Analysis II Home Assignment 4 Subhadip Chowdhury

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1 Analysis II Home Assignment 4 Subhadi Chowdhury Problem 4. f L (R N ) R N ( + x α ) ( + log x β ) < Problem 4.2 ( f ( f ) q ) ( ) q ( () /( q ) q ) dµ = f q. Ω /q f Ω q f q Thus f L q f L. Also f g injection. ɛ Ω q f g q ɛ. So the L (Ω) L q (Ω) with continuous Problem Note that f g + f + g h = max{f, g} = 2 Now L is a vector sace. So f + g L. Then by definition, f + g L. So h L Note that h n = f n g n + f n + g n 2 Now f n f in L (Ω) and g n g in L (Ω) imly that f n g n f g in L (Ω). Thus h n h in L (Ω).

2 4.3.3 Note that f n g n fg c ( f n f g n + g n g f ) (*) for some constant c deending only on. Now gn g a.e. and (g n ) is bounded in L, suose g n k for all n. Hence by DCT, we get g n g. Thus taking n in ( ), we get f n g n fg ck f n f + f g n g So f n g n fg in L. Problem We rove the result by inducting on k. For the base case, k = 2. Then for = + 2, we have ( f f 2 = ) ( f f 2 ( ) ( f. Note that = = = 2. Thus f 2 = f 2 2 f f 2 f f 2 2 ) f 2. ) Suose for k = m N, we have g m i= g i i whenever g i L i and = m Then let q = m+ i= and q i and f i L q i. We have, q = q m+ + m i= (f f 2... f m ) L mi= qi f q f m+ qm+. f f 2... f m mi= qi f m+ qm+. Thus by induction rincile the result is true for all k N. = f. f 2 i= i. q i. Then by induction hyothesis, m f i qi = i= m+ i= f i i Suose α,. Put =, α 2 = q. Then note that f α L f α L α. Similarly, f α L q α. So alying 4.4. with, 2 defined as above and noting that = r + 2, We have ( f r f α. f α 2 = f α /α ) α. ( For the cases α =,, the inequality is trivially true. Problem 4.5 ) α f α q q α = f α f α q We call the set C. Let f n be a sequence in C which converges to f in L (Ω). We want to rove that f C. Now f n f in L (Ω) a subsequence (f nk ) and a function h L such that 2

3 (a) f nk (x) f(x) a.e. on Ω (b) f nk (x) h(x) k, a.e. on Ω Then alying Fatou s lemma to ( f nk q ), we get, f q f C. Thus C contains all its limit oints. hence C is closed By 4.4.2, we have f n f r f n f α f n f α q where α is defined as in Note that by art 2, we have f q C. Thus f n f q 2C. Hence SO f n f in L r (Ω). Problem 4.6 f n f r f n f α (2C) α 4.6. Taking in roblem 4.2, we get that lim su f f () We know that there exist a sequence C n f such that f(x) C n a.e on Ω for all n. Then let Ω n = {x Ω : f(x) > C n }. We have f f µ(ω n )(C n ) f C n (µ(ω)) / Ω n Taking, we then have lim inf Combining above inequalities () and (2), we get our result. f C n for all n. Taking n we then have lim inf f f (2) Fix a k > C. Consider the set S = {x Ω : f(x) > k}. Suose µ(s) = ɛ >. Then C f kɛ / Taing we then have C k,contradiction!! So µ(s) = roving that f C i.e. f L (Ω) Consider f(x) = ln x for x (, ). Then we know that for < lim(ln t) 2 t = lim t + = t 2 t 2 3

4 ɛ > such that for < x < ɛ, ln x < ln x dx <. Then x 2 ɛ = ɛ 2 2 < x dx + ɛ ln x dx + a finite number Thus f L (Ω) for <. But clearly ln x is unbounded on (, ). So f L (Ω). Problem 4.7 Let (u n ) be sequence in L (Ω) converging to u such that au n f in L q (Ω). By theorem 4.9, by assing to a subsequence if necessary, we may assume u n (x) u(x) a.e. Then au n (x) au(x) a.e. Hence f = au a.e. Thus by closed grah theorem, the oerator T : L L q given by u au is continuous. So C > such that au q C u (*) Case : If <, then Putting u = v q Ω au q C q ( for some v L q, we get that ( a q v C q Ω Ω v q Ω u ) q ) q = C q v q for all v L q. Thus the ma S : L q R given by v Ω a q v is a continuous linear ma. ( ) Hence S L q a q L q q i.e. a L q. Case 2: If =, from equation ( ) we have a q C a L q. [Note that L ]. Problem For every integer n, consider the sets X n = {f X L + n (Ω) : f + n Note that if f X then f L for some >. But also f L. Hence f L r for all r. Note that from 4.4.2, f r K for some constant K for all r. Hence f X n for some n N. Thus X n = X n N Then since each X n is closed, by Baire category theorem, there exists N N such that Int(X N ). Let x Int(X N ). Then there exists r > such that x + B(, r) X N L + n. But x L + n. Hence B(, r) L + n λb(, r) L + n for all λ R X λb(, r) L q for q = +. n λ n} 4

