Bounded Linear Operators and Quotient Spaces — Examples

By [L1], $\mathrm{\parallel }(x,y){\mathrm{\parallel }}_{\max }\ge \mathrm{\parallel }x\mathrm{\parallel }$ and $\mathrm{\parallel }(x,y){\mathrm{\parallel }}_{\max }\ge \mathrm{\parallel }y\mathrm{\parallel }$, so $\mathrm{\parallel }{\pi }_X(x,y)\mathrm{\parallel }\le \mathrm{\parallel }(x,y){\mathrm{\parallel }}_{\max }$ and $\mathrm{\parallel }{\pi }_Y(x,y)\mathrm{\parallel }\le \mathrm{\parallel }(x,y){\mathrm{\parallel }}_{\max }$. Thus both projections are bounded with operator norm at most $1$ by [L2] and [L3].

For $x\in X$ and $y\in Y$, [L1] gives $\mathrm{\parallel }{\iota }_X(x){\mathrm{\parallel }}_{\max }=\mathrm{\parallel }(x,0){\mathrm{\parallel }}_{\max }=\mathrm{\parallel }x\mathrm{\parallel }$ and $\mathrm{\parallel }{\iota }_Y(y){\mathrm{\parallel }}_{\max }=\mathrm{\parallel }(0,y){\mathrm{\parallel }}_{\max }=\mathrm{\parallel }y\mathrm{\parallel }$. So $\mathrm{\parallel }{\iota }_X\mathrm{\parallel }\le 1$ and $\mathrm{\parallel }{\iota }_Y\mathrm{\parallel }\le 1$, with equality whenever the relevant domain is nonzero. If $X\mathrm{\ne }\{0\}$, choose $u\in X$ with $\mathrm{\parallel }u\mathrm{\parallel }=1$. Then $\mathrm{\parallel }{\iota }_X(u){\mathrm{\parallel }}_{\max }=1$, and also $\mathrm{\parallel }{\pi }_X(u,0)\mathrm{\parallel }=1=\mathrm{\parallel }(u,0){\mathrm{\parallel }}_{\max }$, so $\mathrm{\parallel }{\iota }_X\mathrm{\parallel }=\mathrm{\parallel }{\pi }_X\mathrm{\parallel }=1$. If $X=\{0\}$, then both ${\iota }_X$ and ${\pi }_X$ are the zero operator, so both norms are $0$. The same argument with $Y$ gives the corresponding statements for ${\iota }_Y$ and ${\pi }_Y$.

Therefore ${\pi }_X\in \mathcal{B}(X\times Y,X)$, ${\pi }_Y\in \mathcal{B}(X\times Y,Y)$, ${\iota }_X\in \mathcal{B}(X,X\times Y)$, and ${\iota }_Y\in \mathcal{B}(Y,X\times Y)$, with the norms stated above.

On either space, $F$ and $B$ are linear by coordinatewise inspection. Also $\mathrm{\parallel }Fx{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $\mathrm{\parallel }Bx{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}$, so both operators are bounded with operator norm at most $1$ by [L2] and [L3].

The vectors $e_0=(1,0,0,\dots )$ and $e_1=(0,1,0,\dots )$ lie in both spaces. They satisfy $\mathrm{\parallel }F{e}_0{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }{e}_0{\mathrm{\parallel }}_{\mathrm{\infty }}=1$ and $\mathrm{\parallel }B{e}_1{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }{e}_0{\mathrm{\parallel }}_{\mathrm{\infty }}=1$, so $\mathrm{\parallel }F\mathrm{\parallel }\ge 1$ and $\mathrm{\parallel }B\mathrm{\parallel }\ge 1$.

Combining steps 1.1 and 2.1 gives $\mathrm{\parallel }F\mathrm{\parallel }=\mathrm{\parallel }B\mathrm{\parallel }=1$ on both $c_0$ and ${\mathrm{\ell }}^{\mathrm{\infty }}$.

