The natural canonical embedding into the bidual is $J_{X}x(f)=f(x)$. It is linear and $\Vert J_{X}x\Vert \le \Vert x\Vert$. A norming functional from the Hahn-Banach theorem gives the reverse inequality, so it is an isometry. A reflexive Banach space is one for which this canonical embedding is onto.

A nonreflexive example is $c_0$. Its dual is ${\ell }^1$ and its bidual is ${\ell }^{\infty }$; the canonical map is the natural inclusion, which misses the constant sequence one. This also shows why merely having an isometric embedding in the bidual is insufficient.

Take $S=T(B_X)$. A bounded linear operator is compact exactly when the closure of this set is compact. In a complete metric space this is equivalent to its being totally bounded.

If $T$ is compact, choose a finite $\epsilon$-net $z_1,\dots ,z_m$ for $T(B_X)$ and let $Y$ be their span. Then $\mathrm{dist}(Tx,Y)<\epsilon$ for $\Vert x\Vert \le 1$, giving $\Vert {\pi }_{Y}T\Vert \le \epsilon$.

Conversely, fix $\delta >0$ and take $Y$ with $\Vert {\pi }_{Y}T\Vert \le \delta /4$. For every $x\in B_X$ choose $y_x\in Y$ with $\Vert Tx-y_x\Vert <\delta /2$. These approximants satisfy $\Vert y_x\Vert \le \Vert T\Vert +\delta /2$, so form a bounded subset of a finite-dimensional space and have a finite $\delta /2$-net. The same centers give a finite $\delta$-net for $T(B_X)$. Thus it is totally bounded and $T$ is compact. In particular

A set is totally bounded if for every $\epsilon >0$ it can be covered by finitely many open balls of radius $\epsilon$.

Here is the needed separated-sequence argument, including the relevant Riesz lemma. For a proper closed subspace $Y$, choose $z\notin Y$, put $d=\mathrm{dist}(z,Y)>0$, and choose $y\in Y$ with $\Vert z-y\Vert <2d$. Then $w=(z-y)/\Vert z-y\Vert$ is a unit vector at distance greater than $1/2$ from $Y$. Starting with one unit vector, apply this to the span of the vectors already chosen. These spans are closed and proper because they are finite-dimensional while $X$ is infinite-dimensional. The resulting unit vectors satisfy $\Vert w_m-w_n\Vert >1/2$ for $m\ne n$. A ball of radius $1/4$ contains at most one of them. Hence the closed unit ball is not totally bounded.

The quotient norm isChanging the representative does not change the infimum. The triangle inequality and absolute homogeneity follow from those of the original norm. Vanishing means $x$ belongs to the closure of $Y$, hence to $Y$ because the subspace is closed. Thus this is a norm.

Use the absolutely convergent series criterion for completeness. If $\sum \Vert {\xi }_n{\Vert }_{X/Y}<\infty$, choose representatives $z_n$ of ${\xi }_n$ satisfying $\Vert z_n\Vert \le \Vert {\xi }_n{\Vert }_{X/Y}+2^{-n}$. The series $\sum z_n$ converges in the Banach space $X$ to some $z$. Since the quotient map is contractive, $\sum {\xi }_n$ converges to $z+Y$. Consequently the quotient is a Banach space.

TakeThe series defines a bounded linear functional of norm one. It corresponds under ${\Phi }_2$ to the constant sequence ${\gamma }_j=1$, so is $\sigma ({\ell }^1,c)$-continuous. Its coefficient sequence is constantly one, which is not in $c_0$, so the continuous-dual result makes it noncontinuous for $\sigma ({\ell }^1,c_0)$. Directly $e_n\to 0$ in that topology but ${\psi }_2(e_n)=1$.

TakeIt is evaluation against the first coordinate vector of $c_0$, hence continuous for $\sigma ({\ell }^1,c_0)$. The preceding continuous-dual result says that a functional continuous for $\sigma ({\ell }^1,c)$ must have coefficient sequence $({\gamma }_{\infty },{\gamma }_1,{\gamma }_2,\dots )$ for a convergent sequence $\gamma$. To represent ${\alpha }_1$, it would require ${\gamma }_{\infty }=1$ but every ${\gamma }_j=0$, impossible.