5 4.8.2 Consider the inclusion ma T : X L given by T u = u. Suose we have a sequence (u n ) converging to u in X (note that X is closed) such that T u n v in L. Then by theorem 4.9, there is a subsequence (u nk ) such that u nk (x) u(x) a.e. Then T u nk (x) u(x) a.e. imlying u = v a.e. So by closed grah theorem, T is continuous. Hence there exist some C such that f C f ; f X Problem 4.9 We may assume WLOG Ω =, by dividing the measure by Ω if necessary. Then j(f)dµ = (af + b)dµ since j is convex l.s.c. Ω Problem 4. su Ω a,b R at+b j(t) t D(j) su a,b R at+b j(t) t D(j) ( = j fdµ Ω ) ( b + a Ω ) fdµ 4.. Since j is convex and integration is linear, we have J is convex. Problem We want to show that u + v α ( u + v ) α u α + v α We claim that ( u + v ) α u α + v α. Then it is enough to show that for x, (x + ) α x α + (*) Let f(x) = (x + ) α x α. Then f (x) = α((x + ) α x α ) < imlying that f is a decreasing function of x [, ). Hence f(x) f() = ( ). Putting x = u, we get our required result. v 4..2 Let f = u α g = v α 5

6 Then We have [u] α + [v] α = f α + g α = f α = = = u α + g α v α f α u α + g α v α f α ( α) u α + g α ( α) v α ( ) α f α ( u + v α ) + ( f α + g α ) α ( u + v α ) ( g α ) α ( u + v α ) since u, v ([u] α + [v] α ) α = ( ) α f α + g α u + v α = ([u + v] α ) α [u] α + [v] α [u + v] α Thus L α is a vector sace but not a norm. Problem Since the inequality is homogeneous, we may divide by b if necessary, to assume wlog that b =. Let f(a) = be a function from {a : a + > } R. ( a + ) 2 ) ( a + 2 a+ 2 2 a Problem Case : a + b a Then, a + b a b a + b a + b = a + b a + b b + b = 2 b Case 2: a + b < a Then a a + b b = b a a + b b. Thus, a + b a b = a a + b + b b + b = 2 b 6

7 4.3.2 Note that su n fn M. Thus by Fatou s lemma, f lim fn. So f L (Ω). Let a = f n f, b = f. Consider the sequence ϕ n = a + b a b. Then ϕ n 2 b = 2 f. Also note that ( a + b a b )(x) = f n (x) f(x) + f n (x) f(x). Then by dominated convergence theorem, ϕ n. Now ( f n f n f ) f ( f n f n f f ) = ( f n f n f ) ϕ n f (*) Note that f n f f n < M for some constant M for all n. Then by art 2, lim f n f = lim Problem 4.4 f n f by ( ) = lim f n f = lim f n f = 4.4. Let us denote the measure by µ. See art Note that for a.e. x, f n (x) f(x) f n (x) f(x) < α for all n bigger than a sufficiently large N. Thus x {x : f k (x) f(x) > α} k N for some N for a.e. x. Thus χ Sn(α)(x) a.e. Also clearly χ Sn(α) < L (Ω) since µ(ω) <. Thus by dominated convergence theorem, χ Sn(α) µ(s n (α)) Also µ[ f n f > α] µ(s n (α)) µ[ f n f > α] Let α = m for some fixed m N. Then by art 2, µ(s n ( )) m Thus given δ >, there exists N m N such that µ(s n ( m )) < δ 2 m for all n N m. Now x S Nm ( m ) f k(x) f(x) m for all k N m Let Σ m = S Nm ( ). Let Σ = Σ m m. Then µ(σ) < δ. Note that x Σ x Σ m for all m. m N Fix any ɛ >. Then there exists M N such that < ɛ Then x Ω\Σ x Σ M M f k (x) f(x) M < ɛ for all k N M 7