The map ${\delta }_{x_0}$ is linear, and for every $f\in C(K,\mathbb{K})$, $\mathrm{\mid }{\delta }_{x_0}(f)\mathrm{\mid }=\mathrm{\mid }f(x_0)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. Hence ${\delta }_{x_0}$ is bounded with operator norm at most $1$ by [L2] and [L3].

Let $\mathbf{1}$ be the constant function $1$ on $K$. Then $\mathrm{\parallel }\mathbf{1}{\mathrm{\parallel }}_{\mathrm{\infty }}=1$ and $\mathrm{\mid }{\delta }_{x_0}(\mathbf{1})\mathrm{\mid }=1$, so $\mathrm{\parallel }{\delta }_{x_0}\mathrm{\parallel }\ge 1$. Therefore $\mathrm{\parallel }{\delta }_{x_0}\mathrm{\parallel }=1$.

Differentiation is linear on polynomials. For each $n\ge 1$, $\mathrm{\parallel }{p}_n{\mathrm{\parallel }}_{\mathrm{\infty }}=1$ on $[0,1]$, while $D{p}_n(x)=n{x}^{n-1}$ and therefore $\mathrm{\parallel }D{p}_n{\mathrm{\parallel }}_{\mathrm{\infty }}=n$.

If $D$ were bounded, [L1] would give a constant $C$ with $n=\mathrm{\parallel }D{p}_n{\mathrm{\parallel }}_{\mathrm{\infty }}\le C\mathrm{\parallel }{p}_n{\mathrm{\parallel }}_{\mathrm{\infty }}=C$ for every $n\ge 1$, impossible. Hence $D$ is unbounded.

Since $T$ is linear by [L1], the set $M=\ker T$ is a linear subspace by [L3], and certainly $M\subseteq \ker T$. Therefore [L5] gives a unique linear map $\bar{T}:X\mathrm{/}M\to Y$ with $\bar{T}(x+M)=Tx$. Its range is exactly $\mathrm{im}T$, so we may read it as a map into $\mathrm{im}T$.

If $\bar{T}(x+M)=0$, then $Tx=0$, so $x\in \ker T=M$ by [L2]. Hence $x+M=M$, the zero coset. Therefore $\bar{T}$ is injective by [L3].

If $Tx=Ty$, then $x-y\in \ker T=M$ by [L2], so $x+M=y+M$. Thus the formula $\mathrm{\parallel }Tx{\mathrm{\parallel }}_{\mathrm{q}\mathrm{u}\mathrm{o}\mathrm{t}}:=\mathrm{\parallel }x+M{\mathrm{\parallel }}_{X\mathrm{/}M}$ is well defined on $\mathrm{im}T$, using the quotient seminorm of [L4]. By construction, $\mathrm{\parallel }\bar{T}(x+M){\mathrm{\parallel }}_{\mathrm{q}\mathrm{u}\mathrm{o}\mathrm{t}}=\mathrm{\parallel }x+M{\mathrm{\parallel }}_{X\mathrm{/}M}$ for every coset, so $\bar{T}$ is an isometry onto $\mathrm{im}T$.

Steps 1.1, 2.1, and 2.2 show that $X\mathrm{/}\ker T$ is linearly isomorphic to the range of $T$, and isometric once the range is given the induced quotient norm.

By [L2], let $m:={\min }_{K}f$ and $M_0:={\max }_{K}f$, and put $c_0:=(M_0+m)\mathrm{/}2$. Then for every $x\in K$, $m\le f(x)\le M_0$, so $\mathrm{\mid }f(x)-c_0\mathrm{\mid }\le (M_0-m)\mathrm{/}2$. Hence $\mathrm{\parallel }f-c_0{\mathrm{\parallel }}_{\mathrm{\infty }}\le (M_0-m)\mathrm{/}2$.

