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Solutions to the exam
Problem 1 (a) Let X be a normed space and Y ⊂ X be a subspace. Show that the closure of Y in X coincides with ⋂
f ∈X∗: Y ⊂ker f
ker f.
(b) Consider the space p^ for some 1 < p < ∞ and the subspace V ⊂p^ consisting of all sequences x = (xn)n ∈ `p^ such that xn 6 = 0 only for finitely many n’s and
n xn^ = 0. Show that^ V^ is dense in^ ` p.
Solution: (a) The inclusion Y¯ ⊂
f ∈X∗: Y ⊂ker f ker^ f^ is clear. For the opposite inclusion we have to show that for any x /∈ Y¯ there exists f ∈ X∗^ such that f (x) 6 = 0 and Y ⊂ ker f. This can be done using the Hahn-Banach separation theorem. Alternatively, consider the quotient normed space X/ Y¯. Then by the Hahn-Banach theorem there exists f˜ ∈ (X/ Y¯ )∗^ such that f˜ (x + Y¯ ) 6 = 0. Composing f˜ with the quotient map X → X/ Y¯ we get the required functional f.
(b) By part (a) it suffices to show that given f ∈ (p)∗^ such that V ⊂ ker f , we have f = 0. Recalling that (p)∗^ = q^ (with 1 = 1/p + 1/q), we have f = (fn)∞ n=1 ∈q^. For n ≥ 2 consider the vector x ∈ V such x 1 = 1, xn = −1 and xk = 0 for k 6 = 1, n. Then the equality f (x) = 0 means that f 1 = fn. Since this is true for all n ≥ 2 and
n |fn| q (^) < ∞, we conclude that fn = 0 for all n, so f = 0.
Problem 2 Let X be a locally compact space (not necessarily second countable) and μ be a Radon measure on it. (a) Show that there exists the largest open set U ⊂ X such that μ(U ) = 0. Its complement X \ U is called the support of μ and denoted by supp μ. (b) Show that the set U = X \ supp μ can be characterized as the largest open set such that if f ∈ Cc(X) and supp f ⊂ U , then
X f dμ^ = 0. Solution: (a) Let {Ui}i∈I be the collection of all open sets of measure zero. Then U = ∪i∈I Ui is the required set. Indeed, we only have to check that μ(U ) = 0. Since μ is inner regular on open sets, it suffices to show that μ(K) = 0 for any compact K ⊂ U. By compactness there exists a finite number of indices i 1 ,... , in such that K ⊂ ∪nk=1Uik. Then μ(K) ≤
∑n k=1 μ(Uik^ ) = 0,^ so^ μ(K) = 0. (b) It is clear that
X f dμ^ = 0 for any^ f^ ∈^ Cc(X) such that supp^ f^ ⊂^ U^. Conversely, assume an open set^ V is such that
X f dμ^ = 0 for any^ f^ ∈^ Cc(X) such that supp^ f^ ⊂^ V^. We claim that^ μ(V^ ) = 0. As in (a), for this it suffices to show that μ(K) = 0 for any compact K ⊂ V. By Urysohn’s lemma we can find f ∈ Cc(X) such that supp f ⊂ V and χK ≤ f. Then μ(K) ≤
X f dμ^ = 0, so^ μ(K) = 0. Thus^ μ(V^ ) = 0 and hence^ V^ ⊂^ U^. Problem 3 Let (X, B, μ) be a σ-finite measure space. (a) Let ν be another σ-finite measure on (X, B) such that ν μ. Show that for any measurable function f : X → [0, +∞] we have (^) ∫
X
f dν =
X
f
dν dμ
dμ.
(b) Show that for 1 ≤ p < ∞ and a measurable function f : X → C we have f ∈ Lp(X, dν) if and only if (dν/dμ)^1 /pf ∈ Lp(X, dμ). What can you suggest as a replacement of this statement for p = ∞? (c) Show that if ν is as above and η is another σ-finite measure on (X, B) such that η ν, then dη dμ
dη dν
dν dμ
μ-a.e.
(d) Let T : X → X be a measurable map such that the measures T∗μ and μ are equivalent, so μ(T −^1 (A)) = 0 if and only if μ(A) = 0. For 1 ≤ p < ∞ define an operator UT on Lp(X, dμ) by
(UT f )(x) = dμ dT∗μ
(T (x))^1 /pf (T (x)).
