Compact Operators and Riesz Schauder Theory — Examples

$D_a$ is well defined and bounded with $\mathrm{\parallel }{D}_{a}x{\mathrm{\parallel }}_p\le \mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }x{\mathrm{\parallel }}_p$, because $\mathrm{\mid }{a}_n{x}_n\mathrm{\mid }\le \mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }{x}_n\mathrm{\mid }$ for every $n$; so $\mathrm{\parallel }{D}_a\mathrm{\parallel }\le \mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}$.

Each truncation $D_a^{(N)}$ is bounded, and its range is contained in the linear span of $e_0,\dots ,e_{N-1}$, so $D_a^{(N)}$ has finite rank and is compact by [A4].

If $a_n\to ̸0$, then there is a real $\epsilon >0$ and a strictly increasing $n$ with $\mathrm{\mid }{a}_{n_k}\mathrm{\mid }\ge \epsilon$ for all $k$: this is the negation of convergence, and the indices are chosen by DC (The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$-indexed chain, A strictly increasing index map satisfies $n_k\ge k$).

For every $N$ one has $\mathrm{\parallel }{D}_a-D_a^{(N)}\mathrm{\parallel }={\sup }_{n\ge N}\mathrm{\mid }{a}_n\mathrm{\mid }$: the upper bound is [step 1.1] applied to the tail sequence, and the lower bound follows by testing $e_n$ for $n\ge N$, where $\mathrm{\parallel }(D_a-D_a^{(N)})e_n{\mathrm{\parallel }}_p=\mathrm{\mid }{a}_n\mathrm{\mid }$.

Under the hypothesis of [step 1.3], the bounded sequence $(e_{n_k})$ has $\mathrm{\parallel }{D}_a{e}_{n_k}-D_a{e}_{n_l}{\mathrm{\parallel }}_p\ge \epsilon$ for all $k\mathrm{\ne }l$ (indeed it is at least $({\epsilon }^p+{\epsilon }^p{)}^{1\mathrm{/}p}$ for $p<\mathrm{\infty }$ and at least $\epsilon$ for $p=\mathrm{\infty }$), so the sequence of images has no convergent subsequence; by the sequential characterization [A4] the operator $D_a$ is not compact.

If $a_n\to 0$, then for every real $\epsilon >0$ there is $N$ with $\mathrm{\mid }{a}_n\mathrm{\mid }<\epsilon$ for all $n\ge N$, so $\mathrm{\parallel }{D}_a-D_a^{(N)}\mathrm{\parallel }\le \epsilon$ by [step 2.1]; hence $D_a$ is a norm limit of the compact operators $D_a^{(N)}$ and is compact by [A4], the target ${\mathrm{\ell }}^p$ being Banach by [A1] and [A2].

Steps [step 3.1] and [step 2.2] are the two directions of the equivalence, so the diagonal operator $D_a$ on ${\mathrm{\ell }}^p(\mathbb{N},\mathbb{K})$ is compact exactly when $a_n\to 0$.

For every $f\in C([a,b],\mathbb{K})$ the function $\mathcal{K}f$ is well defined and continuous, and $\mathrm{\parallel }\mathcal{K}f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2M(b-a)\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$: for real scalars the sharper bound without the factor $2$ is the uniform integral estimate of [A2] applied to $y\mapsto k(x,y)f(y)$; for complex scalars the real and imaginary parts of the integrand are continuous, each has absolute value at most $M\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, and [A2] bounds each real integral by $M(b-a)\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, so the complex triangle inequality gives the displayed factor $2$. Continuity in $x$ follows from the uniform continuity of $k$ on the square and the same real-component estimates. Linearity follows componentwise from real Riemann-integral linearity (Integrable functions on $[a,b]$ form a set closed under sums and scalar multiples, and ${\int }_a^b(\lambda f+\mu g)=\lambda {\int }_a^{b}f+\mu {\int }_a^{b}g$), so this bound also makes $\mathcal{K}$ a bounded linear operator.

For all $f$ and all $x,x^{\mathrm{\prime }}\in [a,b]$ one has $\mathrm{\mid }\mathcal{K}f(x)-\mathcal{K}f(x^{\mathrm{\prime }})\mathrm{\mid }\le 2(b-a)\omega (x,x^{\mathrm{\prime }})\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, where $\omega (x,x^{\mathrm{\prime }})={\sup }_{y\in [a,b]}\mathrm{\mid }k(x,y)-k(x^{\mathrm{\prime }},y)\mathrm{\mid }$ and $\omega (x,x^{\mathrm{\prime }})\to 0$ as $\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }\to 0$ uniformly in $y$, by the uniform continuity of $k$ on the square; the factor $2$ covers the complex real-and-imaginary-part estimate and is harmless in the real case. In particular the family $F:=\{\mathcal{K}f:\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1\}$ is equicontinuous (with complex modulus in the complex case, so both real component families satisfy [A3]) and pointwise bounded with $\mathrm{\mid }\mathcal{K}f(x)\mathrm{\mid }\le 2M(b-a)$.