8 i.e. f k f uniformly on Ω\Σ. Taking A = Σ, we rove the theorem Fix an ɛ >. Then δ > such that A f n < ɛ for all n and A Ω measurable with µ(a) < δ. By art 3, there exists a certain A δ and an integer N δ > such that µ(a δ ) < δ and f n (x) f(x) < for all n N δ for all x Ω\A δ. So in articular, f n < ɛ and f n (x) f(x) < ɛ A δ Ω\A δ ɛ µ(ω\a δ ) for all n N δ. Then ( f n f = f n f + A δ f n f Ω\A δ ) f + A δ f n + ɛ A δ = 2ɛ + f A δ But f n L (Ω) and su n A δ f n <. Also f n (x) f(x) for a.e. x A δ. So by Fatou s lemma, A δ f lim inf A δ f n ɛ. Thus f n f 3ɛ for all n N δ Hence f n f i.e. f n f in L (Ω). Note that (f n f) and f n both are in L (Ω), imly that f L (Ω). Problem (i) For x, Thus f n a.e. lim ne nx = lim n = lim enx + x + n nx2 /2 + n 2 x 3 / = 4.5..(ii) f n = f n = ne nx dx = n e z dz = e n ; n N 8

9 4.5..(iii) f n = e n (i) For x, Thus g n a.e (ii) n lim n e nx = lim g n = g n = = lim enx n + n x + n 2 x2 /2 + n 3 x3 / = ne nx dx = n e z dz = ( e n ) ; n N (iii) g n = ( e n ) (iv) Observe that g n (x)x m dx = n e nx x m = Thus for any olynomial ϕ(x) R[x], we have n n m+ Since the olynomials are dense in L (Ω), we have for all h L (Ω) i.e. g n. n g n ϕ g n h e y y m dy n Γ(m + ) nm+ Problem Let K n = conv ( i=n {f i}) We claim that n=k n = {f}. Indeed since f n s are bounded, by Cantor s intersection theorem, n=k n. if g n=k n, then g(x) conv ( i=n {f i(x)}) for all n i.e. f n (x) g(x) a.e. Thus f(x) = g(x) a.e. So g f in L (Ω) and n=k n = {f}. Then by exercise 3.3.2, f n f weakly in σ(l, L ). 9

10 4.6.2 By theorem 4.9 alied to the sequence (f n f) in L, we get that there is a subsequence (f nk f) such that f nk f a.e. Then by art, f nk f weakly. Indeed, we can say that every subsequence of (f n f) has a further subsubsequence (f nki f)such that f nki f weakly. Thus f n f weakly fix an ɛ >. By Egorov s theorem, A Ω such that A < ɛ and f n f uniformly on Ω\A. Now can write f n f q = f n f q A }{{} Note that by uniform convexity for sufficiently large n, And by roblem 4.2, we have I J Ω\A ɛ q + Ω\A f n f q } {{ } J I q A q fn f < I ɛ q fn f q But f n f is bounded by some constant for all n since f n, f L (Ω) f n f L (Ω) and f n is bounded in L. Thus I + J < C ɛ q + C 2 ɛ q for constants Ci not deending on. So, f n f q f n f q Problem Note that the inequality is homogeneous of degree. WLOG, let a b. Then we can divide both sides by a. Let t = b/a. Then it is enough to show that + t t C ( t + t ) for some constant C deending on for all t. Note that and Let + t t lim t + ( t + t ) + t t lim t ( t + t ) g(t) = = lim t + ( + t) (t) ( + ( )t 2 ) = lim t ( + t) ( t) ( + ( )( t) 2 ) { +t t ( t + t ) t t = Then clearly g is a continuous function on [, ]. So it is bounded above since [, ] is comact. Thus + t t su t ( t + t ) < C for some C deending only on which roves the given inequality. = =

11 4.7.2 Note that by roblem 4.6., we have f n f in L. Also f L f L f L f L. So f n f f Also note that ( fn f ) = f n f < So by 4.6., (f n f) in L. Thus f n f f Put a = f n f and b = f in question. Then f n f n f f < C ( f n f f + f n f f ) The assertion in art 2 follows By art 2, So f n f. lim f n f = Problem Take any ste function g L (, ) such that g = l i= a iχ Ai, where a i R and A i is an interval (b i, b i+ ) with A i = (, ). We have u n g = n f(nx)g(x)dx = f(y)g(y/n)dy n ( l = ) nb i+ nb T i a i f(y)dy + n T i= n ( l ) bi+ f a i dy + l nbi+ a i n i= fg + n fg b i l a i f,[,t ] i= i= l nbi+ a i i= f(y) dy n b i+ b i T +nb T i [since f is eriodic] f(y)dy n b i+ b i T +nb T i since f L bounded term loc means the second term is of the form ; and hence it goes to. Since functions n of above form are dense in L (, ), we get that u n f weakly.

