LEC CAL1 3

Topological properties of linear maps

See Pugh (2015) 5§1.

Definition 172.1(Definition).

If $V,W$ are normed spaces then the operator norm of a linear transformation $T:V\to W$ is

$$\Vert T\Vert :=\sup \left\{\frac{|T\mathbf{v}|}{|\mathbf{v}|}:\mathbf{v}\ne 0\right\}.$$

If $\Vert T\Vert <\infty$, we say $T$ is bounded.

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Remark 172.2(Remark).

For a bounded linear operator $T$ on a normed space the three quantities

$$\underset{|x|\le 1}{\sup }|Tx|,\underset{|x|=1}{\sup }|Tx|,\underset{x\ne 0}{\sup }\frac{|Tx|}{|x|}$$

are equal, even in infinite dimensions (however, note that the supremum need not be attained in infinite dimensions, since the unit ball is not compact).

It is clear from the definition that $|T\mathbf{v}|\le \Vert T\Vert |\mathbf{v}|$ for all $\mathbf{v}\in V$.

Proposition 172.3(Properties of the operator norm, Rudin (1976) 9.7).

1. $\Vert A+B\Vert \le \Vert A\Vert +\Vert B\Vert$,
2. $\Vert AB\Vert \le \Vert A\Vert \Vert B\Vert$,
   whenever the matrices on the left are well defined.

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Proposition 3 (1) turns $\mathcal{L}({\mathbb{R}}^n,{\mathbb{R}}^m)$ with the distance between $A$ and $B$ defined by $\Vert A-B\Vert$ into a metric space.

Proposition 172.4(Rudin (1976) 9.8).

Let $\Omega$ be the set of invertible linear operators on ${\mathbb{R}}^n$. If $A\in \Omega$ and $B\in \mathcal{L}({\mathbb{R}}^n)$, and

$$\Vert B-A\Vert <\frac{1}{\Vert A^{-1}\Vert },$$

then $B\in \Omega$. Thus, $\Omega$ is an open subset of $\mathcal{L}({\mathbb{R}}^n)$. Further, the mapping $A\mapsto A^{-1}$ is continuous on $\Omega$ and obviously injective.

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Remark 172.5(Remark).

The operator norm of a matrix is bounded above by its Euclidean norm:

$$\begin{array}{rl}& |A\mathbf{x}{|}^2=\sum _i{\left(\sum _j{a}_{ij}{c}_j\right)}^2\le \sum _i\left(\sum _j{a}_{ij}^2\cdot \sum _j{c}_j\right)=|\mathbf{x}{|}^2\sum _{i,j}{a}_{ij}^2\\ & ⟹\Vert A\Vert \le {\left(\sum _{i,j}{a}_{ij}^2\right)}^{1/2}.\end{array}$$

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Proposition 172.6(Proposition).

Let $T:V\to W$ be a linear transformation from one normed linear space to another. The following are equivalent:

1. $\Vert T\Vert \le \infty$.

2. $T$ is uniformly continuous.

3. $T$ is continuous.

4. $T$ is continuous at the origin.

5. $T$ is continuous at some $\mathbf{v}\in V$.

Proof.

$(1⟹4)$: For $ϵ>0$, take $\delta =ϵ/\Vert T\Vert$.
$(4⟹1)$: Let $ϵ=1$, and $\delta >0$ be obtained by the continuity of $T$ at $0$. For all $\mathbf{v}\ne 0$,

$$|T\mathbf{v}|=\frac{2|\mathbf{v}|}{\delta }\left|T\left(\frac{\delta v}{2|\mathbf{v}|}\right)\right|\le \frac{2}{\delta }|\mathbf{v}|.$$

Thus, $\Vert T\Vert <\infty$.□

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Proposition 172.7(Proposition).

Every linear transformation $T:{\mathbb{R}}^n\to W$ is continuous and every isomorphism $T:{\mathbb{R}}^n\to W$ is a homeomorphism.

Corollary 172.8(Corollary).

When working with finite dimensional normed spaces, all linear transformations are continuous and all isomorphisms are homeomorphisms.

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Equivalence of norms

We will prove that all norms on a finite dimensional normed linear space are equivalent.

Lemma 172.9(Lemma).

A norm is a continuous function.

Proof.

Consider a norm$\Vert \cdot \Vert :V\to \mathbb{R}$. We will prove that $\Vert \cdot \Vert$ is uniformly continuous, which implies that it is also continuous. Let $ϵ>0$. We need to find a $\delta$ such that $\Vert {\mathbf{v}}_1-{\mathbf{v}}_2\Vert <\delta ⟹|\Vert {\mathbf{v}}_1\Vert -\Vert {\mathbf{v}}_2\Vert |<ϵ$. But, we know that $|\Vert {\mathbf{v}}_1\Vert -\Vert {\mathbf{v}}_2\Vert |\le \Vert {\mathbf{v}}_1-{\mathbf{v}}_2\Vert$. Thus, choosing $\delta =ϵ$ will do.□

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Lemma 172.10(Lemma).

Suppose $(V,\Vert \cdot \Vert )$ is finite dimensional normed linear space. Consider $S=\{\mathbf{x}\in V:\Vert \mathbf{x}\Vert =1\}$. $S$ is closed in $(V,\Vert \cdot \Vert )$.

Since $\Vert \cdot \Vert$ is continuous, and singleton sets are closed, $\Vert \cdot {\Vert }^{-1}(\{1\})$ is closed in $V$ ($f^{-1}$ of a closed set is closed when $f$ is a continuous function).

Proposition 172.11(Proposition).

All norms on a finite dimensional vector space are equivalent.

Proof.

