ANALYSIS HW 3. f(x + y) = f(x) + f(y) for all real x, y. Demonstration: Let f be such a function. Since f is smooth, f exists.

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1 ANALYSIS HW 3 CLAY SHONKWILER () Fid ll smooth fuctios f : R R with the property f(x + y) = f(x) + f(y) for ll rel x, y. Demostrtio: Let f be such fuctio. Sice f is smooth, f exists. The The f f(x + h) f(x) (x) = lim h h f(x) = f(x) + f(h) f(x) = lim = f(h) h h h = c. f (x)dx = cdx = cx + d However, f() = f( + ) = f() + f() = f(), so f() =, meig d =, so f(x) = cx. (b) Wht c you sy if f is oly ssumed to be cotiuous? Aswer: If f is merely cotiuous, the sme result holds. Assume f is cotiuous d f(x + y) = f(x) + f(y). The, for ll Z, f(x) = f(x). Usig this result, we see tht if p, q Z, f( p q x) = pf( q x) = pf(x q ). Therefore, f(x) = f( p p x) = pf(x p ), so f( x p ) = f(x) p. Specificlly, f( p q x) = pf(x q = pf(x) q = p q f(x). The for r R, we c fid Cuchy sequece r j of rtiol umbers tht coverge to r. The, by the result proved bove for rtiol umbers, Now, lim j f(r j x) = lim j r j f(x) which is to sy f(rx) = rf(x) f(x) = f(x ) = xf(), so f is of the form f(x) = mx for some m R.

2 CLAY SHONKWILER (c) Repet this for g(x) stisfyig g(x + y) = g(x)g(y). Aswer: Let g be fuctio such tht g(x + y) = g(x)g(y). For Z, g(x) = g(x), so g( x = g(x ), which implies g( x ) = [g(x)]/. I tur, if p, q Z, g( p q x) = [g( x q ) ] p = [g(x)] p q. Now, if r R, let r j be Cuchy sequece covergig to r. The g(rx) = lim j g(r j x) = lim j g(x) r j = g(x) r. Hece, g(x) = g( x) = g() x. Sice g() is just costt, g hs the form g(x) = c x. Let f(x) be cotiuous fuctio for x. Evlute lim f(x)x dx. Evlutio: Let ɛ >. The f c be pproximted by smooth fuctio h such tht f h uif < ɛ/3. Now, h exists d is bouded o [, ], so let M := mx x [,] f (x). Let N > M ɛ. Now, f(x)x dx f() = (f(x) h(x) + h(x))x dx f() = (f(x) h(x))x dx + h(x)x dx f() ɛ/3x dx + h(x)x dx f() = ɛ/3 x dx + where we itegrted by prts. Hece, ( = ɛ/3 + + h() + + ɛ/3 + h() h(x)x ) dx f() h (x)x + dx f() h (x)x dx f() ɛ/3 + h() M x dx f() = ɛ/3 + h() M + f() < ɛ/3 + h() ɛ/3 f() epsilo/3 + h() f() + ɛ/3 < ɛ/3 + ɛ/3 + ɛ/3 = ɛ lim f(x)x dx = f().

3 ANALYSIS HW Let f : R R be uiformly cotiuous. Are there costts, b such tht f(x) x + b for ll x? Proof or couterexmple. Proof. Let ɛ >. The, sice f is uiformly cotiuous, there exists δ > such tht x y < δ implies The f(x) f(x) f(y) < ɛ. = f(x) f() + f() = (f(x) f(x δ)) + (f(x δ) f(x δ)) (f(x kδ) f()) + f() f(x) f(x δ) f(x kδ) f() + f() < ɛ ɛ + f() = kɛ + f() where k is the smllest iteger such tht x kδ < δ. Hece, k is the smllest iteger such tht k > x δ. We see, the, tht f(x) < kɛ + f() ɛ δ x + f(). 4 Let P j, j =,,... be sequece of poits i R 3. If P j+ P j j 4, show tht these poits coverge. Proof. Let ɛ >. The, sice the series coverges, there exists N N j 4 such tht, if > m > N, m ɛ > j 4 j 4 = m +. The, for y > m > N, P + P m+ = (P + P ) + (P P ) (P m+ P m+ ) P + P +... P m+ P m (m+) 4 < ɛ. Hece, {P j } is Cuchy sequece d, sice R 3 is complete, coverges. 5 () Let f(x) be cotiuous fuctio for x. Fid ll rel umbers c for which Q c (f) := lim c f(x)e x dx exists. If the limit

