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This proves property \(d)\) of the definition of a matrix norm. Properties \(a)\) through \(c)\) are also true, since the Frobenius norm is derived from the Euclidean norm.
**Remark 43**.: Unfortunately, there are situations where the Frobenius norm, also known as the Hilbert-Schmidt norm, is not convenient to use. In practice often the _operator norm_ is used. To understand what we mean by operator norm, let us keep in mind that we want to measure the “size” of \(A\) and compare it with \(||x||\) and \(||Ax||\); thus what we need is the inequality
\[||Ax||\leq||A||||x||\;\;\mbox{or}\;\;\frac{||Ax||}{||x||}\leq||A||\;\;\mbox{ for}\;\;x\neq 0.\]
The expression \(\frac{||Ax||}{||x||}\) can be rewritten if we consider normalizing the vector \(x\), which means we get a new vector \(\overline{x}\) by dividing the vector \(x\) by its length. Thisis done by setting \(\overline{x}=\dfrac{x}{||x||}\). Now, we only need to consider the vectors on the unit sphere as opposed to looking at all nonzero vectors \(x\). Thus,
\[\dfrac{||Ax||}{||x||}=\dfrac{1}{||x||}||Ax||=||(\dfrac{x}{||x||})||=||A(\overline{ x})||.\]
Furthermore,
\[\max_{x\neq 0}\dfrac{||Ax||}{||x||}=\max_{||\overline{x}||=1}||A(\overline{x})||\]
always exists, and there is a particular vector \(u\) for which \(||Au||\) is maximal. The operator norm \(||A||\) is then defined to be
\[||A||=\max_{||x||=1}||Ax||,\]
which defines a matrix norm on \(M_{nn}\). The word _operator norm_ reflects the fact that a matrix transformation arising from a square matrix is also called _linear operator_.
### Convergence in Normed Spaces
A sequence \(\{x_{n}\}\) in a normed space is said to be _convergent_ if for all \(\epsilon>0\) there exists a positive integer \(N\) and an element \(x\in X\) such that
\[||x_{n}-x||<\epsilon\quad\text{whenever}\quad n>N.\]
We write this as \(x_{n}\to x\).
**Theorem 80**.: _If \(X\) is a normed space, then the norm_
\[||\,\cdot\,||:X\to\mathbb{R}\]
_is a continuous function on \(X\)._
Proof
: We need to show that \(x_{n}\to x\) implies \(||x_{n}||\to||x||\). This follows from the alternate triangle inequality
\[|||x_{n}||-||x|||\leq||x_{n}-x||\]
and the fact that since \(x_{n}\to x\), there exists an \(N\) such that \(||x_{n}-x||<\epsilon\) for all \(n>N\).
**Theorem 81**.: _The operation of addition and scalar multiplication in a normed space \(X\) are jointly continuous._
Proof
: We will show if \(x_{n}\to x\) and \(y_{n}\to y\), then \(x_{n}+y_{n}\to x+y\) and if \(a_{n}\to a\) then \(a_{n}x_{n}\to ax\). Since \(x_{n}\to x\) and \(y_{n}\to y\),
\[||(x_{n}+y_{n})-(x+y)||=||(x_{n}-x)+(y_{n}-y)||\leq||x_{n}-x||+||y_{n}-y||\]implies \(x_{n}+y_{n}\to x+y\). Now
\[||a_{n}x_{n}-ax|| = ||a_{n}x_{n}-a_{n}x+a_{n}x-ax||\] \[= ||a_{n}(x_{n}-x)+(a_{n}-a)x||\] \[\leq |a_{n}|\,||x_{n}-x||+|a_{n}-a|\,||x||,\]
so since \(\{a_{n}\}\) is bounded, \(a_{n}\to a\) and \(x_{n}\to x\), we obtain \(a_{n}x_{n}\to ax\).
