Grdute Anlysis I Chpter 5 Question If f is simple mesurle function (not necessrily positive) tking vlues j on j, j =,,..., N, show tht f = N j= j j. Proof. We ssume j disjoint nd,, J e nonnegtive ut J+,, N e negtive. First suppose f is finite, then it is cler tht f = i= i f = i i. If f is infinite, then either f + = or f =. If f + =, then J i= i i =. If f =, then N i=j+ i i =. In either cse, we still hve f = N j= j j. If f does not exist, then oth f + nd f re infinite. Moreover, it is cler tht N j= j j does not exist. Question 3 Let {f k } e sequence of nonnegtive mesurle functions defined on. If f k f nd f k f.e. on, show tht f k f. Proof. i= Since f is the limit of sequence of mesurle functions, it is mesurle. Next, y Ftou s lemm nd the fct tht f k f.e., we hve f lim inf f k lim sup f k f. It is now cler tht lim f k = f. Question 4. If f L(, ), show tht x k f(x) L(, ) for k =,,..., nd xk f(x)dx Proof. Clerly, if x [, ], then x k f(x) f(x). Hence xk f(x) dx f(x) dx < +, so xk f(x) L(, ). Moreover, if f(x) <, x (, ),
then x k f(x) s k. Thus, x k f(x).e. on [,]. Thus, y Leesgue s dominted convergence theorem, we hve Question 6 lim x k f(x)dx = =. Let f(x, y), x, y, stisfy the following conditions: for ech x, f(x, y) is n integrle function of y, nd ( f(x, y)/ x) is ounded function of (x, y). Show tht ( f(x, y)/ x) is mesurle function of y for ech x nd d f(x, y)dy = f(x, y)dy. dx x Proof. First, for ny y [, ], define f(x, y) = f(, y) for x > nd f(x, y) = f(, y) for x <. Clerly for ech x [, ], f(x, y) is still mesurle function of y (on [,]). Let + nd. Define g k (x, y) = f(x+,y) f(x,y), since f(x +, y) nd f(x, y) re mesurle functions of y for ech x, g k (x, y) is mesurle function of y for ech x. Since f(x, y)/ x = lim g k (x, y), f(x, y)/ x is mesurle function of y for ech x. By the men vlue theorem, f(x +, y) f(x, y) = f(x+t,y) x k, t (, ), then f(x+,y) f(x,y) we hve f(x+,y) f(x,y) Theorem to conclude tht lim f(x + k, y) = lim = f(x+t,y). Since f(x, y)/ x M, x M. We cn now pply the Bounded Convergence f(x, y) dy = lim f(x +, y) f(x, y) dy = f(x +, y) f(x, y) dy Since the ove holds for ny sequence we hve d f(x, y)dy. x Question 8 f(x, y)dy. x f(x, y)dy = dx f is mesurle, show tht there is n L p version of Tcheyshev s inequlity: ω(α) α {f>α} f p, α >. p Proof. Clerly, for ny α >, α p ω(α) = {f>α} αp dx f p dx, {f>α} then ω(α) α p {f>α} f p dx.
Question 9 If p > nd f f k p s k, show tht f m k f on. Proof. Since f f k p, given ε >, K when k > K f f k p εp. By the conclusion of Question8, ω k f f k (ε) = { f f k > ε} f f ε p { f f k >ε} k p ε f f k p. Thus, f m p k k f on. Question If p > nd f f k p, nd f k p M for ll k, show tht f p M. Proof. By the conclusion of Question9 we hve f k m f, then there is susequence f kj f.e. in. Since α p is continuous, then f kj p f p.e. in. By Ftou slemm, f p = lim f k j p lim inf f j j k j p. Since f k j p M, lim inf f j k j p M. Thus, f p M. Question For which p > does /x L p (, )? L p (, )? L p (, )? Proof. First we consider for which P > such tht /x L p (, ). dx = lim(r) dx = lim ( [,] x p ε ε x p ε p ε p ), if < p <, we hve lim ( ε p ε p ) =. If p >, lim ( p ε p ε p ) =, if p =, dx = lim dx = lim ln ε = +. Thus, when < p <, /x [,] x ε ε x ε L p (, ). Now we consider for which p > such tht /x L p (, ). [, ) lim (R) N N dx = lim x p N If < p <, lim N when p >, x Lp (, ). [, ) p >, [, ) dx = x p ( p N p ( p dx + x p dx = x p dx =. x p p ). If p >, we hve N p [, ] ) = +. If p =, dx = +. Thus, [, ] x dx. x p dx =, thus, / x p x Lp (, ). From the ove discussion, for ny 3
