Lecture 25

[[lecture-data]]

2024-10-30

5. Chapter 5

Note

Will use the same norm symbol and context determines which norm we are using

Recall that if $T:V\to W$ is a linear function on NLS and our theorem/definition for continuity for linear functions - ie, if ${\sup }_{x\in V\ne 0}\frac{||Tx||}{||x||}<M,M\in \mathbb{R}$then $T$ is continuous. And further, the least such bound $M$ is a lipschitz constant. ie, continuous linear functions are lipschitz continuous.

Example

$\frac{d}{dx}:c^1[a,b]\to c^0[a,b]$ is a linear operator.

Let $||f||={\max }_{t\in [a,b]}|f(t)|$. This satisfies the conditions of the norm:

• positive unless $f$ is uniformly 0
• any scalar comes out with an absolute value around it
• satisfies the triangle inequality

Unfortunately, this is not continuous. Take $f(t)=t^k,k>0$. Then suppose we have some $t$ such that $||t^k||=1$. But then $\left|\left|\frac{d}{dt}{t}^k\right|\right|=||k{t}^k||=k||t^k||=k$ and so $\frac{||\frac{d}{dt}{t}^k||}{||t^k||}$ is unbounded.

Operator Norm

Let $V,W$ be NLS. Let $T:V\to W$ be a linear function. If $T$ is continuous, then the operator norm of $T$ is defined as $||T||:={\sup }_{x\in V\ne 0}\frac{||Tx||}{||x||}$

(see operator norm)

Note: The Operator Norm is indeed a Norm

Let $V,W$ be NLS and suppose that $T,S:V\to W$ both linear, and $\alpha ,\beta \in K$. Define function $\alpha T+\beta S:V\to W$. Then for all $x\in V$, we have that $[\alpha T+\beta S](x)=\alpha T(x)+\beta S(x)$ ie we calculate the output of these functions pointwise.

$\alpha T+\beta S$ is linear (just think about it).

And if $T,S$ are also continuous then

1. $||T||=0⟺T=0$ function
2. $||\alpha T||=|\alpha |\cdot ||T||$
3. $||T+S||\le ||T||+||S||$

ie, the operator norm can create a normed linear space on the space of continuous linear functions. 🤯

$T\ne 0$, then there exists fomr $x\ne 0$ such that $Tx\ne 0$. Then $\frac{||Tx||}{||x||}>0$. Thus the $\sup$ must be greater than 0.

(1) If

(2) ${\sup }_{x\in V\ne 0}\frac{||\alpha T(x)||}{||x||}={\sup }_{x\ne 0}\frac{|\alpha |||T(x)||}{||x||}=|\alpha |{\sup }_{x\ne 0}\frac{||T(x)||}{||x||}$ where we can take $\alpha$ out by homogeneity of $W$‘s norm

(3) note that

\sup_{x \neq 0} \frac{\lvert \lvert [T+S ](x)\rvert \rvert }{\lvert \lvert x \rvert \rvert } \leq \sup_{x \neq 0} \frac{\lvert \lvert T(x)\rvert \rvert + \lvert \lvert S(x) \rvert \rvert }{\lvert \lvert x \rvert \rvert } \\ & \leq \sup_{x \neq 0} \frac{\lvert \lvert T(x)\rvert \rvert }{\lvert \lvert x \rvert \rvert } + \sup_{x \neq 0} \frac{\lvert \lvert S(x)\rvert \rvert }{\lvert \lvert x \rvert \rvert } \end{aligned}$$

Continuous Linear Function Space

Let $V,W$ be NLS. Then define $B(V,W)$ as the normed linear space whose elements are continuous linear functions $V\to W$, and whose norm is the operator norm.

(see continuous linear function space)

Dual Space

Let $V$ be a NLS. Then $B(V,K)$ of continuous linear functionals is called the dual space $V^{\ast }$

(see dual space)

Observations

1. If $T\in B(V,W)$, then for all $x\in V$ we have $||Tx||\le ||T||\cdot ||x||$. This result is trivial by definition of the operator norm
2. If $T\in B(V,W)$ and $S\in B(W,U)$ then $S\circ T\in B(V,U)$. And $||S\circ T||\le ||S||\cdot ||T||$ (careful, each of these norms are different!).
3. Since for all $x\in V,||[S\circ T](x)||=||S(T(x))||\le ||S||\cdot ||Tx||\le ||S||\cdot ||T||\cdot ||x||$ by the above result (1). But then we can see that $\frac{||[S\circ T](x)||}{||x||}\le ||S||\cdot ||T||$ - and since $T,S$ are continuous this is bounded (since this holds for any $x$). And also, $||S\circ T||\le ||S||\cdot ||T||$

Note

Suppose $A\in M_{m,n}(K)$ and $B\in M_{m,n}(K)$ thought of as functions "$A$" and "$B$" both from $K^n\to K^m$. Suppose $C\in M_{p,m}(K)$ is associated with function "$C$" from $K^m\to K^p$. Then $\text{"CA"}=C\circ A=CA$ and "$\alpha A+\beta B$" is $\alpha$"$A$" + $\beta$"$B$" is $\alpha A+\beta B$.

Theorem

Suppose $V,W$ are NLS where $\dim (V)<\infty$. Then $T:V\to W$ is continuous.

Proof

First,

• Let $b_1,\dots ,b_n$ be a basis for $V$. Let $||\cdot |{|}^{\prime }$ be defined $V\to \mathbb{R}$ as such: $\forall y\in V$, there exist unique coefficients $alph{a}_1,\dots ,{\alpha }_n\in K$ such that $y={\sum }_{i=1}^n{\alpha }_i{b}_i$. Let $||y|{|}^{\prime }={\sum }_{i=1}^n|{\alpha }_i|$ (ie, the $L^1$ norm on the column vector space, which is the same as the vector space).
• Let $E:={\max }_{i\in \{1,\dots ,n\}}||T{b}_i|{|}_W$.
• And let $M>0$ be such that for all $y\in V$ we have $||y|{|}^{\prime }\le M||y|{|}_V$ (by norm equivalence).

Then for all $x\in V$, say ${\beta }_1,\dots ,{\beta }_n\in K$ such that $x={\sum }_{i=1}^n{\beta }_i{b}_i$, then

\lvert \lvert Tx \rvert \rvert_{w} &= \left\lvert \left\lvert T \sum \beta_{i} b_{i} \right\rvert \right\rvert _{W}\\ &= \left\lvert \left\lvert \sum \beta_{i} T b \right\rvert \right\rvert _{W} \\ &\leq \sum \beta_{i}\lvert \lvert T b_{i} \rvert \rvert_{W} \\ & \leq \sum_{i=1}^n \lvert \beta_{i} \rvert E \\ &= \lvert \lvert x \rvert \rvert ' E \\ \leq ME \lvert \lvert x \rvert \rvert _{v} \end{aligned}$$ So we can see that $\frac{\lvert \lvert Tx \rvert \rvert_{W}}{\lvert \lvert x \rvert \rvert_{W}} \leq ME$ ie $T$ is [[Concept Wiki/continuity for linear functions\|continuous]]

(see linear functions with finite dimensional domain are continuous)

Next time: thinking about norms and things for continuous linear spaces