---- Let $V$ and $W$ be [[norm|normed]] [[vector space|vector spaces]] over a common base [[field]] $\mathbb{F}$ (either $\mathbb{R}$ or $\mathbb{C}$). > [!definition] Definition. ([[operator norm]]) > Suppose $T:V \to W$ is a [[linear map]]. We define a [[norm]] on $\text{Hom}(V,W)$ as $\|T\|:=\sup_{\|v \| \leq 1} \{ \|Tv\|: v \in V \}.$ We say $T$ is **bounded** if $\|T\|<\infty$. The collection $\mathcal{B}(V,W)$ of bounded linear maps $V \to W$ forms a [[Banach space]] with respect to this [[norm]] iff $W$ is [[Banach space|Banach]], and in general coincides with the set $\text{Hom}\big(V,W\big) \cap C(V,W)$ of [[continuous]] [[linear map|linear maps]] $V \to W$ (per [[characterizing continuity of linear maps]]). ^definition - [ ] can extend to discuss bounded multilinear maps > [!intuition] > In light of the equivalence below, we often think of $\|T\|$ as the smallest number such that $\|Tv\| \leq \|T\| \|v\|$ for all $v$ in the domain of $T$. ^intuition > [!equivalence] > 1. $\|T\|=\sup_{\|v\|<1} \|Tv\|$ > 2. $\|T\|=\sup_{\|v\|=1} \|Tv\|$ > 3. $\|T\|=\inf \{ c \in [0, \infty): \|Tv\| \leq c \|v\| \text{ for all } v \in V \}$ > 4. $\|T\|= \sup_{v \neq 0} \frac{\|Tv\|}{\|v\|}.$ ^equivalence > [!proof]- Proof of Equivalence. > Write $A=\sup_{\|v\| \leq 1}\|Tv\|$ (our original definition), $A_{1}=\sup_{\|v\|<1}\|Tv\|$, $A_{2}=\sup_{\|v\|=1}\|Tv\|$, $A_{3}=\inf\{ c \in [0, \infty): \|Tv\| \leq c \|v\| \ \forall v \in V \}$, $A_{4}=\sup_{v \neq 0} \frac{\|Tv\|}{\|v\|}.$ We want to show $A=A_{1}=A_{2}=A_{3}=A_{4}$. > > **$A=A_{2}$.** Since $\{ \|Tv\|: \|v\|=1 \} \subset \{ \|Tv\|: \|v\| \leq 1 \}$, clearly $A_{2} \leq A$. For the reverse, we will show that for all $a \in \{ \|Tv\|: \|v\| \leq 1 \}$, $a \leq A_{2}$. If $a$ comes from $v=0$ then the result is obvious, so suppose $v \in V$ with $0<\|v\| \leq 1$. Then $v$ is a nonzero scaling of some $u$ on the sphere, namely, $v=\|v\|u$ for some $u$ with $\|u\| = 1$. Then $\|Tv\|= \|Tu\| \cdot \|v\| \leq \|Tu\|\leq A_{2}$. Passing to $\sup$ on the LHS gives the desired inequality. > > **$A=A_{1}.$** Since $\{ \|Tv\|: v < 1 \} \subset \{ \|Tv\|:v \leq 1 \}$, clearly $A_{1} \leq A$. > > Write $v=u \|v\|$ with $\|u\|=1$. Then $\begin{align} > A_{1}= \sup_{\| u \|=1, \|v\|<1} \|T (u \|v\|)\|&=\sup_{\|u\|=1, \|v\|<1} \|v\| \| Tu\| \\ > &= \sup_{\|u\|=1} \underbrace{ \sup _{\|v\|<1} \|v\| \|Tu\| }_{ =\| Tu\| } > \end{align}$ > where we have used the properties in [[supremum]]. We see that this just equals $A_{2}$, which we've already shown equals $A$. > > > **$A=A_{4}.$** We claim $\{ \|Tv\|: \|v\|=1 \}$ $=$ $\left\{ \frac{\|Tv\|}{\|v\|} : v \neq 0\right\}$, from which the result follows. The inclusion $\subset$ is immediate. For the reverse inclusion $\supset$, consider $v \neq 0$. Write $v=\|v\|u$ for some $u$ with $\|u\|=1$, then $\frac{\|Tv\|}{\|v\|}=\frac{\|v\| \cdot \big\|T u\big\|}{\|v\|}=\|Tu\|$. > > **$A=A_{3}$.