For any real constant $c$, both $\mathrm{\mid }{M}_0-c\mathrm{\mid }$ and $\mathrm{\mid }m-c\mathrm{\mid }$ are bounded above by $\mathrm{\parallel }f-c{\mathrm{\parallel }}_{\mathrm{\infty }}$. Since the distance between $M_0$ and $m$ is $M_0-m$, at least one of those two numbers is at least $(M_0-m)\mathrm{/}2$. Therefore $\mathrm{\parallel }f-c{\mathrm{\parallel }}_{\mathrm{\infty }}\ge (M_0-m)\mathrm{/}2$ for every $c$.

Steps 1.1 and 2.1 give ${\inf }_{c\in \mathbb{R}}\mathrm{\parallel }f-c{\mathrm{\parallel }}_{\mathrm{\infty }}=(M_0-m)\mathrm{/}2$, so [L1] yields $\mathrm{\parallel }f+M\mathrm{\parallel }=\mathrm{osc}(f)\mathrm{/}2$. The constant subspace is closed because a uniform limit of constant functions is constant, so [L3] confirms that this is an honest norm on the quotient.

Let $(x^{(m)})$ be a Cauchy sequence in $c_0$ for the supremum norm, with coordinates $x^{(m)}=(x_n^{(m)}{)}_{n\ge 0}$. For each $n$, the scalar sequence $(x_n^{(m)}{)}_m$ is Cauchy because $\mathrm{\mid }{x}_n^{(m)}-x_n^{(\mathrm{\ell })}\mathrm{\mid }\le \mathrm{\parallel }{x}^{(m)}-x^{(\mathrm{\ell })}{\mathrm{\parallel }}_{\mathrm{\infty }}$. Since the scalars are real, let $z_n:={\lim }_m{x}_n^{(m)}$. Choosing $M_0$ with $\mathrm{\parallel }{x}^{(m)}-x^{(\mathrm{\ell })}{\mathrm{\parallel }}_{\mathrm{\infty }}<1$ for $m,\mathrm{\ell }\ge M_0$ and letting $\mathrm{\ell }\to \mathrm{\infty }$ gives $\mathrm{\mid }{z}_n\mathrm{\mid }\le \mathrm{\parallel }{x}^{(M_0)}{\mathrm{\parallel }}_{\mathrm{\infty }}+1$ for every $n$, so $z:=(z_n)$ is bounded. Given $\epsilon >0$, choose $M$ such that $\mathrm{\parallel }{x}^{(m)}-x^{(\mathrm{\ell })}{\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon \mathrm{/}2$ whenever $m,\mathrm{\ell }\ge M$. Letting $\mathrm{\ell }\to \mathrm{\infty }$ coordinatewise gives $\mathrm{\mid }{x}_n^{(m)}-z_n\mathrm{\mid }\le \epsilon \mathrm{/}2$ for every $n$ and every $m\ge M$, hence $\mathrm{\parallel }{x}^{(m)}-z{\mathrm{\parallel }}_{\mathrm{\infty }}\le \epsilon \mathrm{/}2$. Fix such an $m$ and choose $N$ with $\mathrm{\mid }{x}_n^{(m)}\mathrm{\mid }<\epsilon \mathrm{/}2$ for all $n\ge N$, since $x^{(m)}\in c_0$. Then $\mathrm{\mid }{z}_n\mathrm{\mid }<\epsilon$ for $n\ge N$, so $z\in c_0$. The displayed uniform estimate also gives $\mathrm{\parallel }{x}^{(m)}-z{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. Therefore $c_0$ is Banach.

The series defining $\phi$ converges absolutely and $\mathrm{\mid }\phi (z)\mathrm{\mid }\le {\sum }_{n=0}^{\mathrm{\infty }}{2}^{-n-1}\mathrm{\mid }{z}_n\mathrm{\mid }\le \mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}$ for every $z\in c_0$, so $\phi$ is a bounded linear functional with $\mathrm{\parallel }\phi \mathrm{\parallel }\le 1$. If $z^{(m)}\in H$ and $z^{(m)}\to z$ in $c_0$, then $\mathrm{\mid }\phi (z)\mathrm{\mid }=\mathrm{\mid }\phi (z-z^{(m)})\mathrm{\mid }\le \mathrm{\parallel }z-z^{(m)}{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$, so $z\in H$. Thus $H$ is a closed linear subspace. Also $\phi (x)=1$.

