Proof of Elementary Properties of Bounded Linear Maps and Functionals on Real Inner Product Spaces

We use the notation and claims of Elementary Identities in a Real Inner Product Space and The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity in each of the spaces $E$, $F$, $G$, together with Elementary Order Arithmetic in an Ordered Field, Elementary Arithmetic in an Ordered Field and Properties of the Absolute Value in an Ordered Field for real numbers. For a linear map $T:E\to F$ we have $T0_{E}=0_{F}$: indeed $T0_{E}=T(0\cdot 0_{E})=0\cdot T0_{E}=0_{F}$ by claims 3 and 4 of Elementary Identities in a Vector Space and condition 2 of Linear Map; and $T(x-x')=Tx-Tx'$ by the two conditions of Linear Map and claim 5 of Elementary Identities in a Vector Space. For $T\in\mathcal{L}(E,F)$ let $B_{T}$ be the set of nonnegative real numbers $C$ with $|Tx|_{F}\le C|x|_{E}$ for every $x\in E$, as in Bounded Linear Maps and Bounded Linear Functionals on Real Inner Product Spaces, and the Operator Norm §operator-norm, so that $\lVert T\rVert=\inf B_{T}$.

Claim 1. Let $T\in\mathcal{L}(E,F)$ and $x\in E$. If $|x|_{E}=0$, then $x=0_{E}$ (Elementary Identities in a Real Inner Product Space §vanishing), so $Tx=0_{F}$ and $|Tx|_{F}=0=\lVert T\rVert\,|x|_{E}$. If $|x|_{E}>0$, then for every $C\in B_{T}$ we have $|Tx|_{F}\le C|x|_{E}$, hence $|Tx|_{F}\,|x|_{E}^{-1}\le C$ (claim 5 of Elementary Arithmetic in an Ordered Field with the positive multiplier $|x|_{E}^{-1}$); thus $|Tx|_{F}\,|x|_{E}^{-1}$ is a lower bound of $B_{T}$, so it is at most the greatest lower bound $\lVert T\rVert$, and multiplying by $|x|_{E}$ gives $|Tx|_{F}\le\lVert T\rVert\,|x|_{E}$. For the second assertion, a real $C$ with $0\le C$ and $|Tx|_{F}\le C|x|_{E}$ for all $x$ belongs to $B_{T}$, so $\lVert T\rVert=\inf B_{T}\le C$.

Claim 2. Let $S=\{|Tx|_{F}:x\in E,\ |x|_{E}\le 1\}$. It contains $|T0_{E}|_{F}=0$, and $\lVert T\rVert$ is an upper bound of $S$: for $|x|_{E}\le 1$, claim 1 gives $|Tx|_{F}\le\lVert T\rVert\,|x|_{E}\le\lVert T\rVert$, using $0\le\lVert T\rVert$ (a lower bound of $B_{T}$ is $0$, and the greatest lower bound is at least it). Now let $u$ be any upper bound of $S$; then $0\le u$ since $0\in S$. For $x\in E$ with $x\ne 0_{E}$, the vector $y=|x|_{E}^{-1}x$ satisfies $|y|_{E}=|x|_{E}^{-1}|x|_{E}=1$ by Elementary Identities in a Real Inner Product Space §homogeneity, so $|Ty|_{F}\le u$; and $Tx=T(|x|_{E}\,y)=|x|_{E}\,Ty$, so $|Tx|_{F}=|x|_{E}\,|Ty|_{F}\le u\,|x|_{E}$. For $x=0_{E}$ the inequality $|Tx|_{F}\le u|x|_{E}$ is trivial. Hence $u\in B_{T}$ and $\lVert T\rVert\le u$. So $\lVert T\rVert$ is the least upper bound of $S$ in the sense of Upper Bound and Least Upper Bound.

Claim 3. Suppose $T$ is bounded. For $x,x'\in E$, $d_{F}(Tx,Tx')=|Tx-Tx'|_{F}=|T(x-x')|_{F}\le\lVert T\rVert\,|x-x'|_{E}=\lVert T\rVert\,d_{E}(x,x')$ by claim 1, and $0\le\lVert T\rVert$; so $T$ is Lipschitz with constant $\lVert T\rVert$. A Lipschitz map is continuous on $E$ by A Lipschitz Map is Uniformly Continuous, and continuity on $E$ includes continuity at $0_{E}$ relative to $E$ by Continuous Map Between Metric Spaces. Finally suppose $T$ is continuous at $0_{E}$ relative to $E$. Taking $\varepsilon=1$ in Continuous Map Between Metric Spaces, there is $\delta>0$ such that $|y|_{E}=d_{E}(y,0_{E})<\delta$ implies $|Ty|_{F}=d_{F}(Ty,T0_{E})<1$. Let $x\in E$ with $x\ne 0_{E}$ and put $y=\tfrac{\delta}{2}|x|_{E}^{-1}x$; then $|y|_{E}=\tfrac{\delta}{2}|x|_{E}^{-1}|x|_{E}=\delta/2<\delta$ by Elementary Identities in a Real Inner Product Space §homogeneity (the scalar $\tfrac{\delta}{2}|x|_{E}^{-1}$ being positive and hence equal to its absolute value) and claim 8 of Elementary Order Arithmetic in an Ordered Field, so $|Ty|_{F}<1$, while $Ty=\tfrac{\delta}{2}|x|_{E}^{-1}Tx$ gives $|Ty|_{F}=\tfrac{\delta}{2}|x|_{E}^{-1}|Tx|_{F}$. Hence $|Tx|_{F}<2\delta^{-1}|x|_{E}$ by claim 10 of Elementary Order Arithmetic in an Ordered Field (multiply by the positive number $2\delta^{-1}|x|_{E}$). Together with the trivial case $x=0_{E}$, $C=2\delta^{-1}$ witnesses that $T$ is bounded.