The sequence \(\{x_{n}\}\) is called a _Cauchy sequence_ if, given \(\epsilon>0\), there exists a positive integer \(N\) such that
\[||x_{n}-x_{m}||<\epsilon\quad\text{whenever}\quad m,n>N.\]
When every Cauchy sequence in a normed space converges, then we say the normed space is _complete_. We have a special term for them:
**Definition 67**.: A complete normed vector space is called a _Banach space_.
The spaces \(\mathbb{R}^{n}\), \(\mathbb{C}^{n}\), \(\ell^{2}\), and \(C[a,b]\) with the sup norm are all examples of Banach spaces. One can also prove that every finite dimensional normed vector space is a Banach space. All of the theorems we have seen about the convergence of sequences are valid in these spaces as well. For example the limit of a convergent sequence in a normed space is unique, and any convergent sequence in a normed space is a Cauchy sequence. Every normed space is a metric space; there is a metric induced by the norm and consequently a normed space may be identified as compact if and only if it is sequentially compact. Furthermore, in a vector space we can add vectors together so we have an idea of infinite series of real and complex numbers. Let \(\{x_{n}\}\) be a sequence of elements in a normed linear space \(X\); then we say \(\sum_{k=1}^{\infty}x_{k}\) converges if the sequence of partial sums \(s_{n}=\sum_{k=1}^{n}x_{k}\) converges.
**Definition 68**.: A series \(\sum_{k=1}^{\infty}x_{k}\)_absolutely convergent_ if the series \(\sum_{k=1}^{\infty}||x_{n}||\) (of real numbers) is convergent.
This is just a generalization to Banach spaces of the definition of absolute convergence of sequence of real numbers. If a series is absolutely convergent in a normed linear space \(X\), it need not converge in \(X\). This is illustrated in the following example:
**Example 79**.: Consider the set \(\mathcal{P}[0,1]\) of all polynomials with the supremum norm. In this normed linear space, consider the series \(\sum_{n=0}^{\infty}\dfrac{x^{n}}{n!}\) for \(x\in[0,1]\). The series
\[\sum_{n=0}^{\infty}\dfrac{||x||^{n}}{n!}\leq\sum_{n=0}^{\infty}\dfrac{1}{n!}\]converges to a constant less than \(e\). But the constants can be considered as polynomials, thus the series \(\sum_{n=0}^{\infty}\frac{x^{n}}{n!}\) converges absolutely. However, if we consider the sequence of partial sums \(s_{n}=\sum_{k=0}^{n}\frac{x^{k}}{k!}\), we see that \(s_{n}\to e^{x}\) as \(n\to\infty\) in the norm. Since \(e^{x}\notin\mathcal{P}[0,1]\), the series is not convergent in \(\mathcal{P}[0,1]\).
The following theorem shows that absolute convergence implies convergence in Banach spaces.
**Theorem 82**.: _A normed vector space \(X\) is a Banach space if and only if every absolutely convergent series in \(X\) is convergent._
Proof
: Suppose \(X\) is a Banach space and that \(\sum_{k=1}^{\infty}x_{k}\) is an absolutely convergent series in \(X\). Then by definition \(\sum_{k=1}^{\infty}||x_{k}||\) converges. Using the triangle inequality we get
\[||s_{n}-s_{m}||=||\sum_{k=m}^{n}x_{k}||\leq\sum_{k=m}^{n}||x_{k}||<\epsilon\]
for all sufficiently large \(m,n\) with \(m
\[||x_{n}-x_{m}||<\frac{1}{2^{k}}\quad\text{for all}\quad m,n>N_{k}.\]
From this we get,
\[||x_{n_{k+1}}-x_{n_{k}}||<\frac{1}{2^{k}}\quad\text{for}\quad k=1,2,\dots\]
It follows that
\[\sum_{k=1}^{\infty}||x_{n_{k+1}}-x_{n_{k}}||\leq\sum_{k=1}^{\infty}\frac{1}{2 ^{k}}=1,\]
so the series \(\sum_{k=1}^{\infty}||x_{n_{k+1}}-x_{n_{k}}||\) is convergent. By hypothesis the series \(\sum_{k=1}^{\infty}(x_{n_{k+1}}-x_{n_{k}})\) is convergent. This implies that \(\{x_{n_{k}}\}\) converges. Hence the Cauchy sequence \(\{x_{n}\}\) has a convergent subsequence \(\{x_{n_{k}}\}\), but by the property of Cauchy sequences this means the Cauchy sequence \(\{x_{n}\}\) converges. Hence \(X\) is a complete normed space.
### Exercises
1. Show that \(d(x,y)=|e^{x}-e^{y}|\) is a metric on \(\mathbb{R}\).
2. Let \(S\) denote the collection of all real sequences \(x=\{x_{n}\}\). Show that the expression
\[d(x,y)=\sum_{i=1}^{\infty}\left(\frac{1}{2^{i}}\right)\frac{|x_{i}-y_{i}|}{1+| x_{i}-y_{i}|}\]
defines a metric on \(S\).
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