Show that UT is an isometry, that is, ‖UT f ‖p = ‖f ‖p for all f. 1
2
Solution: (a) The equality holds for f = χA for any measurable set A by definition of the Radon-Nikodym derivative. Hence it holds for all simple positive measurable functions. Then it holds for all positive measurable functions by the monotone convergence theorem.
(b) Using (a), we have ∫
X
|f |pdν =
X
|f |p^ dν dμ
dμ =
X
dν dμ
) 1 /p f
p dμ,
which solves the first part of the problem. For the second, consider the set A = {x : dνdμ (x) > 0 }. Then ν(X \ A) = 0, while for measurable sets B ⊂ A we have ν(B) = 0 if and only if μ(B) = 0. It follows that for any measurable function its essential supremums on A with respect to ν and μ coincide. Therefore f ∈ L∞(X, ν) if and only if χAf ∈ L∞(X, μ).
(c) Using (a) again, for any measurable set A we have ∫
A
dη dν
(x)
dν dμ
(x)dμ(x) =
X
χA
dη dν
(x)
dν dμ
(x)dμ(x) =
X
χA
dη dν
dν =
X
χAdη = η(A).
This shows that the measurable function x 7 → dηdν (x) dνdμ (x) satisfies the defining property of the Radon-Nikodym derivative of η with respect to μ.
(d) Using once again (a) and that
g dT∗μ =
g ◦ T dμ, we have ∫
X
|UT f |pdμ =
X
dμ dT∗μ
(T (x))|f (T (x))|pdμ(x) =
X
dμ dT∗μ
(x)|f (x)|pd(T∗μ)(x) =
X
|f |pdμ.
Problem 4 Let (X, B, μ) be a measure space, μ(X) < ∞. (a) Assume V ⊂ L∞(X, μ) is a subspace such that ‖f ‖∞ ≤ C‖f ‖ 2 for all f ∈ V for some C > 0. Show that dim V ≤ C^2 μ(X).
Hint: take an orthonormal system {fk}nk=1 in V and try to get an estimate on n. (b) Show that if V ⊂ L^2 (X, dμ) is an infinite-dimensional closed subspace, then V 6 ⊂ L∞(X, μ).
Solution: (a) Following the hint take an orthonormal system {fk}nk=1 in V. Then for any c 1 ,... , cn ∈ C the inequality ‖f ‖∞ ≤ C‖f ‖ 2 for the function f = c 1 f 1 + · · · + cnfn means that
(1) |c 1 f 1 (x) + · · · + cnfn(x)|^2 ≤ C^2 (|c 1 |^2 + · · · + |cn|^2 )
for a.e. x. Choose a dense countable subset Ω ⊂ Cn. Then we can find a measurable subset Y ⊂ X such that μ(X \ Y ) = 0 and (1) holds for all x ∈ Y and (c 1 ,... , cn) ∈ Ω. But then by continuity (1) holds for all x ∈ Y and c 1 ,... , cn ∈ C. For x ∈ Y , taking ck = fk(x) we get
|f 1 (x)|^2 + · · · + |fn(x)|^2 ≤ C^2.
Integrating over X (or, what is the same, over Y ) and using that the functions fk are of L^2 -norm one, we get n ≤ C^2 μ(X). Since any finite dimensional subspace of V has an orthonormal basis, it follows that dim V ≤ C^2 μ(X). (b) Assume by contradiction that V ⊂ L∞(X, μ). Since
(2) ‖f ‖ 2 ≤ μ(X)^1 /^2 ‖f ‖∞
for any f ∈ L∞, the space V is complete with respect to the L∞-norm. Indeed, if a sequence {fn}n ⊂ V is Cauchy with respect to the L∞-norm, it converges to a function f ∈ L∞(X, μ). But then by (2) we have fn → f with respect to the L^2 -norm, so f ∈ V. Consider now the identity operator (V, ‖ · ‖∞) → (V, ‖ · ‖ 2 ). By (2) this operator is bounded, so it is a bounded linear isomorphism of Banach spaces. By the open mapping theorem its inverse is bounded as well. Hence there exists C such that ‖f ‖∞ ≤ C‖f ‖ 2 for all f ∈ V. By (a) it follows that V is finite dimensional, which is a contradiction.