4 4 CLAY SHONKWILER Q c (f) exists, compute it. Aswer: Let M = mx x [,] f(x). If c, the c f(x)e x dx c M e x dx = c M( e ) = c M(e + ). Now, e vishes, s does c M (sice c ), so, for c, Q c (f) exists d is equl to. Now, if c >, the we pproximte f by C fuctio h. The ) + ) h (x)e x dx = lim c ( h() e h() ) + c h (x)e x dx = lim c h() c e h() + c h (x)e x dx, lim c (( h(x)e x dx = lim c e h() h() where we itegrted by prts. Now, c h() will get rbitrrily lrge, sice c >, so this limit does ot exist. Hece, we coclude tht Q c (f) exists oly whe c d tht, whe it does exist, Q c (f) =. (b) Wht if you oly ssume tht f L (, )? Aswer: If we oly ssume f L (, ), we c pproximte f rbitrrily close by C fucio h, d the bove lysis demostrtes tht Q c (h) exists oly whe c d tht, whe it exists, Q c (h) =. 6 Let B be the ope uit disk i R. Give exmple of fuctio f L (B ) tht is ot i L (B ). Justify your ssertios. Exmple: Defie f(r, θ) = r. The, defie f (r, θ) = q. Let ɛ > r+ d let N > π rɛ. The, if > N, f f = r q = r+ q r+ r r + r q r+ r r + r = r +r r < ɛ/π. Hece, f k is Cuchy, meig there exists N N such tht, if m, > N, f f m < ɛ/π. Tht mes tht, for the sme m,, f f m drdθ < R R ɛ/πdrdθ = ɛ,

5 ANALYSIS HW 3 5 so f k is Cuchy i L. Now, we wt to demostrte tht f k coverges i L to f. π lim q [ ] π drdθ = lim r+ r + dθ = π + dθ = 4π + 4π = 4π = f L. Hece, we see tht, ideed, f(r, θ) = r L (B ). However, [ ] π / f L = r drdθ = [ π [log r] dθ ] /. However, sice log() is udefied, it is goig to be impossible to costruct Cuchy sequece of cotiuous fuctios tht coverge to f i the L orm, meig f / L (B ). 7 For ech of the followig, give exmple of sequece of cotiuous fuctios. () See ttched sheet. (b) See ttched sheet. (c) See ttched sheet. (d) See ttched sheet. (e) Exmple: Let f (x) =, let ɛ > d let N N such tht x N > ɛ. If > N, the f L = f (x) dx = dx = [ ] x x = < ɛ. Hece, f coverges to zero i the L orm. O the other hd, [ ] / f L = f (x) dx = [ ] / x dx = [ / [log x] ]. However, sice log() is udefied, f clerly does ot coverge to zero i the L orm. (f). Exmple: Let f (x) = 4 x x x (, ] x [, ] x >

6 6 CLAY SHONKWILER The let ɛ > d let N N such tht N > ɛ. If > N, the f L = f (x)dx = f (x)dx + f (x)dx + f (x)dx = 4 x dx + x [ ] + = x x = x + + < ɛ/ + ɛ/ = ɛ, so f coverges to zero i the L orm. O the other hd f k L = [ f (x) dx ] / [ = 6 x + ] ( x) / [ [ = log x ] 6 + ] ( x) /. However, sice logx is ot defied t zero, this sequece cot coverge to zero. (g). Exmple: Let g (x) = x log x (, ] (log )x x [, ] (+) x log x [, + ] x + Let ɛ >. The, if N N such tht N > mx{, e 3 ɛ, e ɛ } d > N, the g L = [ g (x) dx ] / [ x = dx + (log ) [ = + ( 3(log ) < < (log ) [ ] ɛ / (log ) (log ) ] / [ ɛ 3 + ɛ 3 + ɛ 3 = ɛ = ɛ. dx + + (log ) x (log ) ) + 4+ ] (+) (+)x+x / dx (log ) ] / 6(log )

