5.13. EXERCISES 153

then for every ε > 0, there exists g ∈ G such that ∥ f −g∥∞< ε where ∥h∥

∞≡

maxx∈B(0,R) |h(x)|. Thus, from multi-variable calculus, every continuous function f

is uniformly close to an infinitely differentiable function on any closed ball centeredat 0.

14. Suppose now that f ∈ C0 (Rp) . This means that f is everywhere continuous andthat lim∥x∥→∞ | f (x)| = 0. Show that for every ε > 0 there exists g ∈ G such thatsupx∈Rp | f (x)−g(x)| < ε . Thus you can approximate such a continuous functionf uniformly on all ofRp with a function which has infinitely many continuous partialderivatives. I assume the reader has had a beginning course in multi-variable calcu-lus including partial derivatives. If not, a partial derivative is just a derivative withrespect to one of the variables, fixing all the others.

15. In Problem 23 on Page 124, and V ≡ span( fp1 , ..., fpn) , fr (x) ≡ xr,x ∈ [0,1] and− 1

2 < p1 < p2 < · · · with limk→∞ pk = ∞. The distance between fm and V is

1√2m+1 ∏

j≤n

∣∣m− p j∣∣

(p j +m+1)= d

Let dn = d so more functions are allowed to be included in V . Show that ∑n1pn

= ∞

if and only if limn→∞ dn = 0. Explain, using the Weierstrass approximation theoremwhy this shows that if g is a function continuous on [0,1] , then there is a function∑

Nk=1 ak fpk with

∣∣g−∑Nk=1 ak fpk

∣∣ < ε . Here |g|2 ≡∫ 1

0 |g(x)|2 dx. This is Müntz’s

first theorem. Hint: dn → 0, if and only if lndn → −∞ so you might want toarrange things so that this happens. You might want to use the fact that for x ∈[0,1/2] ,−x≥ ln(1− x)≥−2x. See [10] which is where I read this. That product is

∏ j≤n

(1−(

1− |m−p j|(p j+m+1)

))and so ln of this expression is

n

∑j=1

ln

(1−

(1−

∣∣m− p j∣∣

(p j +m+1)

))

which is in the interval[−2

n

∑j=1

(1−

∣∣m− p j∣∣

(p j +m+1)

),−

n

∑j=1

(1−

∣∣m− p j∣∣

(p j +m+1)

)]

and so dn → 0 if and only if ∑∞j=1

(1− |m−p j|

(p j+m+1)

)= ∞. Since pn → ∞ it suffices

to consider the convergence of ∑ j

(1− p j−m

(p j+m+1)

)= ∑ j

(2m+1

(p j+m+1)

). Now recall

theorems from calculus.

16. For f ∈ C ([a,b] ;R) , real valued continuous functions, let | f | ≡(∫ b

a | f (t)|2)1/2

( f , f )1/2 where ( f ,g) ≡∫ b

a f (x)g(x)dx. Recall the Cauchy Schwarz inequality|( f ,g)| ≤ | f | |g| . Now suppose 1

2 < p1 < p2 · · · where limk→∞ pk = ∞. Let Vn =span(1, fp1 , fp2 , ..., fpn) . For ∥·∥ the uniform approximation norm, show that for ev-ery g ∈C ([0,1]) , there exists there exists a sequence of functions, fn ∈Vn such that

5.13.14.15.16.EXERCISES 153then for every € > 0, there exists g € Y such that || f —g||,, < € where ||A||,, =MAX, -B(0,R) |h(a:)|. Thus, from multi-variable calculus, every continuous function fis uniformly close to an infinitely differentiable function on any closed ball centeredat 0.Suppose now that f € Co(IR’). This means that f is everywhere continuous andthat lim).|-,.. | f (z)| = 0. Show that for every € > 0 there exists g < Y such thatsupeere Lf (@) —g(x)| < €. Thus you can approximate such a continuous functionf uniformly on all of R? with a function which has infinitely many continuous partialderivatives. I assume the reader has had a beginning course in multi-variable calcu-lus including partial derivatives. If not, a partial derivative is just a derivative withrespect to one of the variables, fixing all the others.In Problem 23 on Page 124, and V = span(fp,,---s fp.) > fr (x) = ",x € [0,1] and—5 < pi < po<--: with limy_,.. p, = 00. The distance between f,, and V isl |m—=pj|Let d, =d so more functions are allowed to be included in V. Show that Y., rs = 00if and only if lim,_,..d, = 0. Explain, using the Weierstrass approximation theoremwhy this shows that if g is a function continuous on [0,1], then there is a functionYh anf, with lg — Ley ap, | < &€. Here |g|° = fo |g (x)|°dx. This is Miintz’sfirst theorem. Hint: d, — 0, if and only if Ind, —> —o so you might want toarrange things so that this happens. You might want to use the fact that for x €(0, 1/2] ,—x > In (1 —x) > —2x. See [10] which is where I read this. That product isIj<n (1 _ (1 — ey) and so In of this expression is~ jyin(1— paila (pj+m+1)which is in the intervaloy (,__lmepil_ \ ff, _lreeil8 ( (pjtm+1) }° X (pj +m+1)ip yo, (y— lmand so d, — 0 if and only if (: (pjtm+1)= oo, Since p, — © it sufficesto consider the convergence of Y; ( 1- Sy) =yj ( ats) . Now recallj jtheorems from calculus.1/2For f € C({a,b];R), real valued continuous functions, let |f| = (i /(0)") =(f, py! > where (f,2) = i? f (x) g(x)dx. Recall the Cauchy Schwarz inequality(f,8)| < Ifllg|. Now suppose 5 < pi < p2-++ where limp. pe = 0. Let Vy =span (1, fp, .fpo>+-+sfpn) - For ||-|| the uniform approximation norm, show that for ev-ery g € C([0,1]), there exists there exists a sequence of functions, f, € V, such that