Claim 4. Let $S,T\in\mathcal{L}(E,F)$ and $\lambda\in\mathbb{R}$. The maps $S+T$ and $\lambda T$ are linear, by the vector space axioms of $F$ applied pointwise. By The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §triangle and claim 1, $|(S+T)x|_{F}\le|Sx|_{F}+|Tx|_{F}\le(\lVert S\rVert+\lVert T\rVert)|x|_{E}$ for all $x$, so $S+T$ is bounded and, $\lVert S\rVert+\lVert T\rVert$ being nonnegative, claim 1 gives $\lVert S+T\rVert\le\lVert S\rVert+\lVert T\rVert$. By Elementary Identities in a Real Inner Product Space §homogeneity, $|(\lambda T)x|_{F}=|\lambda|\,|Tx|_{F}\le|\lambda|\,\lVert T\rVert\,|x|_{E}$, so $\lambda T$ is bounded with $\lVert\lambda T\rVert\le|\lambda|\,\lVert T\rVert$. If $\lambda=0$, then $\lambda T$ is the zero map $x\mapsto 0_{F}$, whose norm is $0$ (it satisfies $|0_{F}|_{F}=0\le 0\cdot|x|_{E}$, so $0\in B$ and $\inf B\le 0$, while $0\le\inf B$), so equality holds. If $\lambda\ne 0$, then $T=\lambda^{-1}(\lambda T)$ and the inequality just proved gives $\lVert T\rVert\le|\lambda^{-1}|\,\lVert\lambda T\rVert=|\lambda|^{-1}\lVert\lambda T\rVert$ (claim 4 of Properties of the Absolute Value in an Ordered Field applied to $\lambda\lambda^{-1}=1$), that is, $|\lambda|\,\lVert T\rVert\le\lVert\lambda T\rVert$; so $\lVert\lambda T\rVert=|\lambda|\,\lVert T\rVert$. If $\lVert T\rVert=0$, then $|Tx|_{F}\le 0$ for all $x$ by claim 1, so $Tx=0_{F}$ by Elementary Identities in a Real Inner Product Space §vanishing; conversely the zero map has norm $0$ as shown. The eight conditions of Vector Space over a Field for $\mathcal{L}(E,F)$ hold because they hold pointwise in $F$: for instance $((S+T)+R)x=(Sx+Tx)+Rx=Sx+(Tx+Rx)=(S+(T+R))x$; the zero vector is the zero map, which is bounded as shown, and the additive inverse of $T$ is $(-1)T$.

Claim 5. $S\circ T$ is linear (compose the conditions of Linear Map), and $|S(Tx)|_{G}\le\lVert S\rVert\,|Tx|_{F}\le\lVert S\rVert\,\lVert T\rVert\,|x|_{E}$ by claim 1 twice and claim 5 of Elementary Arithmetic in an Ordered Field; as $\lVert S\rVert\lVert T\rVert\ge 0$, claim 1 gives $\lVert S\circ T\rVert\le\lVert S\rVert\lVert T\rVert$. The identity map is linear and satisfies $|\mathrm{id}_{E}x|_{E}=|x|_{E}\le 1\cdot|x|_{E}$, so $\mathrm{id}_{E}\in\mathcal{L}(E)$ with $\lVert\mathrm{id}_{E}\rVert\le 1$. If $E\ne\{0_{E}\}$, pick $x\ne 0_{E}$; claim 1 gives $|x|_{E}\le\lVert\mathrm{id}_{E}\rVert\,|x|_{E}$ with $|x|_{E}>0$, so $1\le\lVert\mathrm{id}_{E}\rVert$.