7 ANALYSIS HW 3 7 Therefore, g k coverges to zero i the L orm. However, for y > e, g L = g (x)dx = x log + = log [ x ] (log )x dx log [log x] + log = log + log log + log > + log 3 log = log >, (+) x log so g k clerly does ot coverge to zero i the L orm. [ dx ( + ) x ] + 8 Let γ : R R 3 defie smooth curve. Show tht γ(t)dt γ(t) dt. Proof. Let v = γ(t)dt. The v = v, v = v, γ(t)dt = v, γ(t) v γ(t) dt = v γ(t) dt where we rrive t the third lie by the Cuchy-Schwrz Iequlity, which tells us tht v, γ(t) v γ(t). Hece, we c coclude tht γ(t)dt = v γt dt. Let f(x), where x b, be smooth fuctio. () If f(c) = for some c b, show tht f(x) 9 f (t) dt [ ] / b f (t) dt d hece, usig the uiform orm f uif := mx x b f(x), f uif f (t) dt [ / b f (t) dt].

8 8 CLAY SHONKWILER Proof. By the Fudmetl Theorem of Clculus, Also, by Holder, f(x) = ( f (t) dt x c f (t)dt x c f (t) dt f (t) dt. ) / ( / dt f (t) ) = ( / b f (t) dt). (b) If f(t)dt =, show tht the bove iequlity still holds. Proof. Let F (x) = x f(t)dt. The F () = d F (b) = f(t)dt =, so, by the Me Vlue Theorem, there exists c [, b] such tht = F (c) = f(c). Now, by (), f(x) f (t) dt [ / b f (t) dt]. (c) Use the result i prt (b) to show tht for y smooth f f uif α [ / [ f (t) + f(t) ]dt β [ f (t) + f(t) ]dt], where the costts α d β deped oly o the regio of itegrtio. Proof. Let g = f f, where f = regio. The, g(t) b = ( b f(t) ) b b f(t)dt = f(t)dt b = f(t)dt =. f(t)dt is the verge of f i the dt f(t)dtdt b (b ) f(t)dt Hece we c use the result i prt (b) to coclude tht g uif g (t) dt [ / b g (t) dt].

9 ANALYSIS HW 3 9 Now, f uif = g + f uif g uif + f g (t) dt + b f(t) dt = ( ) b f (t) + b f(t) dt (α f (t) + α f(t) ) dt = α [ f (t) + f(t) ]dt. Defie the ormed lier spce L, (, ) s the completio of the C [, ] fuctios i the orm f L, := ( f (t) + f(t) )dt. () Is f(x) = x i this spce? Why? Aswer: Yes. Let f k (x) = x + k. The ech f k C [, ]. Let ɛ >. The, if N >, the > N ɛ implies f (x) x = x + x x + x = < ɛ. Hece, f k coverges to x = f(x) i [, ], which mes tht f f. Therefore, f d f re Cuchy sequeces, which is to sy tht there exists N N such tht, if, m > N, f (x) f m (x) < ɛ/ d there exists N 3 N such tht, if, m > N 3, f (x) f m(x) < ɛ/ for ll x [, ]. Let N = mx{n, N 3 }. Now, the, if > N, f f m L, = ( f (t) f m(t) + f (t) f m (t) )dt < (ɛ/+ɛ/)dt = ɛ. So f k is cuchy i the L, orm. Its limit must be f(x) = x, so f(x) = x L, (, ). (b) Show tht ll the fuctios i this spce re cotiuous. Proof. Fix x [, ]. Let f L, (, ) d let f be Cuchy sequece of C fuctios tht coverge to f. Let ɛ >. The there exists N N such tht > N implies ɛ/3 > f f L, = ( f (t) f (t) dt + or f(t) f (t) dt < ɛ/3 f(t) f (t) dt