In the real case $\mathbb{K}=\mathbb{R}$ the closure of $F$ in the supremum metric is compact by [A3] and [step 2.1], hence $\mathcal{K}$ is compact by [A5].

In the complex case every sequence $(f_j)$ with $\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$ has a subsequence for which $\mathcal{K}{f}_j$ converges uniformly: the real functions $x\mapsto \mathrm{Re}\mathcal{K}{f}_j(x)$ form an equicontinuous pointwise bounded sequence in $C([a,b],\mathbb{R})$ by [step 2.1] and [A2], so by [A3] they have a uniformly convergent subsequence; within that subsequence the imaginary parts, which are again equicontinuous and pointwise bounded, have a further uniformly convergent subsequence; along that further subsequence $\mathcal{K}{f}_j$ converges uniformly because the complex modulus is at most the sum of the moduli of the real and imaginary parts by [A2].

In the complex case the closure of $F$ is sequentially compact: given a sequence $(g_j)$ in the closure, [A4] chooses $f_j$ with $\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$ and $\mathrm{\parallel }\mathcal{K}{f}_j-g_j{\mathrm{\parallel }}_{\mathrm{\infty }}<1\mathrm{/}(j+1)$ for every $j$, and [step 3.2] applied to $(f_j)$ gives a subsequence along which $\mathcal{K}{f}_j$ converges, hence $g_j$ converges to the same limit; by [A3] the closure of $F$ is compact, and $\mathcal{K}$ is compact by [A5].

In both cases $\mathbb{K}=\mathbb{R}$ and $\mathbb{K}=\mathbb{C}$ the operator $\mathcal{K}$ is compact, which is the assertion.

$I_X$ is compact if and only if ${\bar{B}}_X$ is compact, because $I_X({\bar{B}}_X)={\bar{B}}_X$ is already closed.

In ${\mathrm{\ell }}^2(\mathbb{N},\mathbb{K})$ the vectors $e_0,\dots ,e_n$ are linearly independent for every $n$: if ${\sum }_{k\le n}{c}_k{e}_k=0$, then pairing with $e_j$ gives $c_j=⟨{\sum }_k{c}_k{e}_k,e_j⟩=0$ for every $j\le n$.

Hence $I_X$ is compact if and only if $X$ admits an ordered basis of finite length, by [step 1.1] and [A2].

The space ${\mathrm{\ell }}^2(\mathbb{N},\mathbb{K})$ admits no ordered basis of finite length: if it admitted a spanning list of length $n$, then by [A3] every linearly independent subset would have at most $n$ elements, contradicting the independent list $e_0,\dots ,e_n$ of [step 1.2] of length $n+1$.

Therefore the identity of ${\mathrm{\ell }}^2(\mathbb{N},\mathbb{K})$ is not compact, by [step 2.1] and [step 2.2]; this is the promised witness, and [step 2.1] is the asserted equivalence.

$P$ is well defined and linear with $\mathrm{\parallel }Px{\mathrm{\parallel }}_2=\mathrm{\mid }{x}_1\mathrm{\mid }\le \mathrm{\parallel }x{\mathrm{\parallel }}_2$, because $\mathrm{\mid }{x}_1{\mathrm{\mid }}^2\le {\sum }_n\mathrm{\mid }{x}_n{\mathrm{\mid }}^2$; hence $P$ is a bounded operator with $\mathrm{\parallel }P\mathrm{\parallel }\le 1$.

$P$ is nonzero: $P(e_1)=e_1\mathrm{\ne }0$, and $\mathrm{\parallel }{e}_1{\mathrm{\parallel }}_2=1$.

The range of $P$ is $\mathrm{span}\{e_1\}$: every $Px=x_1{e}_1$ lies in it, and conversely $c{e}_1=P(c{e}_1)$; this span is one dimensional, hence finite dimensional, and closed by [A2].

$P$ is compact by [A2], since it is bounded with finite-dimensional range.

