Hand in y our solutions un til F rida y , June 29, 8.30 am (PO b o x 183 in V3-128)

Hand in y our solutions un til F rida y , June 29, 8.30 am (PO b o x 183 in V3-128)

total points: 20

Prof. Dr. Moritz Kaßmann Fakultät für Mathematik

Sommersemester 2018 Universität Bielefeld

Partial Differential Equations III Exercise sheet XII, June 23

Exercise XII.1 (4 points)

Let(µn)be a sequence of probability measures and µa probability measure on a metric spaceX. Show that the following assertions are equivalent:

(i) µn converges weakly to µ, i.e.,´

fdµn→´

fdµ for everyf ∈Cb(X).

(ii) lim sup

n→∞

µ_n(A)≤µ(A) for every closed setA.

(iii) lim inf

n→∞ µ_n(O)≥µ(O) for every open set O.

(iv) lim

n→∞µn(B) =µ(B) for every Borel setB withµ(∂B) = 0.

Hint: You do not need to invoke any very deep result from measure theory. One option to establish the equivalencies, is to prove the following implications:

(ii)⇔(iii),(ii) + (iii)⇒(iv),(iv) =⇒ (ii),(i)⇔(ii).

For Exercises XII.2 and XII.3, letE be a Dirichlet form onL²(X, m)and(T_t)and(G_λ)be the corresponding semigroup resp. resolvent onL²(X, m). Givenα >0, we call a function u∈L²(X, m) α-excessive if for everyt >0 the inequalitye^−αtTtu≤u holds truem-a.e.

Exercise XII.2 (4 points) Prove the following observations:

(i) If uis α-excessive, then u≥0.

(ii) Forf ∈L²(X, m), f ≥0, the function Gαf isα-excessive.

(iii) If u1 ≥0 andu2≥0 areα-excessive, then so areu1∧u2 andu1∧1.

(iv) Let u ∈ L²(X, m), u ≥ 0, be α-excessive. Assume v ∈ D(E) with u ≤ v. Then u∈D(E) and E_α(u, u)≤ E_α(v, v).

Hint: The proofs of (i), (ii) and (iii) are simple.

Exercise XII.3 (4 points)

Letu∈D(E) andα >0. Show that the following statements are equivalent:

(i) u is α-excessive

(ii) E_α(u, v)≥0for every nonnegativev ∈D(E).

Exercise XII.4 (8 points)

The classical Sobolev inequality reads kfk

L

2d

d−2 ≤ck∇fk_L2 (f ∈C_c^∞(R^d)). (1) With the help of the Hölder inequality one can deduce from (1) the following inequality:

kfk¹⁺

2 d

L² ≤ck∇fk_L2kfk

2 d

L¹ (f ∈C_c^∞(R^d)). (2) Alternatively, one can derive from (1)

kfk¹⁺

2 d

L^{2(1+ 2d)}

≤ck∇fk_L2kfk

2 d

L² (f ∈C_c^∞(R^d)). (3) (i) Derive (2) and (3) from (1).

(ii) Derive (1) from (2). (Alternatively, you may decide to derive (1) from (3).)

Hint: Part (i) is simple. Part (ii) is not simple. You may restrict yourself to the case d ≥ 3 and to nonnegative functions f. One option is to apply (2) to the function f_k = (f −2^k)₊∧2^k for k∈ Z. If you setB_k = {x ∈R^d|2^k ≤f(x) <2^k+1},q = _d−2^2d , a_k= 2^qk|{f ≥2^k}|,b_k= ´

Bk

|∇f|², then you can observe

X

k∈Z

bk≤ k∇fk²_L2, (2^q−1)X

k∈Z

ak≥ kfk^q_Lq,

which turns out to be helpful.

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