Let$\Vert \cdot \Vert$ be a norm on a finite dimensional vector space $V$ over $\mathbb{R}$. We will show that $\Vert \cdot \Vert$ is equivalent to $\Vert \cdot {\Vert }_{\infty }$, and the theorem will follow from the transitivity of the equivalence of norms.

Note that the infinity norm is not unique, and depends upon a choice of basis. Fix a basis $\{{\mathbf{e}}_1,{\mathbf{e}}_2,\dots ,{\mathbf{e}}_n\}$ for $V$, and define the infinity norm with respect to this basis. Let $\mathbf{x}\in V$. Then,

$$\begin{array}{rl}\Vert \mathbf{x}\Vert & =\Vert \sum x_i{\mathbf{e}}_i\Vert \\ & \le \sum \Vert x_i{\mathbf{e}}_i\Vert \\ & =\sum |x_i|\Vert {\mathbf{e}}_i\Vert \\ & \le \Vert \mathbf{x}{\Vert }_{\infty }\sum \Vert {\mathbf{e}}_i\Vert \end{array}$$

So, we have found a constant $c_2=\sum \Vert {\mathbf{e}}_i\Vert >0$ such that $\Vert \cdot \Vert \le c_2\Vert \cdot {\Vert }_{\infty }$.

To get $c_1$, consider the isomorphism $\phi :{\mathbb{R}}^n\to (V,\Vert \cdot {\Vert }_{\infty })$. We have shown that this is a homeomorphism, so the image $\phi (S)$ in $(V,\Vert \cdot {\Vert }_{\infty })$ of the unit sphere in ${\mathbb{R}}^n$ is compact. Two ways to go from here:

Way 1: Injectivity implies $0\notin \phi (S)$. $\phi (S)$ being compact also implies $\phi (S)$ is bounded, so $\Vert \mathbf{u}{\Vert }_{\infty }\le d$ for all $\mathbf{u}\in \phi (S)$ for some $d>0$. Consider the identity map $\psi :(V,\Vert \cdot {\Vert }_{\infty })\to (V,\Vert \cdot \Vert )$. It is continuous, and hence $S^{\prime }=\psi (\phi (S))$ is compact. Injectivity again implies $0\notin S^{\prime }$. Now, $\{0\}$ and $S^{\prime }$ are disjoint compact sets, and hence the distance between them is nonzero. Thus, $c\le \Vert \mathbf{v}\Vert$ for all $\mathbf{v}\in S^{\prime }$ for some $c>0$.

Now, consider $\mathbf{v}\in V$. There exists $\lambda >0$ and ${\mathbf{v}}^{\prime }\in \phi (S)$ such that
$\mathbf{v}=\lambda {\mathbf{v}}^{\prime }$. $\Vert \mathbf{v}{\Vert }_{\infty }=\lambda \Vert {\mathbf{v}}^{\prime }{\Vert }_{\infty }\le \lambda d$, and $\Vert \mathbf{v}\Vert =\lambda \Vert {\mathbf{v}}^{\prime }\Vert \ge \lambda c$. Thus,

$$\begin{array}{r}\frac{c}{d}\Vert \mathbf{v}{\Vert }_{\infty }\le \Vert \mathbf{v}\Vert .\end{array}$$

Way 2: Same as the previous one, but instead consider the real function $V\to \mathbb{R}$ defined by $\mathbf{v}\mapsto \Vert \mathbf{v}\Vert$. This is continuous, and by the extreme value theorem must attain a minimum value on $\phi (S)$, which cannot be zero since any norm of a nonzero vector can’t be zero. Proceed as we did before.□

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Heine Borel in finite dimensional normed linear spaces

The Heine Borel theorem in ANA1 was stated for ${\mathbb{R}}^k$. It shouldn’t be surprising that it in fact holds in all finite dimensional normed linear spaces.

Theorem 172.12(Theorem).

Let $(V,\Vert \cdot \Vert )$ be a finite dimensional normed linear space. Let $X\subseteq V$. Then, $X$ is compact $⟺$ $X$ is closed and bounded in $V$.

Proof.

Compact$⟹$ closed and bounded: true in any metric space.

For the converse, suppose $n=\dim V$. Choose a basis $v_1,\dots ,v_n$ and let $T$ be the natural isomorphism from ${\mathbb{R}}^n$ to $V$ for this basis. This induces a norm $\Vert \cdot {\Vert }^{\prime }$ on $V$. By Proposition 11, $\Vert \cdot \Vert$ and $\Vert \cdot {\Vert }^{\prime }$ are equivalent. Suppose there exist $c_1$ and $c_2$ such that

$$c_1\Vert \cdot {\Vert }^{\prime }\stackrel{1}{\le }\Vert \cdot \Vert \stackrel{2}{\le }{c}_2\Vert \cdot {\Vert }^{\prime }.$$

If a sequence in ${\mathbb{R}}^n$ converges, the images of the sequence must converge in $(V,\Vert \cdot {\Vert }^{\prime })$, and thus they must also converge in $(V,\Vert \cdot \Vert )$ by the second inequality from above. Thus, by the sequence criterion for continuity, $T$ is continuous. Similarly, by the first inequality, $T^{-1}$ is continuous.

Now, if $K\subseteq V$ is closed and bounded then $T^{-1}(K)\subseteq {\mathbb{R}}^n$ is closed and bounded, hence compact by the usual Heine Borel in ${\mathbb{R}}^n$. Since $T$ is continuous, $K=T(T^{-1}(K))$ is compact.□

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Obviously, this is NOT true in general for infinite dimensional normed linear spaces.

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References

Pugh, C. C. (2015). Real Mathematical Analysis. Springer International Publishing. https://doi.org/10.1007/978-3-319-17771-7

Rudin, W. (1976). Principles of Mathematical Analysis (3d ed). McGraw-Hill.