12 4.8.2 Let kt n (k + )T for k Z. u n f = = n = n T f(nx) f dx n ( k T f(y) f dy T n ) f(y) f dy + f(y) f dy kt f(y) f dy + n f f,[,t ] So u n f T T f(y) f dy ( T ) lim u n f = f(y) f dy T (i) u n (x) = sin(nx) u n L (, ) for all. 2π By art, u n sin ydy = weakly in σ(l, L ). 2π ( ) 2π Also u n sin sin y dy 2π (ii) u n dx = n 2 (α +β ) < for all <. Hence u n L (, ) for all <. Also su u n = max{α, β}. Hence u n L (, ) for all. u n α+β weakly in σ(l, L ). 2 ( Also u n f f dy Problem 4.9 ) = ( α + β 2 ) 4.9. Since L is uniformly convex for < <, we have by ro. 3.32, f n f strongly in L Consider f n (x) = sin(2πnx) +. Then given g L (, ), we have So, f n. sin(2πnx)g(x)dx + g = n n sin(2πy)g(y)dy + g n g + g g 2

13 Problem Clearly by the given inequality, f L Af L q. Now take any sequence u n u strongly in L. Then by theorem 4.9, we can get a subsequence u nk and a function h L such that u nk (x) u(x) a.e. and u nk (x) h(x) for all k, a.e. Then for some h L q, we have Au nk (x) Au(x) a.e. and Au nk (x) h (x) for all k, a.e. Thus by dominated convergence theorem alied to (Au nk ) q we get that Au nk Au q By uniqueness of limit we then get that Au n Au strongly in L q. Hence A is continuous from L strong to L q strong From 4.8.3(ii), we know that u n (x) = f(nx) α+β 2 where f is defined as in the roblem 4.8.3(ii). Now by given hyothesis Au n Au; i.e. a(u n ) a( α+β a(u n (x)) = a(f(nx)) = { a(α) a(β) Since L q (, ), integrating Au n and Au against, we get k 2 ) weakly in σ(lq, L q ). Now < x < k+ 2 n n k+ 2 < x < k+ n n n + β (a(α) + a(β)) = a(α ) 2n 2 i.e + β (a(α) + a(β)) = a(α ) 2 2 for all α, β R. Clearly the only solution to the functional equation is a(x) = (a() a())x + a() which is an affine function. Problem Take g = χ (a,b). Then R u n g = b a u (x + n)dx = b+n a+n u (x)dx = b+n u (x)dx a+n u (x)dx as n. Thus for any ste functions(with comact suort) g, we have u n g. Since ste functions(with comact suort) is dense in L (R). we get that u n weakly Take g = χ I. Let I = (a, b). Fix a δ >. Let E = [ u > δ]. Then R u n g = I u (x + n)dx = I+n u (y)dy = (I+n) E u (y)dy + u (y)dy (I+n) E c Now since E <, there exists N such that for n > N, we have (I + n) E < δ. Then u n g u δ + δ (I + n) E c u δ + δ I R Taking δ, we get that R u ng. Thus for ste function g with comact suort, R u ng. Since ste functions with comact suort are dense in L (R), we get that u n. 3