The range is not dense: for every scalar $c\in \mathbb{K}$ one has $\mathrm{\parallel }{e}_2-c{e}_1{\mathrm{\parallel }}_2^2=\mathrm{\parallel }{e}_2{\mathrm{\parallel }}_2^2+\mathrm{\mid }c{\mathrm{\mid }}^2\mathrm{\parallel }{e}_1{\mathrm{\parallel }}_2^2=1+\mathrm{\mid }c{\mathrm{\mid }}^2\ge 1$ by the orthogonality $⟨e_1,e_2⟩=0$, so $\mathrm{dist}(e_2,\mathrm{span}\{e_1\})\ge 1$ and $e_2\mathrm{\notin }\overline{\mathrm{span}\{e_1\}}$ by [A3]; since the range is closed by [step 2.1], it is therefore a proper closed subspace.

So $P$ is a nonzero compact operator with nondense range, which refutes the general statement; the claims are [step 3.1], [step 1.2] and [step 3.2].

$C([a,b],\mathbb{R})$ with the supremum norm is a Banach space, being complete by [A1] and normed by [A2].

$C([a,b],\mathbb{C})$ with the supremum norm is a Banach space: a sequence $(f_j)$ is Cauchy for the supremum norm exactly when the real sequences $(\mathrm{Re}{f}_j)$ and $(\mathrm{Im}{f}_j)$ are Cauchy, by the two inequalities of [A2]; those have continuous limits $u,v$ by [A1], and then $\mathrm{\parallel }{f}_j-(u+iv){\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }\mathrm{Re}{f}_j-u{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }\mathrm{Im}{f}_j-v{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$; the norm axioms hold by [A2].

The transpose of $I-\mathcal{K}$ is $I-{\mathcal{K}}^{\ast }$, by [A3].

In either scalar field, $\mathcal{K}$ is a compact operator on the Banach space $C([a,b],\mathbb{K})$, by [step 1.1], [step 1.2] and [A3].

Claim 1: by [A4] applied to the compact operator $\mathcal{K}$ on the Banach space $C([a,b],\mathbb{K})$, the equation $f-\mathcal{K}f=g$ is solvable exactly when $\phi (g)=0$ for every $\phi$ in the kernel of $(I-\mathcal{K}{)}^{\ast }$, which is $\ker (I-{\mathcal{K}}^{\ast })$ by [step 1.3].

Claim 2: the equation $f-\mathcal{K}f=g$ has exactly one solution for every $g$ exactly when $I-\mathcal{K}$ is a bijection, which by [A4] is equivalent to injectivity of $I-\mathcal{K}$, that is, to the homogeneous equation $f=\mathcal{K}f$ having only the zero solution.

The two displayed claims are [step 3.1] and [step 3.2].

Each $P_N$ is linear with $\mathrm{\parallel }{P}_{N}x{\mathrm{\parallel }}_2^2={\sum }_{n<N}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2\le \mathrm{\parallel }x{\mathrm{\parallel }}_2^2$, hence bounded with $\mathrm{\parallel }{P}_N\mathrm{\parallel }\le 1$, and its range is contained in the linear span of $e_0,\dots ,e_{N-1}$, a finite-dimensional subspace; so $P_N$ is compact by [A2].

For every $x\in H$ one has $\mathrm{\parallel }x-P_{N}x{\mathrm{\parallel }}_2^2={\sum }_{n\ge N}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2\to 0$: given $\epsilon >0$, the convergence of the nonnegative sum gives a finite $F\subseteq \mathbb{N}$ with ${\sum }_{n\mathrm{\notin }F}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2<\epsilon$, and for $N>\max F$ the tail $\{n:n\ge N\}$ is contained in $\mathbb{N}\setminus F$, so the tail sum is below $\epsilon$.

$\mathrm{\parallel }I-P_N\mathrm{\parallel }=1$ for every $N$: the upper bound $\mathrm{\parallel }(I-P_N)x{\mathrm{\parallel }}_2^2={\sum }_{n\ge N}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2\le \mathrm{\parallel }x{\mathrm{\parallel }}_2^2$ follows from [A1], and the lower bound holds because $(I-P_N)e_N=e_N$ with $\mathrm{\parallel }{e}_N{\mathrm{\parallel }}_2=1$.

By [step 1.2] the sequence $(P_N)$ converges to $I$ in the strong operator topology; by [step 1.3] the convergence is not in operator norm; and the limit $I$ is not compact by [A2].

With [step 1.1] and [step 1.2] this shows that a strong operator limit of compact operators need not be compact, refuting the general statement.