114 CHAPTER 6. MULTI-VARIABLE CALCULUS
Proof: If this is not so, there exists ε > 0 and pairs of points, xn and yn satisfying|xn−yn|< 1/n but |f(xn)− f(yn)| ≥ ε. Since C is sequentially compact, there exists x ∈Cand a subsequence,
{xnk
}satisfying xnk → x. But
∣∣xnk −ynk
∣∣ < 1/k and so ynk → x also.Therefore, from Theorem 6.7.10 on Page 108,
ε ≤ limk→∞
∣∣f(xnk
)− f(ynk
)∣∣= |f(x)− f(x)|= 0,
a contradiction. This proves the theorem.
6.11 The Space L (Fn,Fm)
Definition 6.11.1 The symbol, L (Fn,Fm) will denote the set of linear transformationsmapping Fn to Fm. Thus L ∈L (Fn,Fm) means that for α,β scalars and x,y vectors in Fn,
L(αx+βy) = αL(x)+βL(y) .
It is convenient to give a norm for the elements of L (Fn,Fm) . This will allow theconsideration of questions such as whether a function having values in this space of lineartransformations is continuous.
6.11.1 The Operator NormHow do you measure the distance between linear transformations defined on Fn? It turnsout there are many ways to do this but I will give the most common one here.
Definition 6.11.2 L (Fn,Fm) denotes the space of linear transformations mapping Fn toFm. For A ∈L (Fn,Fm) , the operator norm is defined by
||A|| ≡max{|Ax|Fm : |x|Fn ≤ 1}< ∞.
Theorem 6.11.3 Denote by |·| the norm on either Fn or Fm. Then L (Fn,Fm) with thisoperator norm is a complete normed linear space of dimension nm with
||Ax|| ≤ ||A|| |x| .
Here Completeness means that every Cauchy sequence converges.
Proof: It is necessary to show the norm defined on L (Fn,Fm) really is a norm. Thismeans it is necessary to verify
||A|| ≥ 0 and equals zero if and only if A = 0.
For α a scalar,||αA||= |α| ||A|| ,
and for A,B ∈L (Fn,Fm) ,
||A+B|| ≤ ||A||+ ||B||