14 4.2.3 Suose, by contradiction, there is a subsequence u nk which weakly converge. Then for any g L, u nk g. Take g L defined by g(x) = { ( ) i n i < x < n i+ o.w. Then u nk g = ( ) k which do not converge as k. Thus we have a contradiction! Hence such a subsequence does not exist. Problem A B Since χ E L for any measurable E with E <, this imlication is clear by definition of weak convergence. [or weak* convergence in case =.] B A Note that, since simle functions are dense in L (Ω), the vector sace sanned by characteristic functions χ A of subsets A of Ω with A < are dense in L (Ω) for <. Now for any element g of the vector sace, we have Ω f ng Ω fg. Thus Ω f nh Ω fh for all h L (Ω) i.e. f n f Again the imlication A B is trivial. For the other direction observe that every simle function belongs to L (Ω) if Ω <. Hence the vector sace sanned by characteristic functions χ A of subsets A of Ω with A < are dense in L (Ω). Thus A follows by the same argument as in art By roosition 3.5, and definition of weak convergence, A B. To see that the oosite imlication is not true consider Ω = R. Consider the sequence f n = χ (n,n+). then clearly f n = < and f E n = for any E R with E <. But clearly by roblem E f n does not weakly converge Given ɛ >, we can find a subset ω Ω such that ω c f < ɛ and ω <. Now f n + f n = f n f = f + f ω c ω Ω Ω ω c ω f n f f f n ω c ω c ω ω (*) since ω <. Also we have, for any F Ω with F, f n = f n + f n f + F F ω F ω c F ω f n F ω c (f n f) F (f n f) F ω c f n + ω c f 2 ω c f < 2ɛ ω c since f n, f. Thus we get F (f n f) for F measurable subset of Ω, F. Since vector sace sanned by χ F for such F s is dense in L (Ω), we get that f n f. 4

15 Problem Clearly by definition C is convex. Suose (u n ) is a sequence in C which converge to u in L. Then there is a subsequence (u nk ) such that u nk (x) u(x) a.e.[since < ] Thus u(x) f(x) a.e. Hence C is closed. Now C is convex and strongly closed C is weakly closed Assume first that f L (Ω). Call the set mentioned in art 2 C. Then clearly, C C. Now suose, u C. Thus fϕ uϕ ϕ L (Ω) with fϕ L (Ω) and ϕ a.e. Since Ω is σ finite we can find Ω n with Ω n = Ω and Ω n <. Let Ω n = Ω n { f < n}. Then Ω n = Ω. Let A = {f < u}. Then choose ϕ = χ A Ωn So f ua.e. imlying C C. A Ω n f u A Ω n = A = The second definition C clearly shows that C is closed in σ(l, L ) since it is the intersection of all closed sets in σ(l, L ) of the form { } u L (ω) fϕ uϕ and fϕ L (Ω) for a fixed ϕ L (Ω) Note that a weak* closed and bounded set is comact in weak* toology, since we can scale down the bounded set into a closed subset of the unit ball in weak* toology which is comact. Hence being a closed subset of a comact set, the scaled down set will be comact giving us that the original set is comact. Now given set C is a weak*(σ((l ), L ))closed and bounded set; hence it is comact. Problem Given ϕ L (R N ), we have by ro. 4.6, ϕ(ρ n ζ n u) = ζ n u( ˇρ n ϕ) Hence, v n ϕ vϕ = = ζ n u( ˇρ n ϕ) ζϕu ζ n u( ˇρ n ϕ ϕ) + uϕ(ζ n ζ) ζ n u ˇρ n ϕ ϕ + u (ζ n ζ)ϕ u ˇρ n ϕ ϕ + u (ζ n ζ)ϕ first term by thm and second term by DCT. 5

16 Let B = B(x, R) and let χ := χ B(x,R+). Then let w n = ρ n (ζ n uχ). Note that w n = v n on B since by roosition 4.8, su(w n v n ) B(, n ) + B(x, R + ) c ) Thus v n v = w n χv B B ρ n (ζ n uχ) χζu R N ρ n (ζ n uχ) ρ n (χζu) + ρ n (χζu) χζu R N R N (ζ n uχ) (χζu) + ρ n (χζu) χζu R N R N Problem Let u (x) = { u(x) x Ω x Ω c Then u L (R N ). Let Define Ω n = {x Ω dist(x, Ω c ) > 2/n; x < n} ζ n := χ Ωn, ζ := χ Ω Then note that ζ n ζ a.e. on R N and ζ n for all n. Define v n = ρ n (ζ n u ) and v = ζu = u. Then by roblem 4.24, v n v in L (R N ). Thus v n u = u on L (Ω). Also v n C c (Ω). and B v n v = B v n u for every ball B. Thus we may find an subsequence (v nk ) which converges a.e. to u (x) on B. By a diagonal rocess, we can find a further subsequence (v nki ) which converges a.e. to u on R N. Renaming this subsequence (u n ), we see that u n satisfies all given roerties By our construction u a.e. ρ n (ζ n u ) a.e By art, C c (Ω) is dense in L (Ω). 6

