f\rVert_\infty\) , with equality for \(f=1\) , so \(\lVert\phi_n\rVert=1\) for every \(n\) . For continuous \(f\) , \(\phi_n(f)\) is the average of \(f\) over \([0,\tfrac1n]\) , which tends to \(f(0)\) . So \(\phi_n\) converges pointwise to the evaluation functional at \(0\) , which has norm \(1\) ; this is consistent with Corollary 4.5.

On \(L^1[0,1]\) . \(\lvert\phi_n(f)\rvert\le n\int_0^{1/n}\lvert f\rvert\le n\lVert f\rVert_1\) , and \(f=n\mathbf 1_{[0,1/n]}\) has \(\lVert f\rVert_1=1\) and \(\phi_n(f)=n\) . So \(\lVert\phi_n\rVert=n\) , which is unbounded. Since \(L^1[0,1]\) is a Banach space, the uniform boundedness principle (Theorem 4.4), read contrapositively, yields some \(f\in L^1[0,1]\) with \(\sup_n\lvert\phi_n(f)\rvert=\infty\) . One such function is \(f(x)=x^{-1/2}\) , for which \(\phi_n(f)=n\cdot 2n^{-1/2}=2\sqrt n\) . The same formula thus defines a bounded, convergent sequence of functionals on one space and an unbounded one on another: averages over shrinking intervals converge for continuous functions and can diverge for integrable ones.

The kernel \(1+xy\) is real and symmetric, so \(K\) is self-adjoint, and \(Kf=\langle f,1\rangle\cdot 1+\langle f,x\rangle\,x\) has range in the span of \(1\) and \(x\) . Every eigenvector for a nonzero eigenvalue lies in that span, so it suffices to find the matrix of \(K\) there:

For \(f=a+bx\) this gives \(Kf=(a+\tfrac12b)+(\tfrac12a+\tfrac13b)x\) , so the coefficient vector \((a,b)\) is multiplied by

with trace \(\tfrac43\) and determinant \(\tfrac13-\tfrac14=\tfrac1{12}\) . Its eigenvalues are the roots of \(\lambda^2-\tfrac43\lambda+\tfrac1{12}=0\) :

These are the nonzero eigenvalues of \(K\) ; all other spectrum is the eigenvalue \(0\) , whose eigenspace is the orthogonal complement of the span of \(1\) and \(x\) . Since \(K\) is self-adjoint and compact, Theorem 10.6 gives \(\lVert K\rVert=\tfrac16(4+\sqrt{13})\approx 1.2676\) . As a check, the sum of the squares of the eigenvalues is \(\tfrac{2(16+13)}{36}=\tfrac{29}{18}\) , and \(\lVert k\rVert_{L^2}^2=\int_0^1\!\!\int_0^1(1+xy)^2=1+\tfrac12+\tfrac19=\tfrac{29}{18}\) .

Completing the square,

so minimizing \(J\) over \(K\) is the same as finding the point of \(K\) nearest to \(f\) . The cone \(K\) is convex, and closed because coordinatewise limits of nonnegative numbers are nonnegative and norm convergence implies coordinatewise convergence. By Theorem 7.3 there is exactly one nearest point \(P_Kf\) , which is therefore the unique minimizer of \(J\) .

To identify it, let \(u^\star=f^+\) , the sequence with coordinates \(\max\{f_k,0\}\) ; it is in \(\ell^2\) and in \(K\) . Then \(f-u^\star\) has coordinates \(\min\{f_k,0\}\le 0\) , vanishing wherever \(u^\star_k>0\) . For any \(v\in K\) ,

since \(u^\star_k=0\) and \(v_k\ge 0\) when \(f_k<0\) . This is the variational inequality of Theorem 7.3, so \(P_Kf=f^+\) . The minimum value is \(J(f^+)=\tfrac12\lVert f^-\rVert^2-\tfrac12\lVert f\rVert^2=-\tfrac12\lVert f^+\rVert^2\) , where \(f^-=f-f^+\) .