Directly, $e_1-e_n\to 0$ in $\sigma ({\ell }^1,c)$ because its pairing is ${\gamma }_{\infty }-{\gamma }_{n-1}\to 0$, whereas ${\psi }_1(e_1-e_n)=1$ for $n\ge 2$.

If the two ambient topologies are not comparable, neither is contained in the other. By the definition of inclusion of topologies, there is a ${\tau }_1$-open set absent from ${\tau }_2$, and a ${\tau }_2$-open set absent from ${\tau }_1$. These are the requested $U_1,U_2$, giving alternative (b). Together with the comparable case, this proves that at least one of the two alternatives always holds; it does not assume that the alternatives are mutually exclusive.

The weak-star topology on $X^{\ast }$ is the coarsest topology making every evaluation $\varphi \mapsto \varphi (x)$ continuous. Finite intersections of evaluation inequalities form a neighborhood basis. Distinct functionals differ on some $x$, so disjoint disks around their evaluation values give disjoint neighborhoods; thus the topology is Hausdorff. The Banach-Alaoglu theorem states that the closed unit ball of $X^{\ast }$ is compact in this topology.

If $f\in X^{\ast \ast }$ is weak-star continuous, some $x_1,\dots ,x_n$ and $\epsilon >0$ make $|f(\varphi )|<1$ whenever $|\varphi (x_i)|<\epsilon$ for every $i$. If all evaluations vanish, every scalar multiple of $\varphi$ lies in this neighborhood, forcing $f(\varphi )=0$. Thus $f$ vanishes on the kernel of the map $E(\varphi )=(\varphi (x_1),\dots ,\varphi (x_n))$. It factors linearly through the finite-dimensional range of $E$. Extending that factor algebraically to ${\mathbb{K}}^n$ gives scalars $a_i$ withConsequently $f=J_{X}x$ for $x={\sum }_i{a}_i{x}_i$.

For the two induced topologies, let $B$ be the common norm-closed unit ball and write ${\tau }_j=\sigma (Y,X_j)$. Under either topology $B$ is compact by Banach-Alaoglu, and the ambient space is Hausdorff. If the two topologies are comparable, the identity map from $B$ with the finer topology to $B$ with the coarser one is a continuous bijection from a compact space to a Hausdorff space. Such a map is a homeomorphism, since it carries closed compact sets to closed sets. Thus in the comparable case the topologies agree on $B$, giving alternative (a).

The extension already agrees with the usual limit on $c$. Alternatively, applying the domination inequality to both $x$ and $-x$ givesWhen the sequence converges, both endpoints are its limit, proving $\phi (x)=\lim {\alpha }_n$. These bounds also give $\Vert \phi \Vert =1$ and positivity on nonnegative sequences, but do not by themselves assert shift invariance.

A sublinear functional on a real vector space satisfies $p(x+y)\le p(x)+p(y)$ and $p(tx)=tp(x)$ for every $t\ge 0$. The dominated real Hahn-Banach theorem says that a linear functional on a subspace satisfying $f\le p$ extends to a linear functional on the whole space still dominated by $p$.

The limit-superior functional is finite on bounded sequences, positively homogeneous, and subadditive, since for every tail $\sup ({\alpha }_n+{\beta }_n)\le \sup {\alpha }_n+\sup {\beta }_n$. Passing to the tail limits proves sublinearity. On the subspace $c$ of convergent real sequences, the ordinary limit is linear and equals $p$. Hahn-Banach therefore extends it to $\phi$ on ${\ell }^{\infty }$ with

For $x-Sx$, choose the consecutive shifts $j_r=r$, $1\le r\le k$. The average telescopes:Its absolute value is at most $2\Vert x{\Vert }_{\infty }/k$. Taking the limit superior and then the infimum over $k$ proves $q(x-Sx)\le 0$.

For the requested functional, use the allowed sublinearity of $q$. On convergent sequences, every finite shifted average has the same limiting value, so $q$ equals the ordinary limit there. Hahn-Banach extends this limit to a linear $\psi$ satisfying $\psi \le q\le p$. Apply domination to $x-Sx$ and to $(-x)-S(-x)$ to obtain respectively $\psi (x-Sx)\le 0$ and $-\psi (x-Sx)\le 0$. HenceThis is a Banach limit; the two-sided limit bounds also show positivity, norm one, and agreement with ordinary limits.