17 Problem If f L (Ω) then A su f ϕ f <. ϕ Conversely, if A <, Then for all ϕ C c (Ω), we have ϕ f A fϕ A ϕ ϕ Fix any comact subset K of Ω. Let ψ C c (Ω) be a function such that ψ and ψ on K. Let u L (Ω) be any function. Then by roblem 4.25, we can find a sequence (u n ) in C c (Ω) such that u n u and u n u a.e. on Ω and u n u in σ(l, L ). Then taking ϕ = u n ψ we get fψu n A ψu n A u By DCT, taking limit as n, we then have fψu A u Choose u = sign(f). Then K f. f.u.ψ A for all comact subset K of Ω. Thus f A < i.e. f L (Ω). In case f L (Ω), by above roof we have A f A f = A If f + L (Ω) then B su (max{f, }ϕ) ϕ ϕ Conversely if B <, we can roceed as in art, to get that fψu B u su ϕ ϕ f + ϕ f + <. for all u L (Ω) Put u = χ {f } K. Then K f + B for every comact subset K of Ω. Hence f + B. The roof clearly shows that f + = B if f L (Ω) Note that the sequence constructed in roblem 4.25 is actually in C c (Ω). Hence the roof is same [ [ ] fϕ = ϕ Cc (Ω) A = f = f = a.e. ] fϕ ϕ Cc (Ω), ϕ B = f + f > a.e. 7

18 Problem 4.27 Fix a ϕ Cc (Ω) such that uϕ =. Fix any ɛ >. Then u(ϕ (ϕ (uϕ)) + ɛϕ ) = + ɛ > v(ϕ (ϕ (uϕ)) + ɛϕ ) Taking ɛ, we then have vϕ uϕ vϕ = λ uϕ Conversely, taking ɛ < and relacing ϕ (ϕ (uϕ)) + ɛϕ by ϕ (ϕ (uϕ)) ɛϕ we get that as ɛ, vϕ uϕ vϕ = λ uϕ. Thus (v λu)ϕ = for all ϕ C c (Ω). Hence v = λu a.e. Clearly by construction λ. Problem 4.28 We can aroximate ρ by a sequence of Cc (Ω) functions (ψ i ) such that ρ ψ i. Define ψn(x) i = n N ψ i (nx). Note that ψn i = n N ψ i (nx)dx R N... dx N = ψ i (y)dy R N... dy N ρ =. Fix K R N a comact set and ɛ >. Then there exist n large enough such that suψn i B(, δ) where δ is such that f(x y) f(y) < ɛ for all x K and for all y B(, δ). Then for n sufficiently large, (ψn i f)(x) f(x) dx = f(x y) f(x) ψn(y)dy i dx f(x y) f(x) ψn(y)dydx i ( ) = ψn(y)dy i f(x y) f(x) dx ɛ ψ i ɛ ψ i ɛ B(,δ) Hence taking ɛ, we get that ψ i n f f in L. Taking limit i, we then have ρ n f f in L. Problem 4.29 Let χ n := χ K+B(, 2n ) and take u n = ρ 2n χ n where ρ(x) = {e x 2 x < x > and ρ n(x) = Then all it remains to check is that u n satisfies (a), (b), (c) and (d). Note that u n χ n ρ 2n = 8 ρ n N ρ(nx).

19 So (a) is true. Next on K, ρ 2n χ n = ρ 2n = So (b) holds. Also So (c) holds. Lastly, Now So So (d) follows. D α ρ 2n = C(2n) α Problem 4.3 su(u n ) K + B(, 2n ) + B(, 2n ) K + B(, n ) D α u n (x) = D α ρ 2n (x) χ n D α ρ 2n D α (Cn N ρ(2nx)) = Cn N (2n) α (D α ρ)(2nx) (D α ρ)(2nx) n N dx = C(2n) α (D α ρ)(2y) dy = C α n α Let + = = q + q. Then r = q. Then set α = q and β = q. Note that ( α)r = ( q )r = r r =, and similarly ( β)r = q. Then ϕ (y) = f(x y) α L q, ϕ 2 (y) = g(y) β L. Also for ϕ 3 (y) = f(x y) α g(y) β, we have ϕ r 3(y) = f(x y) g(y) q L ϕ 3 L r. We have, (f g)(x) = f(x y)g(y)dy = f(x y)g(y) dy By Holder s inequality y f(x y)g(y) is integrable. In fact, f(x y) α g(y) β ( f(x y) α g(y) β) (f g)(x) ϕ q ϕ 2 ϕ 3 r = f α g β q ϕ 2 r f g r f αr g βr q f(x y) ( α)r g(y) ( β)r dy = f αr g βr q f(x y) g(y) q dy f αr g βr q f g q q = f r g r q f g r f g q By revious art f g L (R N ) if r =. Also we know that convolution of a L function and a L function is continuous. Hence f g C(R N ) L (R N ). For <, q <, note that we can construct sequences (f n ), (g n ) in C c (Ω) so that f n f in L and g n g in L q. But clearly, su(f n g n ) su(f n ) + su(g n ) (f n g n )(x) = as x. But f n g n f g. Hence (f g)(x) as x. 9

