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9.1. BOUNDED CONTINUOUS FUNCTIONS 221

Then F has a unique fixed point, that is a point x ∈ X such that Fx = x. In addition tothis, if ∥Fx0− x0∥< R(1− r) and F is only defined on B(x0,R) then F has a unique fixedpoint in this ball. Here B(x0,R) signifies the set of all x such that ∥x− x0∥ ≤ R. Also, thesequence {Fnx0} converges.

Proof: Pick any x0 ∈ X . Consider the sequence {Fnx0} . I will argue that this is aCauchy sequence. To see this, suppose n,m ≥ M with n > m and consider the followingwhich comes from the triangle inequality for the norm, ∥x+ y∥ ≤ ∥x∥+∥y∥.

∥Fnx0−Fmx0∥ ≤n−1

∑k=m

∥∥∥Fk+1x0−Fkx0

∥∥∥Now

∥∥Fk+1x0−Fkx0∥∥≤

r∥∥∥Fkx0−Fk−1x0

∥∥∥≤ r2∥∥∥Fk−1x0−Fk−2x0

∥∥∥ · · · ≤ rk ∥Fx0− x0∥ .

Using this in the above, ∥Fnx0−Fmx0∥ ≤

n−1

∑k=m

∥∥∥Fk+1x0−Fkx0

∥∥∥≤ n−1

∑k=m

rk ∥Fx0− x0∥ ≤rm

1− r∥Fx0− x0∥ (9.1)

since r < 1, this is a Cauchy sequence. Hence it converges to some x. Therefore,

x = limn→∞

Fnx0 = limn→∞

Fn+1x0 = F limn→∞

Fnx0 = Fx.

The third equality is a consequence of the following consideration. If zn→ z, then

∥Fzn−Fz∥ ≤ r∥zn− z∥

so also Fzn→ Fz. In the above, Fnx0 plays the role of zn and its limit plays the role of z.The fixed point is unique because if you had two of them, x, x̂, then

∥x− x̂∥= ∥Fx−Fx̂∥ ≤ r∥x− x̂∥

and so x = x̂.In the second case, let m = 0 in 9.1 and you get the estimate

∥Fnx0− x0∥ ≤1

1− r∥Fx0− x0∥< R.

It is still the case that the sequence {Fnx0} is a Cauchy sequence and must therefore con-verge to some x ∈ B(x0,R) which is a fixed point as before. The fixed point is uniquebecause of the same argument as before. ■

Now there is another norm which works just as well in the case where T ≡ [a,b] , aninterval. This is described in the following definition.

Definition 9.1.5 For f ∈ BC ([a,b] ;Fn) , let c ∈ [a,b] ,γ a real number. Then

∥f∥γ≡ sup

t∈[a,b]

∣∣∣f(t)e−|γ(t−c)|∣∣∣

Then this is a norm. The above Definition 9.1.1 corresponds to γ = 0.

9.1. BOUNDED CONTINUOUS FUNCTIONS 221Then F has a unique fixed point, that is a point x € X such that Fx =x. In addition tothis, if ||F xo —xo|| <.R(1—r) and F is only defined on B(xo,R) then F has a unique fixedpoint in this ball. Here B(xo,R) signifies the set of all x such that \|x —xo|| < R. Also, thesequence {F"x} converges.Proof: Pick any x9 € X. Consider the sequence {F"xo}. I will argue that this is aCauchy sequence. To see this, suppose n,m > M with n > m and consider the followingwhich comes from the triangle inequality for the norm, ||x-+y|| < ||x|| + ly].n—1|F'xo— P"axoll < Yo ||P x0 — Fok=mNow ||F**!x9 — F*xo|| <r||Ftx0 — Fly9|| <r [Flo — FF0|| we k ||F-xo — xol|l -Using this in the above, ||F”x9 — F’"xo|| <n—1 n—1 yny Feta — Pho] < Y rf |Fx0 —x0l| < 7 ||Fx0—20| (9.1)k=m k=m —rsince r < 1, this is a Cauchy sequence. Hence it converges to some x. Therefore,x= lim F"xo = lim F”*!xq = F lim F"xo = Fx.n— co n—oo n—ooThe third equality is a consequence of the following consideration. If z, — z, then|F2n — Fz\| <r|lzn —2llso also F'z, — Fz. In the above, Fx plays the role of z, and its limit plays the role of z.The fixed point is unique because if you had two of them, x,%, then|x —£|| = ||Fx— FS] <r||x—3|and so x = ¥.In the second case, let m = 0 in 9.1 and you get the estimate1\||F"x0 — xo|| < Tor \|F-xo —xo|| < R.It is still the case that the sequence {Fx} is a Cauchy sequence and must therefore con-verge to some x € B(xo,R) which is a fixed point as before. The fixed point is uniquebecause of the same argument as before. HiNow there is another norm which works just as well in the case where T = [a,b], aninterval. This is described in the following definition.Definition 9.1.5 For f € BC ({a,b];F"), let c € [a,b], y a real number. ThenI|f||,= sup tije totela,b]Then this is anorm. The above Definition 9.1.1 corresponds to y = 0.