Kenneth Kuttler

150 CHAPTER 5. FUNCTIONS ON NORMED LINEAR SPACES

5.13 Exercises1. Consider the metric space C ([0,T ] ,Rn) with the norm ∥f∥ ≡ maxx∈[0,T ] ∥f (x)∥∞

.Explain why the maximum exists. Show this is a complete metric space. Hint: If youhave {fm} a Cauchy sequence in C ([0,T ] ,Rn) , then for each x, you have {fm (x)}a Cauchy sequence in Rn. Recall that this is a complete space. Thus there existsf (x) = limm→∞fm (x). You must show that f is continuous. This was in the sectionon the Ascoli Arzela theorem in more generality if you need an outline of how thisgoes. Write down the details for this case. Note how f is in bold face. This means itis a function which has values in Rn. f (t) = ( f1 (t) , f2 (t) , · · · , fn (t)).

2. For f ∈C ([0,T ] ,Rn) , you define the Riemann integral in the usual way using Rie-mann sums. Alternatively, you can define it as∫ t

0f (s)ds =

(∫ t

0f1 (s)ds,

∫ t

0f2 (s)ds, · · · ,

∫ t

0fn (s)ds

)Then show that the following limit exists in Rn for each t ∈ (0,T ) .

limh→0

∫ t+h0 f (s)ds−

∫ t0 f (s)ds

h= f (t) .

You should use the fundamental theorem of calculus from one variable calculus andthe definition of the norm to verify this. As a review, in case we don’t get to it intime, for f defined on an interval [0,T ] and s ∈ [0,T ] , limt→sf (t) = l means thatfor all ε > 0, there exists δ > 0 such that if 0 < |t− s|< δ , then ∥f (t)− l∥

∞< ε .

3. Suppose f :R→ R and f ≥ 0 on [−1,1] with f (−1) = f (1) = 0 and f (x)< 0 for allx /∈ [−1,1] . Can you use a modification of the proof of the Weierstrass approximationtheorem for functions on an interval presented earlier to show that for all ε > 0 thereexists a polynomial p, such that |p(x)− f (x)| < ε for x ∈ [−1,1] and p(x) ≤ 0 forall x /∈ [−1,1]?

4. A collection of functions F of C ([0,T ] ,Rn) is said to be uniformly equicontinu-ous if for every ε > 0 there exists δ > 0 such that if f ∈ F and |t− s| < δ , then∥f (t)−f (s)∥

∞< ε . Thus the functions are uniformly continuous all at once. The

single δ works for every pair t,s closer together than δ and for all functions f ∈F .As an easy case, suppose there exists K such that for all f ∈F , ∥f (t)−f (s)∥

∞≤

K |t− s| . Show that F is uniformly equicontinuous. Now suppose G is a collectionof functions of C ([0,T ] ,Rn) which is bounded. That is, ∥f∥= maxt∈[0,T ] ∥f (t)∥∞

<M < ∞ for all f ∈ G . Then let F denote the functions which are of the formF (t) ≡ y0 +

∫ t0 f (s)ds where f ∈ G . Show that F is uniformly equicontinuous.

Hint: This is a really easy problem if you do the right things. Here is the wayyou should proceed. Remember the triangle inequality from one variable calcu-lus which said that for a < b

∣∣∣∫ ba f (s)ds

∣∣∣ ≤ ∫ ba | f (s)|ds. Then

∥∥∥∫ ba f (s)ds

∥∥∥∞

maxi

∣∣∣∫ ba fi (s)ds

∣∣∣≤maxi∫ b

a | fi (s)|ds≤∫ b

a ∥f (s)∥∞ds. Reduce to the case just con-

sidered using the assumption that these f are bounded.

5. Suppose F is a set of functions in C ([0,T ] ,Rn) which is uniformly bounded anduniformly equicontinuous as described above. Show it must be totally bounded.

150 CHAPTER 5. FUNCTIONS ON NORMED LINEAR SPACES5.13 Exercises1. Consider the metric space C({0,7],IR") with the norm || f|| = max,<jo,7) || f (*)|..-Explain why the maximum exists. Show this is a complete metric space. Hint: If youhave {f,,,} a Cauchy sequence in C([0,7], IR”), then for each x, you have { f,, (x) }a Cauchy sequence in R”. Recall that this is a complete space. Thus there existsf (x) =lim,,..~ f,, (x). You must show that f is continuous. This was in the sectionon the Ascoli Arzela theorem in more generality if you need an outline of how thisgoes. Write down the details for this case. Note how f is in bold face. This means itis a function which has values in R”. f (t) = (fi (t), fo (t),--+ , fn (t)).2. For f © C([0,7],R”), you define the Riemann integral in the usual way using Rie-mann sums. Alternatively, you can define it as[ f0)as= (fa (}ds, [ risyase. [ta(o)as)Then show that the following limit exists in R” for each t € (0,7).tim 1 _F (s)ds— Jo f(s) dsh0 h= F(t).You should use the fundamental theorem of calculus from one variable calculus andthe definition of the norm to verify this. As a review, in case we don’t get to it intime, for f defined on an interval [0,7] and s € [0,7], lim;,; f (¢) = means thatfor all € > 0, there exists 6 > 0 such that if 0 < |t—s| < 6, then ||f (+) —I]|,, < e.3. Suppose f: RR — Rand f > 0 on [—1, 1] with f (—1) = f (1) =Oand f (x) < 0 for allx ¢ [—1,1]. Can you use a modification of the proof of the Weierstrass approximationtheorem for functions on an interval presented earlier to show that for all € > 0 thereexists a polynomial p, such that |p (x) — f (x)| <€ for x € [—1,1] and p(x) <0 forall x ¢ [—1, 1]?4. A collection of functions .¥ of C([0,7],R”) is said to be uniformly equicontinu-ous if for every € > O there exists 6 > 0 such that if f € F and |t—s| < 6, then\|f (t) — f (s)||.. < €. Thus the functions are uniformly continuous all at once. Thesingle 6 works for every pair f,s closer together than 6 and for all functions f € F.As an easy case, suppose there exists K such that for all f € F, ||f (t)-f (s)|. <K |t — s|. Show that .¥ is uniformly equicontinuous. Now suppose @ is a collectionof functions of C ([0, 7] ,IR") which is bounded. That is, || f|] = max,cjo,7) ||f (||. <M <. for all f © Y. Then let ¥Y denote the functions which are of the formF(t) =yo+ Jo f (s)ds where f € Y. Show that ¥ is uniformly equicontinuous.Hint: This is a really easy problem if you do the right things. Here is the wayyou should proceed. Remember the triangle inequality from one variable calcu-lus which said that for a <b | f° f(s)ds| < f? |f(s)|ds. Then || [2 f(s) ds|| =max; ei (s)ds < max; f° |fi(s)|ds < f° \| f (s)||,,ds. Reduce to the case just con-sidered using the assumption that these f are bounded.5. Suppose ¥ is a set of functions in C((0,7],.R”) which is uniformly bounded anduniformly equicontinuous as described above. Show it must be totally bounded.