Claim 6. $\ker T$ contains $0_{E}$ and is closed under sums and scalar multiples by linearity, so it is a linear subspace. If $(x_{m})$ is a sequence in $\ker T$ converging to $x\in E$, then for every $m$, $|Tx|_{F}=|Tx-Tx_{m}|_{F}=|T(x-x_{m})|_{F}\le\lVert T\rVert\,|x-x_{m}|_{E}$; given $\varepsilon>0$, choosing $m$ with $|x-x_{m}|_{E}<\varepsilon/(\lVert T\rVert+1)$ gives $|Tx|_{F}<\varepsilon$. As this holds for every $\varepsilon>0$, $|Tx|_{F}=0$ (if $|Tx|_{F}>0$, take $\varepsilon=|Tx|_{F}$), so $x\in\ker T$. Hence $\ker T$ is closed by Sequential Characterization of Closed Subsets of a Metric Space.

Claim 7. The eight conditions of Vector Space over a Field for $\mathbb{R}$ over itself, with addition as vector addition and multiplication as scalar multiplication, are the associativity and commutativity of addition, the existence of $0$ and of additive inverses, the associativity of multiplication, $1s=s$, and the two distributive laws, all axioms of the field $\mathbb{R}$. The map $\langle s,t\rangle=st$ satisfies conditions (a), (b), (c) of Real Inner Product Space §inner-product by commutativity, distributivity and associativity of multiplication. For (d): $0\le s\cdot s$ holds if $0\le s$ by claim 5 of Elementary Arithmetic in an Ordered Field (multiply $0\le s$ by $s$), and if $s\le 0$ then $0\le -s$ by claim 4 of Elementary Order Arithmetic in an Ordered Field and $s\cdot s=(-s)(-s)\ge 0$ likewise; and $s\cdot s=0$ forces $s=0$ because a field has no zero divisors ($s\ne 0$ would give $s=s^{-1}(s\cdot s)=0$). The norm of $s$ in this space is, by Real Inner Product Space §norm, the unique nonnegative $r$ with $r^{2}=s\cdot s$; since $0\le|s|$ and $|s|^{2}=s^{2}$ (claim 1 of Properties of the Absolute Value in an Ordered Field: $|s|$ is $s$ or $-s$), it is $|s|$, and the distance is $|s-t|=d_{\mathbb{R}}(s,t)$. Next, the two conditions defining a linear functional in Bounded Linear Maps and Bounded Linear Functionals on Real Inner Product Spaces, and the Operator Norm §functional are exactly the two conditions of Linear Map for a map from $E$ to this vector space $\mathbb{R}$; and the boundedness condition $|\ell(x)|\le C|x|_{E}$ of Bounded Linear Maps and Bounded Linear Functionals on Real Inner Product Spaces, and the Operator Norm §functional is the boundedness condition of Bounded Linear Maps and Bounded Linear Functionals on Real Inner Product Spaces, and the Operator Norm §bounded for $F=\mathbb{R}$, because the norm of $\ell(x)$ in $\mathbb{R}$ is $|\ell(x)|$. Consequently the set $B_{\ell}$ of Bounded Linear Maps and Bounded Linear Functionals on Real Inner Product Spaces, and the Operator Norm §functional coincides with the set $B_{T}$ of Bounded Linear Maps and Bounded Linear Functionals on Real Inner Product Spaces, and the Operator Norm §operator-norm for $T=\ell$, $F=\mathbb{R}$, and the two norms, both defined as $\inf$ of this set, agree.

Claim 8. By claim 7, a bounded linear functional $\ell$ on $E$ is an element of $\mathcal{L}(E,\mathbb{R})$ with $\mathbb{R}$ the real inner product space of claim 7, whose norm is the absolute value and whose distance is $d_{\mathbb{R}}$, and $\lVert\ell\rVert$ is its operator norm. Claims 1, 2, 3 and 6 applied with $F=\mathbb{R}$ are therefore exactly the assertions made, with $|\ell(x)|$ for $|Tx|_{F}$, $(\mathbb{R},d_{\mathbb{R}})$ for $(F,d_{F})$ and $\ker\ell=\{x\in E:\ell(x)=0\}$ for $\ker T$. For the last assertion let $z\in E$ and $\ell_{z}(x)=\langle x,z\rangle_{E}$. It is a linear functional by conditions (b) and (c) of Real Inner Product Space §inner-product, and $|\ell_{z}(x)|\le|z|_{E}|x|_{E}$ by The Cauchy-Schwarz Inequality in a Real Inner Product Space, so it is bounded and $\lVert\ell_{z}\rVert\le|z|_{E}$ by claim 1. If $z=0_{E}$ then $\ell_{z}=0$ and $\lVert\ell_{z}\rVert=0=|z|_{E}$. If $z\ne 0_{E}$, then claim 1 at $x=z$ gives $|z|_{E}^{2}=|\ell_{z}(z)|\le\lVert\ell_{z}\rVert\,|z|_{E}$, and dividing by $|z|_{E}>0$ gives $|z|_{E}\le\lVert\ell_{z}\rVert$.