20 Problem Let χ r := χ B(,r) B(,r). Then f r (x) = χ r (x y)f(y)dy = (χ r f)(x) Now f L and χ r L. Hence f χ r L. Also χ r L f χ r C(R N ) by roblem 4.3. Hence f r C(R N ) L (R N ). Also by 4.3, f r (x) as x for < < Note that χ r =, and su(χ r ) = B(, r) Hebce χ n f /n f in L as n f r f in L as r. Problem 4.32 is a sequence of mollifiers. Hence By definition of convolution it is immediate that f g = g f. Next note that (f g) h(u) = (f g)(x)h(u x)dx ( ) = f(y)g(x y)dy h(u x)dx = f(y)g(x y)h(u x)dydx = f(y)g(x y)h(u x)dxdy ( ) = f(y) g(x y)h(u x)dx dy ( ) = f(y) g(x + y y)h(u x y)dx dy = f(y)(g h)(u y)dy = (f (g h))(u) Thus (f g) h = f (g h). Problem Note that comactly suorted continuous functions are uniformly continuous. Hence given ɛ >, there exist δ > such that h < δ imlies ϕ(h + x) ϕ(x) < ɛ for any x R. Thus in fact ϕ(x + n + h) ϕ(x + n) < ɛ for any x R and any n N. So for h < δ, we have τ h ϕ n ϕ n < ɛ. su(ϕ) / for any n. Hence ɛ > ; δ > such that τ h f f < ɛ f F and h R with h < δ. 2

21 Fix ɛ = 3. Consider the oen balls B(f, ɛ) L (R) for each f F. Then clearly they form an oen cover of the closure of F. If the closure is comact, this cover has a finite subcover. Let it be k B(ϕ ni, ɛ). Take N >> max n..., n k so that su(ϕ n ) ( ) k su(ϕ ni ) =. Then i= ϕ N ϕ ni = ( 2 ϕ ) Hence the closure is not comact. Problem 4.34 = C, some constant. Hence if ɛ < C/2, ϕ N k B(ϕ ni, ɛ). Contradiction! Fix an ɛ >. Then there is a finite covering of F by oen ɛ balls. Let F n B(f i, ɛ). i= i= i= Given any f F, f B(f i, ɛ) for some i {,..., n}. Hence f f i +ɛ max{ f i : i n} + ɛ. Hence F is bounded By lemma 4.3, for each i {,..., n}; lim τ hf i f i = h Hence for the same ɛ as above, there exists δ i > such that τ h f i f i < ɛ i and h R N with h < δ i Put δ = min{δ i : i n}. Take any f F, suose f B(f k, ɛ). Then τ h f f τ h f τ h f k + τ h f k f k + f k f = 2 f k f + τ h f k f k 3ɛ for all h < δ Note that f i L (R N ) Ω i R N bounded,oen, such that f i L (R N )\Ω i < ɛ where ɛ is as above. Let Ω = n Ω i. Take any f F and let f B(f k, ɛ). Then i= f L (R N \Ω) f k L (R N \Ω) + f f k L (R N \Ω) f k L (R N \Ω k ) + f f k L (R N ) = 2ɛ Comaring with corollary 4.27, we find that this roblem is the converse of the corollary, and together they characterise all comact sets in L (R N ). Problem 4.35 Clearly F Ω is bounded in L (Ω). Now take f F and write f = G u for an u B. Then τ h f f = (τ h G G) u C τ h G G by lemma 4.3. Hence by theorem 4.26, F Ω has comact closure in L (Ω) for any measurable Ω with finite measure. 2