For every $y\in H$ one has $\phi (x-y)=1$, hence $1=\mathrm{\mid }\phi (x-y)\mathrm{\mid }\le \mathrm{\parallel }x-y{\mathrm{\parallel }}_{\mathrm{\infty }}$ by step 2.1. Therefore $\mathrm{dist}(x,H)\ge 1$.

For $N\ge 1$, define $z^{(N)}\in c_0$ by $z_n^{(N)}:=(1-2^{-N}{)}^{-1}$ for $0\le n<N$ and $z_n^{(N)}:=0$ for $n\ge N$. Then $\phi (z^{(N)})=(1-2^{-N}{)}^{-1}{\sum }_{n=0}^{N-1}{2}^{-n-1}=1$, so $y^{(N)}:=x-z^{(N)}$ lies in $H$, and $\mathrm{\parallel }x-y^{(N)}{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }{z}^{(N)}{\mathrm{\parallel }}_{\mathrm{\infty }}=(1-2^{-N}{)}^{-1}\to 1$. Hence $\mathrm{dist}(x,H)\le 1$.

Steps 3.1 and 3.2 give $\mathrm{dist}(x,H)=1$. Suppose some $y\in H$ satisfied $\mathrm{\parallel }x-y{\mathrm{\parallel }}_{\mathrm{\infty }}=1$, and put $z:=x-y$. Then $\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}=1$ and $\phi (z)=1$ by step 2.1. Also $1=\phi (z)={\sum }_{n=0}^{\mathrm{\infty }}{2}^{-n-1}{z}_n\le {\sum }_{n=0}^{\mathrm{\infty }}{2}^{-n-1}\mathrm{\mid }{z}_n\mathrm{\mid }\le {\sum }_{n=0}^{\mathrm{\infty }}{2}^{-n-1}\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}=1.$

So equality holds throughout in step 4.1, forcing $\mathrm{\mid }{z}_n\mathrm{\mid }=1$ and $z_n\ge 0$ for every $n$. That contradicts $z\in c_0$, since a sequence converging to $0$ cannot have all coordinates equal to $1$ in modulus.

Thus $H$ is a closed subspace of the Banach space $c_0$, the distance from $x$ to $H$ is $1$, and no $y\in H$ attains it. This refutes the claim.

By construction, $P$ is linear, $P^2=P$, and $\mathrm{ran}(P)=M$. So $W:=\ker P$ satisfies $c_0=M\oplus W$ algebraically: every vector decomposes as $Px+(x-Px)$, and the intersection is trivial because $P$ acts as the identity on $M$ and vanishes on $W$.

The projection $P$ is not bounded for the supremum norm. Indeed, $u_n:=e_n\mathrm{/}n$ satisfies $\mathrm{\parallel }{u}_n{\mathrm{\parallel }}_{\mathrm{\infty }}=1\mathrm{/}n\to 0$, while $P(u_n)=e_0$ for every $n\ge 1$, so $\mathrm{\parallel }P(u_n){\mathrm{\parallel }}_{\mathrm{\infty }}=1$. No bounded linear map can behave this way at $0$.

Suppose, toward a contradiction, that $W$ were a topological complement of $M$. Then [L2] gives a bounded projection onto $M$ along $W$, and that projection is unique because the decomposition $x=m+w$ with $m\in M$, $w\in W$ determines the projection value $m$ pointwise. But $P$ already has exactly that range and kernel by step 1.1, so the bounded projection would have to equal $P$, contradicting step 2.1.

Therefore $c_0=M\oplus W$ is an algebraic decomposition that is not a topological direct sum. This refutes the claim.