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Solutions to the Exercises of Lecture 5 of the Internet Seminar 2014/15

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Solutions to the Exercises of Lecture 5 of the Internet Seminar 2014/15 Solutions to the Exercises of Lecture 5 of the Internet Seminar 2014/15 Exercise 5.1 Claim: Let V be a normed vector space, and let a: V × V ! K be a sesquilinear form. Then a is bounded if and only if a is continuous. (Note that this is slightly more general than the statement given in the exercise, since V needs not be a Hilbert space.) Proof. First assume that a is bounded, i.e., there exists a constant M ≥ 0 such that N ja(u; v)j ≤ M kukV kvkV for all u; v 2 V . Let ((un; vn))n2N 2 (V × V ) and u; v 2 V such that (un; vn) ! (u; v) in V × V as n ! 1 (which is equivalent to un ! u; vn ! v in V as n ! 1). Since (un)n2N is convergent there exists C ≥ 0 such that kunkV ≤ C for all n 2 N. Hence we compute ja(u; v) − a(un; vn)j ≤ ja(u; v) − a(un; v)j + ja(un; v) − a(un; vn)j = ja(u − un; v)j + ja(un; v − vn)j ≤ M ku − unkV kvkV + M kunkV kv − vnkV ≤ M ku − unkV kvkV + MC kv − vnkV for all n 2 N, which tends to 0 as n ! 1. Thus we have shown that a is continuous. In order to show the converse, assume that a is not bounded. Then there exist sequences N (un)n2N; (vn)n2N 2 V such that ja(un; vn)j > n kunkV kvnkV 1 un 1 vn for all n 2 . Now let un := p · and vn := p · for n 2 (note that un; vn 6= 0 N e n kunkV e n kvnkV N for every n 2 N). Then uen; ven ! 0 in V as n ! 1, hence (uen; ven) ! (0; 0) in V × V as n ! 1 . Furthermore, ja(un; vn)j ja(uen; ven)j = > 1 (1) n kunkV kvnkV for all n 2 N. Thus (uen; ven) ! (0; 0) as n ! 1, but a(uen; ven) does not converge to 0 as n ! 1 because of (1). Hence a is not continuous. Remark: Note that if V is a Banach space, then a sesquilinear form on V is bounded if and only if it is componentwise continuous, as we shall show now. Claim: Let V be a Banach space and let a: V × V ! K be a sesquilinear form. Then the following statements are equivalent (i) a is continuous; (ii) a is continuous at (0; 0); (iii) a is bounded; 1 (iv) every component of a is continuous, i.e., the mapping au : V ! K; v 7! a(u; v), is continuous for every u 2 V and the mapping av : V ! K; u 7! a(u; v), is continuous for every v 2 V ; Proof. The implications (i) to (ii) and (i) to (iv) are easy to see and the implication (iii) to (i) has already been shown above. Moreover, a closer look at the proof of Exercise 5.1 reveals that in order to show that a is bounded, one only needs that a is continous at (0; 0). Thus we already proved the implication (ii) to (iii). Hence we only have to show v the implication (iv) to (iii). To this end assume that au and a are continuous for all v 0 ∗ u; v 2 V . Since a is sesquilinear, we conclude that a 2 V for all v 2 V and au 2 V for all u 2 V . Thus for every u 2 V there exists a constant Mu ≥ 0 such that jau(v)j ≤ Mu kvkV for all v 2 V . Let B = fv 2 V : kvkV ≤ 1g. Then v sup ja (u)j = sup ja(u; v)j = sup jau(v)j ≤ sup Mu kvkV = Mu v2B v2B v2B v2B for all u 2 V . Applying the uniform boundedness principle (Banach-Steinhaus) (see for v instance Theorem 2.2 in [Bre11]) we conclude that supv2B ka kV 0 is finite, i.e., there exists a constant M ≥ 0 such that v ja (u)j ≤ M kukV for all u 2 V and all v 2 V with kvkV ≤ 1. Hence v v kvk ja(u; v)j = kvkV a(u; ) = kvkV a V (u) ≤ M kvkV kukV kvkV for all u; v 2 V; v 6= 0, which finishes the proof. Note that the same result holds if we consider Banach spaces X and Y and a sesquilinear or bilinear form b: X × Y ! K, the proof being the same as above. Exercise 5.2 a) Let V , H be complex Hilbert spaces, j 2 L(V; H) with dense range and a: V × V ! C a bounded and coercive form, i.e. there are constants M > 0 and α > 0 with ja(u; v)j ≤ MkukV kvkV and 2 Re a(u) ≥ αkukV for u; v 2 V: Let A denote the operator associated to the form a and T the semigroup generated by −A. α Claim: For all " 2 (0; kjk2 ] one can estimate kT (t)k ≤ e−"t for all t ≥ 0: 2 α 2 Proof. Let " 2 (0; kjk2 ); then C" := α−"kjk > 0: Define a sesquilinear form b: V ×V ! C by b(u; v) := a(u; v) − "(j(u) j j(u))H for u; v 2 V: By Cauchy-Schwarz and j 2 L(V; H) we obtain the boundedness estimate jb(u; v)j ≤ ja(u; v)j + "j(j(u) j j(u))H j 2 2 ≤ MkukV kvkV + "kjk kukV kvkV = (M + "kjk )kukV kvkV and the coercivity estimate 2 2 2 2 2 Re b(u) = Re a(u) − "kj(u)kH ≥ αkukV − "kjk kukV = C"kukV for u; v 2 V: Let B denote the operator associated to the form b. By Theorem 5.9 −B gen- erates a holomorphic contraction semigroup S and Remark 5.6 yields the representation −A = −B − " and we obtain kT (t)k = e−"tkS(t)k ≤ e−"t α α for all t ≥ 0. The estimate for " = kjk2 can be obtained by choosing ("n)n2N ⊂ (0; kjk2 ) α with "n ! kjk2 for n ! 1 and using the estimate above for each "n: d 1 b) Claim: Let Ω ⊂ R be an open set which lies in a strip of width d > 0: Let " 2 (0; d2 ]: Then t∆D −"t ke kL(L2(Ω)) ≤ e for all t ≥ 0: 1 1 Proof. Let a: H0 (Ω) × H0 (Ω) ! C be the classical Dirichlet form from the lecture given by Z a(u; v) := ru · rvdx: Ω Since Ω ⊂ Rd is an open set lying in a strip, we obtain by Poincar´e'sinequality (Theorem 5.12.) Z Z Z Z 2 2 2 2 a(u) = jruj dx ≤ jruj dx + juj dx ≤ (1 + cp) jruj dx = (1 + cp)a(u) Ω Ω Ω Ω 1 2 1 for u 2 H (Ω); where c 2 (0; d ) denotes the Poincar´econstant. So k · k 1 := a(·) 2 0 p H0 (Ω) 1 defines an equivalent norm on H0 (Ω) (one can check the norm properties, the definiteness follows directly from the estimate above). With respect to this norm, a is coercive with constant α = 1 and bounded because of the estimate 1 1 Z Z 2 Z 2 2 2 ja(u; v)j = ru · rvdx ≤ jruj dx jrvj dx = kuk 1 kuk 1 H0 (Ω) H0 (Ω) Ω Ω Ω 3 1 for all u; v 2 H0 (Ω): Furthermore, Poincar´e'sinequality yields the bounded embedding 1 p j : H0 (Ω) ,! L2(Ω) with operator norm kjk = cp ≤ d: Finally, we can apply part a) of this exercise and obtain t∆D −"t ke kL(L2(Ω)) ≤ e 1 for all t ≥ 0; if we choose " 2 (0; d2 ]: Exercise 5.3 Let −∞ < a < b < 1. By Theorem 4.9 we know that H1(a; b) ⊆ C[a; b] and so we always use the continuous representative for a function in H1(a; b). Claims: (a) Every function u 2 H1(a; b) is H¨oldercontinuous of index 1=2, i.e., there is a constant c 2 (0; 1) such that ju(t) − u(s)j ≤ cjt − sj1=2 for every t; s 2 (a; b). 1 (b) H (a; b) is compactly embedded in C[a; b], i.e., every bounded sequence (un)n2N in 1 H (a; b) has a uniformly convergent subsequence w.r.t. the norm k · kC[a;b]. (c) Let H2(a; b) := fu 2 H1(a; b): u0 2 H1(a; b)g, then u00 := (u0)0 is defined for all u 2 H2(a; b). Then we have that H2(a; b) ⊆ C1[a; b] if C1[a; b] carries the norm 0 kukC1[a;b] = kukC[a;b] + ku kC[a;b]: Proof. (a) Let u 2 H1(a; b); s; t 2 (a; b); s ≤ t. By H¨older'sinequality we have Z t Z t 0 0 ju(t) − u(s)j = u (x)dx ≤ ju (x)jdx s s 1 1 Z t 2 Z t 2 0 2 1 ≤ ju (x)j dx 1dx ≤ kukH1(a;b)jt − sj 2 : s s 1 (b) Let (un)n2N be a bounded sequence in H (a; b), i.e., there is a constant C 2 (0; 1) such that kunkH1(a;b) ≤ C for every n 2 N. Let " > 0 be arbitrary and define "2 δ := C2 > 0. Then it holds by part (a) for every s; t 2 (a; b) with jt − sj ≤ δ 1 1 jun(t) − un(s)j ≤ kunkH1(a;b)jt − sj 2 ≤ Cδ 2 = " 8n 2 N; i.e., the sequence (un)n2N is equi-continuous.
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