22 Problem [(d) + (e)] [(a) + (b) + (c)] Suose (d) and (e) hold. Then given ɛ >, there exists N > such that su f < ɛ t N f F [ f >t] [ f >t] Now given E measurable with E < ɛ, we have N f = f + E E [ f >N] f < ɛ f F, t N E [ f N] Thus given ɛ >, there exists δ > such that (b) holds. f [ f >N] We also know that given ɛ >, there exists N N such that su f < ɛ n N f F Ω\Ω n f < ɛ f F n N Ω\Ω n So we may take ω = Ω N and thus (c) holds. Clearly (d) imlies (a) is true. [(a) + (b) + (c)] [(d) + (e)] f + N E 2ɛ Suose (a), (b) and (c) hold. Then (a) imlies there is an M N such that f < M. Fix an ɛ >. Then (b) gives us an δ >. Let E = [ f > M ] Then E < δ since otherwise, f > M > f which δ E is not ossible. Thus (b) su f < ɛ f F [ f > M δ ] Thus given ɛ >, there exists a N > such that su f < ɛ for all t > N imlying lim su t f F f F [ f >t] [ f >t] f = By (c) we can roduce an ω satisfying roerties in (c). Then note that since ω <, and Ω n are nondecreasing with union the whole of Ω; we can find a N N such that ω Ω n < ɛ/m for all n > N. Then f f + f f + ɛ 2ɛ f F Ω\Ω n Ω\ω Ω n ω Ω\ω Thus we have (e) i.e. lim su f = f F Ω\Ω n 22

23 Problem We have I u n (x)ϕ(x)dx = = = I n n n n nf(nx)ϕ(x)dx f(t)ϕ(t/n)dt f(t) (ϕ(t/n) ϕ()) dt + ϕ() n n f(t)dt But the first term converges to by Lebesgue s theorem and the second term goes to by given condition: I u n (x) dx = since f L (R). Hence (u n ) is bounded. Note that for all δ >, δ u n = + n n nδ f = f(t) dt f + f > f < Hence there exist f > ɛ > such that for all δ >, there exists E = (, δ) such that E u n > ɛ for sufficiently large n. Hence no subsequence of (u n ) is equi-integrable Suose there exists such a function u L (I). Then by art, I uϕ = ϕ C([, +]) Then by corollary 4.24, u a.e. But for ϕ = χ (,), we have n u n ϕ = f f > Contradiction! Hence u does not exist By Dunford-Pettis thm, we have [ No subsequence of (u n ) is equi-integrable] [no subseqence of (u n ) weakly converges] which agrees with the fact that art 2 and art 3 hold simultaneously. The thm states that art 3 is a direct imlication of art 2. 23

24 I u n (x) dx = = [ n 2,n 2 ] n 2 n 2 + u n (x) dx + f(t) dt + [n 2 < t <n] [n 2 < x <] f(t) dt u n (x) dx since f L. Thus u n in L. Hence there is a subsequence (u nk ) such that u nk (x) a.e. Problem By definition su(u n ) = n ( n 2 ) = n. Also I n u n = j= Suose ϕ C ([, ]). Then n lim u n ϕ ϕ = lim I I j= n = lim n j= n = lim j= n lim j= j n + n 2 j n n = n j= n n 2 = u n = j n + n n n 2 ϕ lim j n ϕ(x j) n j= ) n 3 n 3 nj+ nj nj+ nj nj+ nj n = lim O( n n) 3 j= ( ( t ϕ n 2 ϕ where x j (j/n, (j + )/n) ( yj n 2 )) dt where y j (nj, n(j + )) ((t y j )ϕ (z j )) dt where z j (t, y j ) ( n(j + ) nj ϕ (z j )) dt = lim O( n ) = But C ([, ]) is dense in C([, ]). Hence lim u I nϕ = ϕ for all ϕ C([, ]). I The sequence (u n ) is not equi-integrable because for E n = su(u n ); E n but = So (b) does not hold in definition of equi-integrability. I u n = u n E n 24

25 If (u nk ) u, we have by art 2 and corollary 4.24, u a.e. Choose a further subsequence (u n k ) such that k su(u n k ) <. Let ϕ = χ A where ( ) A = I\ su(u n k ) so that A >. We have u I n ϕ = for all k and hence = ϕ = A. Contradiction! k I k Consider a subsequence (u nk ) such that su(u nk ) < k Let B k = j k (su(u n j )) and B = k B k. Clearly B k = j=k n j as k, and thus B =. If x B, there exists some k such that u nk (x) = for all k k. Thus u nk (x) a.e. as k. 25