** One inequality is easy: assuming $v \neq 0$, set $c':=\sup_{v \neq 0} \frac{\|Tv\|}{\|v\|}$, then $\|Tv\| \leq c' \|v\|$ for all $v \in V$, whence $\inf \{ c \in [0, \infty): \|Tv\| \leq c \|v\| \text{ for all } v \in V \} \leq c' \leq A_{4}$ > and we have just shown $A_{4}=A$. It follows that $A \geq A_{3}$. For the reverse inequality, let $c \in [0, \infty)$ satisfy $\|Tv\| \leq c \|v\|$ for all $v \in V$. Then for every $v$ with $\|v\| \leq 1$, $\|Tv\| \leq c$. Hence $\sup_{\|v\| \leq 1} \|Tv\| \leq c$. Since $c$ was arbitrary, taking the [[infimum]] on the RHS preserves the inequality. > > [!basicproperties] > - (*submultiplicativity*) $\|S \circ T\|\leq \|S\| \|T\|$ > - (*embedding*) BLO $T$ maps isomorphically (in $\mathsf{Top}$) onto its range if and only if there exists $r>0$ such that $\|Tx\|\geq r \|x\|$ for all $x \in X$. > > > [!proof]- Proof. > > Suppose $T$ is an [[isomorphism]] in $\mathsf{Top}$ onto its image, say with $T ^{-1} : \operatorname{im }T \to V$ [[operator norm|bounded]]. If $\|T ^{-1}\|=0$ then $V=W=0$ and $T=0$ and any $r$ will do. So assume $\|T^{-1}\| \neq 0$. Then $\|x\|=\|T^{-1} Tx\| \leq \|T^{-1}\| \|T x\|$ implying $\|Tx\| \geq \frac{\|x\|}{\|T ^{-1}\|}$. Define $r:= \frac{1}{\|T^{-1}\|}$ to get the result. > > > > Conversely suppose there is $r>0$ such that $\|Tx\| \geq r \|x\|$ for all $x \in X$. Then $\operatorname{ker }T=(0)$, so $T$ is an algebraic isomorphism onto its image. For $y \in \operatorname{im }T$, write $y=Tx$. Then $\|T^{-1}y\|=\|x\| \leq \frac{1}{r} \|Tx\|=\frac{1}{r} \|y\|.$ > > Hence $\|T ^{-1}\| \leq \frac{1}{r}$. > > > [!basicexample] > Let $C([0,3])$ be the normed vector space of continuous functions $[0,3] \to \mathbb{F}$, with $\|f\|=\sup_{[0,3]}|f|$. Define $T:C([0,3]) \to C([0,3])$ by $(Tf)(x)=x^{2}f(x).$ Then $T$ (manifestly a [[linear operator]]) is bounded; specifically, $\|T\|=9$. ^basic-example > [!justification] Justification of $B(V,W)$ Banach when $W$ is > We show that $B(V,W)$ is a [[Banach space]] iff $W$ is a [[Banach space]] (there is no requirement on $V$). > > Suppose $W$ is a [[Banach space]]. Let $T_{1},T_{2},\dots$ be a [[Cauchy sequence]] in $B(V,W)$. If $f \in V$, then the equivalence gives $\| T_{j}f - T_{k}f\| \leq \|T_{j}-T_{k}\| \| f\|,$ > and from this inequality it follows that $T_{1}f,T_{2}f,\dots$ is a [[Cauchy sequence]] in $W$. Since $W$ is [[complete]], this sequence converges to some element of $W$ which we denote $Tf$. Letting $f$ range over $V$, this defines a [[linear map|linear map]] $T:V \to W$, which we claim lives in $B(V,W)$ and which we claim $T_{1},T_{2},\dots$ converges to. To begin, $T \in B(V,W)$ because for all $f$, $\|Tf\| \leq \sup_{k \in \mathbb{N}} \|T_{k}f\|\leq \sup_{k \in \mathbb{N}} ( \|T_{k}\| \| f\|) \leq \|f\| \sup_{k \in \mathbb{N}} \|T_{k}\| < \infty, $ > where the last inequality holds because every [[Cauchy sequence]] in a [[normed vector space]] is bounded (see properties in [[Cauchy sequence]]). Finally, we have to show $T=\lim (T_{k})$, which we will do by showing $\lim \|T_{k}-T\|=0$. Fix $\varepsilon>0$. Let $n$ be large enough that $\|T_{j}-T_{k}\|<\varepsilon$ for all $j,k \geq n$. Suppose $j \geq n$ and suppose $f \in V$. Then for all $j,k \geq n$, we have $\|T_{j}f-T_{k}f\| \leq \|T_{j}-T_{k}\| \|f\| < \varepsilon \|f\|$. It follows that > > ${ \|(T_{j}-T)f\| }_{ } \leq \lim_{k \to \infty} \|T_{j}f - T_{k}f\| < \varepsilon \|f\|.$ > Now take sup over $f$ with $\|f\|= 1$ to get $\|T_{j}-T\|<\varepsilon$, completing the proof. > [^1]: For example, we can construct the sequence $(f_{k})$ by choosing each $f_{k}$ so that $\|Tf_{k}\|>k$ (since the [[supremum]] is $\infty$, for any real number $r \in \mathbb{R}$ there exists $f$ such that $\|Tf\|>r$). ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```