In the infimum defining $q$, choose $k=1,j_1=1$. A finite shift leaves the limit superior unchanged, so

A subset is totally bounded if for every $\epsilon >0$ it has a finite cover by balls of radius $\epsilon$. If its closure is compact, the cover by all such balls has a finite subcover, proving total boundedness.

Conversely, cover $S$ by finitely many balls of radius $2^{-r}$ at stage $r$. Starting with all indices, at least one ball contains infinitely many of the currently indexed terms. Retain those indices as ${\sigma }_r$. These sets are infinite and nested, and two terms indexed by ${\sigma }_r$ have distance less than $2^{1-r}$. Choose increasing $n_r\in {\sigma }_r$; then $(x_{n_r})$ is Cauchy, because all its terms after stage $r$ lie in that same ball. If $M$ is complete it converges. To cover sequences in the closure as well, approximate their $n$th terms by points of $S$ at distance less than $1/n$ and apply this argument. The corresponding subsequence in the closure has the same limit. Sequential compactness therefore proves total boundedness implies relative compactness in a complete ambient space.

The Arzelà-Ascoli theorem says a subset of $C[0,1]$ is relatively compact exactly when it is uniformly bounded and equicontinuous. Pointwise boundedness together with equicontinuity is an equivalent version here. In the specified derivative-bounded family, the mean-value estimate gives $|f(t)-f(s)|\le |t-s|$, uniformly over the family, while $\Vert f{\Vert }_{\infty }\le 1$. Thus this family is relatively compact.

For $V$, the image of the unit ball has sup norm at most one and derivative $f$ of sup norm at most one; it lies in that family, so $V$ is compact.

For $W$, take $f_n=n{1}_{[0,1/n]}$, each of $L^1$ norm one. Then $W{f}_n(t)=\min (nt,1)$. Every subsequence converges pointwise to zero at $t=0$ and to one at every $t>0$. This limit is discontinuous, so no subsequence converges uniformly to a continuous function. Thus $W$ is not compact. The same integration formula has different compactness properties because the domain unit balls impose different controls.

A subset is relatively compact in $M$ when its closure in $M$ is compact.

Suppose $T_n$ are compact and $\Vert T_n-T\Vert \to 0$. For any $\epsilon >0$ take $n$ with $\Vert T_n-T\Vert <\epsilon$. A finite $\epsilon$-net of $T_n$ of the unit ball is then a finite $2\epsilon$-net of its $T$ image. That image is totally bounded, so $T$ is compact by the stated criterion. Hence $\mathcal{K}(X)$ is norm closed in $B(X)$.

Extract a subsequence whose $T$ images converge, and apply continuity of the bounded operator $S$ to those images. This proves $ST$ is compact. The reverse composition $TS$ is compact as well, since $S$ sends a bounded sequence to a bounded sequence to which compactness of $T$ applies.

Given a bounded sequence, first pass to a subsequence on which the images under $T_1$ converge. From that subsequence extract another on which the images under $T_2$ converge. Both image sequences, and hence their sum, converge on the latter subsequence. Thus $T_1+T_2$ is compact.

A bounded linear operator is compact when the closure of the image of its unit ball is compact. Equivalently every bounded sequence has a subsequence whose images converge; on a Banach target it also suffices that the image of its unit ball be totally bounded.

For a finite-rank operator that image is bounded in a finite-dimensional subspace, whose closed bounded sets are compact. Thus every finite-rank operator is compact.

Compact. Its column indexed by $m$ has entries $n!/m!$ for $1\le n<m$, and zero otherwise. The sum of that column is $c_m=(1!+\cdots +(m-1)!)/m!\le 2/m$ for $m\ge 2$, by the elementary induction $1!+\cdots +k!\le 2k!$. For operators from ${\ell }^1$ the supremum of the absolute column sums is the operator norm: one inequality is the triangle inequality, and coordinate unit vectors give the reverse. Keeping just the first $N$ input columns is finite rank and its remainder norm is at most ${\sup }_{m>N}{c}_m\le 2/(N+1)$. Hence it converges to $T$ in norm, proving compactness.

Compact. This operator is $BD$, where $D$ is the compact diagonal operator in (a) and $B$ is the bounded backward shift in (b). The composition property applies. Equivalently its tails have the same vanishing-coefficient finite-rank approximation.