10 CLAY SHONKWILER which is to sy tht f(t) f (t) < ɛ for ll t [, ]. Now, sice ech f j C [, ], there exists δ > such tht x y < δ implies f j (x) f j (y) < ɛ/3. Now, the, if > N d x y < δ, f(x) f(y) = (f(x) f (x)) + (f (x) f (y)) + (f (y) f(y)) f(x) f (x) + f (x) f (y) + f (y) f(y) < ɛ/3 + ɛ/3 + ɛ/3 = ɛ. Hece, f is cotiuous t x. Sice our choice of x ws rbitrry, we see tht f is cotiuous. (c) Defie the ormed lier spce L, (, ) s the completio of the C [, ] fuctios i the orm [ / f L, := ( f (t) + f(t) )dt]. Show tht L, (, ) is proper subspce of L, (, ). Proof. If f L, (, ), the, sice the smooth fuctios re dese i the C fuctios, we c pproximte f by sequece h k of smooth fuctios. We see by prt 9(c) bove tht h k L, β α h k L,, so h k L, (, ). Hece, f L, (, ), so L, (, ) L, (, ). O the other hd, we sw bove tht f(x) = x L, (, ). However, f L, is ot defied, s ( f (t) + f(t) ) ( ) dt = 4x + x 4 x dt = log x + d log() is udefied. L, (, ). Hece, f(x) = x / L, (, ), so L, (, ) Let f : [, ] R be cotiuous fuctio. () Show tht lim λ f(x)si(λx)dx =. Proof. Let ɛ >. Sice the smooth fuctios re dese i the cotiuous fuctios, we c choose smooth fuctio h such tht f h uif < ɛ/3. Sice [, ] is compct, h d h re bouded by M d M, respectively.

11 ANALYSIS HW 3 Now, let N > 3M ɛ d N > 3M ɛ. Let N = mx{n, N }. If λ > N, the f(x) si(λx)dx = (f h) si(λx)dx + h si(λx)dx < ɛ 3 dx + h(x) si(λx) dx [ ] = h(x) cos(λx) λ + h (x) cos(λx) λ dx ɛ/3 + M λ + M λ dx < ɛ/3 + ɛ/3 + ɛ/3 = ɛ. I other words, lim λ f(x) si(λx)dx =. (b) Compute lim λ si(λx) dx. (c) If φ : R R is cotiuous with period P, show tht P lim λ f(x)φ(λx)dx = φ f(x)dx, where φ = P φ(t)dt is the verge of φ over oe period. (d) Wht c you sy bout the vlidity of ll of prts () d (c) if oe oly kows tht f L (, )? Aswer: Prt () will still hold vlid, s the smooth fuctios re dese i L (, ), so we could mke the sme pproximtio tht we did for y f L (, ). If p, q, r > stisfy p + q + r =, prove the three-term Holder iequlity f(x)g(x)h(x) dx f p g q h r for ll cotiuous fuctios f(x), g(x), h(x) tht re zero outside of compct set. Proof. Sice g h hs compct support, the two-term Holder iequlity tells us tht [ ] /α f(x)g(x)h(x) f p g(x)h(x) α dx for α = q + r. Now, r q + r + q q + r = q + r q + r =,

12 CLAY SHONKWILER so, gi by the two-term Holder iequlity, [ g(x)h(x) α dx ( g(x) α ) q+r r dx Now, d Therefore, α q + r r α q + r q = qr q + r q + r r = qr q + r q + r q [ g(x)h(x) α dx ] /α [ [ ( g(x) α ) q+r r ] r q+r [ ( h(x) α ) q+r = qr r = q = qr q = r. q ] q q+r dx. ] r [ q+r ] dx ( h(x) α ) q+r q ] α q+r q dx Hece, f(x)g(x)h(x) f p [ = [ g(x) q dx ] q [ h(x) r dx ] r = g q h r. g(x)h(x) α dx] /α f p g q h r. DRL 3E3A, Uiversity of Pesylvi E-mil ddress: shokwil@mth.upe.edu