Question Give n exmple of ounded continuous f on (, ) such tht lim x f(x) = ut f / L p (, ) for ny p > xmple: Put f(x) =, f(x) < on (, ) nd lim ln(x+4) x f(x) = (ln(x+4)). For ny p >, since lim p, then such tht (ln(x+4))p x (x+4) (x+4) for ll x. Hence ( ln(x+4) )p dx Question 3 for x nd ( (ln(x+4)) p x+4 ln(x+4) )p dx > dx = +. Thus f(x) / (x+4) Lp (, ) for ny p >. () Let {f k } e sequence of mesurle functions on. Show tht f k converges solutely.e. in if f k < +. () If {r k } denotes the rtionl numers in [, ] nd { k } stisfies k < +, show tht k x r k converges solutely.e. in [, ]. Proof. () For ech k, since f k is mesurle, f k is lso mesurle. By the monotone convergence theorem, ( f k ) = f k < +, so fk L(). Thus, f k < +.e. in nd hence f k converges solutely.e. in. () Put f k = k x r k, f k is mesurle on [, ], nd then f k is nonnegtive nd mesurle on [, ]. Note tht But [,] k= [,] f k = k= [,] k x r k = k= k x r k. [,] rk x r k = x r k + x r k = ( rk + r k ), r k for k =,,.... Hence, k= [,] f k k < +. Thus y prt (), k x r k converges solutely.e. in [, ]. k= 4
Question 4 Prove the following result (which is ovious if < + ), descriing the ehvior of p ω() s +. If f L p (), then lim + p ω() =. (If f, ε >, choose δ > so tht {f δ} f p < ε. Thus, p [ω() ω(δ)] {<f δ} f p < ε for < < δ. Now let.) Proof. Since >, we only need to consider f which is nonnegtive. Put f, f k = { if f k f else Clerly, f p k f p nd f p k f p. Since f L p (), f p k f p, for ny given ε >, K such tht f p k f p = { f k } f p < ε when k > K. Put δ = k. Then {f δ} f p < ε. For ny < < δ, p [ω() ω(δ)] { f δ} f p {f δ} f p < ε. Let, then lim + p ω() < ε. Thus, we hve lim + p ω() =. Question 5 Suppose tht f is nonnegtive nd mesurle on nd tht ω is finite on (, ). If α p ω(α)dα is finite, show tht lim + p ω() = lim + p ω() =. Proof. Put <. Since ω(α) decreses, α p ω(α)dα ω() α p dα = p p ω(). p p Hence, nd p ω() pp p α p ω(α)dα, α p ω(α)dα α p ω(α)dα = α p ω(α)dα α p ω(α)dα, when, we hve αp ω(α)dα αp ω(α)dα, so lim + p ω() =. 5
Hence Similrly, α p ω(α)dα ω() p ω() α p ω(α)dα = pp p α p dα = p p ω(). p p α p ω(α)dα α p ω(α)dα. α p ω(α)dα. Since α p ω(α)dα is finite, when, we hve αp ω(α)dα αp ω(α)dα. Thus, lim + p ω() =. Question 7 If f nd ω(α) c(+α) p for ll α >, show tht f L r, < r < p. Proof. By conclusion of Question6, we hve f r = r α r ω(α)dα cα r (+α) p dα = rc α r dα, so we just need to prove α r dα < (+α) p (+α) p r +. α r dα = (+α) p α r (+α) p (+α)r (+α) r dα = ( α dα. Put x = +α, then (+α) +p r since + p r >. So we hve f r < +, f L r. dα = (+α) +p r +α )r dα (+α) +p r dx < +, x +p r Question 8 If f, show tht f L p if nd only if + kp ω( k ) < +. Proof. ( = ) By Question 6, f p dx = p α p ω(α)dα. Let Γ = { k } e the prtition of (, ), then α p ω(α)dα R Γ = + ( k ) p ω( k )( k+ k ) = + ( k ) p ω( k ). Since f L p, α p ω(α)dα < +. Hence, + kp ω( k ) < +. ( = ) Since ω(α) is decresing, ω(α) is finite on (, ). By Question6, f p dx = α p dω(α) = + k α p d( ω(α)). Since α p is continu- k ous on [ k, k ] nd ω(α) is monotone incresing on [ k, k ], clerly, we hve k α p d( ω(α)) sup α p V [ ω(α); k, k ] = ( k ) p (ω( k ) k α [ k, k ] 6
ω( k )). So + ) + k k α p d( ω(α)) + ( k ) p ω( k ) < +. Hence, f L p. Question ( k ) p (ω( k ) ω( k )) = ( p Let y = T x e nonsingulr liner trnsformtion of R n. if f(y)dy exists, show tht f(y)dy = dett f(t x)dx. T Proof. First consider f = χ, here is mesurle suset of. Since T is nonsingulr liner trnsformtion, T is Lipschitz trnsformtion nd T is mesurle. f(y)dy = χ dy = = T det(t ) = dett χ T T (x)dx = dett χ T (T x)dx = dett f(t x)dx. T Next if f is simple mesurle function, put f = N i χ i (y), then we hve f(y)dy = i= i χ i (y)dy = N iχ i (y)dy = N dett T i= i χ i (T x)dx = dett T i= i= i χ i (T x)dx = dett f(t x)dx. T i= If f is nonnegtive function, there is incresing sequence of simple functions f k tht converges to f. By the monotone convergence theorem, f(y)dy = lim f k = lim f kdx = lim dett f T k(t x)dx = dett lim f T k(t x)dx = dett f(t x)dx. T For mesurle function f, f = f + f, f + nd f re oth nonnegtive, then the conclusion is true for f + nd f. So f(y)dy = dett f(t x)dx. T Question If f = for every mesurle suset A of mesurle set, show A tht f =.e. in. Proof. Put = {f }, nd = {f < }, then we hve = fdx nd hence f =.e. on. Similrly, we cn conclude tht f =.e. on. It is now cler tht f =.e. on the set. 7