245C Take-home final Due date: June 10 You may use any resources available to you, including textbooks, the

internet, or other students, in order to get hints or other inspiration; however, the work you hand in should be your own (i.e. not an identical copy of someone else’s work). If you cite any result from a textbook other than Folland, please state the result in full. Throughout the final, two functions are considered equal if they agree almost everywhere.

• Q1. Let R^{n} be a Euclidean space endowed with Lebesgue measure dm. Let 1 ≤ p ≤ ∞, and let K : R^{n} → R^{+} be a measurable function. Show that the following three statements are equivalent:

• (a) K is absolutely integrable (i.e. K ∈ L^{1}(R^{n})).

• (b) For every f ∈ L^{p}(R^{n}), the convolution integral f∗K(x) := ∫R^{n} f(x− y)K(y) dy is absolutely convergent for almost every x, and f ∗ K lies in L^{p}(R^{n}).

• (c) For every f ∈ L^{p}(R^{n}), the convolution integral f∗K(x) := ∫R^{n} f(x− y)K(y) dy is absolutely convergent for almost every x, and the map from f to f ∗ K is a bounded map from L^{p}(R^{n}) to L^{p}(R^{n}).

• A1. The implication of (b) and (c) from (a) follows from Young’s inequality (or Minkowski’s inequality, or Schur’s test). The implication of (b) from (c) follows from the uniform boundedness principle. To see this, first observe from (b) that K must at least be locally integrable, for if K had an infinite integral on some ball, then it is easy to find a function f ∈ L^{p} for which K ∗ f was infinite on a set of positive measure. Now if one lets K_{R} be the localization of K to a ball of radius R, then the operators f ↦ f ∗ K_{R} are bounded on L^{p}, and converge as R → ∞ to f ∗ K by (b) and dominated convergence, so by the uniform boundedness principle the map f ↦ f ∗K_{R} is also bounded on L^{p}. Finally, to deduce (c) from (a), we know that

∫_{R_{n}} | ∫_{R_{n}} f(x − y)K(y) dy|^{p} dy ≤ M^{p} ∫_{R_{n}} |f(x)|^{p} dx

for all f ∈ L^{p}(R^{n}) and some M > 0. Applying this with f equal to the indicator function of the ball of radius R, we obtain

∫_{R_{n}} | ∫_{|x−y|≤R} K(y) dy|^{p} dx ≤ M^{p}cR^{n}.

If |x| < R/2, then ∫_{|x−y|≤R} K(y) dy ≥ ∫_{|y|<R/2} K(y) dy by the triangle inequality (here we exploit the non-negativity of K; the claim is false otherwise). Thus

c(R/2)^{n}| ∫_{|y|<R/2} K(y)|^{p} ≤ M^{p}cR^{n}.

Dividing by R^{n} and then using monotone convergence as R → ∞ we conclude the absolute integrability of K.

• Q2. Let 1 < p < ∞. For any measurable function f : R^{n} → R^{∗} (where R^{∗} is the extended reals), let ∥f∥_{L_{∗}^{p,∞}(R^{n})} be the quantity

∥f∥_{L_{∗}^{p,∞}(R^{n})} := sup{m(E)^{−1/p′} ∫_{E} |f(x)| dx : E ⊂ R^{n} measurable, 0 < m(E) < +∞}.

Here p′ is the dual exponent of p, defined by 1/p + 1/p′ = 1.

• (a) Show that ∥f∥_{L_{∗}^{p,∞}(R^{n})} is finite if and only if f lies in weak L^{p}, and furthermore there exist constants C_{p}, c_{p} > 0 such that

c_{p}∥f∥_{L^{p,∞}(R^{n})} ≤ ∥f∥_{L_{∗}^{p,∞}(R^{n})} ≤ C_{p}∥f∥_{L^{p,∞}(R^{n})}

where ∥f∥_{L^{p,∞}(R^{n})} (which is called [f]_{p} in Folland) is the weak L^{p} “norm” of f:

∥f∥_{L^{p,∞}(R^{n})} := sup{λm({x ∈ R^{n} : |f(x)| > λ})^{1/p} : λ > 0}.

• (b) Show that if we give weak L^{p} the L_{∗}^{p,∞}(R^{n}) norm, and we identify any two functions in weak L^{p}(R^{n}) which agree almost everywhere, then weak L^{p} becomes a Banach space.

• A2. (a) Suppose first that f is in weak L^{p}. For any set E with 0 < m(E) < +∞, we recall from the theory of distribution functions that

∫_{E} |f(x)| dx = ∫_{0}^{∞} m({x ∈ E : f(x) > α}) dα.

On the other hand, we have the trivial estimate

m({x ∈ E : |f(x)| > α}) ≤ m(E)

as well as the estimate

m({x ∈ E : |f(x)| > α}) ≤ α^{−p}∥f∥_{L^{p,∞}(R^{n})}^{p}

and hence

∫_{E} |f(x)| dx ≤ ∫_{0}^{∞} min(m(E), α^{−p}∥f∥_{L^{p,∞}(R^{n})}^{p}) dα.

A short calculation shows that the integral on the right is in fact \frac{p}{p−1}m(E)^{1/p′}∥f∥_{L^{p,∞}(R^{n})}, which gives the second inequality in (a). To get the first inequality, let E be any subset of {x ∈ R^{n} : |f(x)| > α} of positive finite measure, then

αm(E) ≤ ∫_{E} |f(x)| ≤ m(E)^{1/p′}∥f∥_{L_{∗}^{p,∞}(R^{n})}

and hence

m(E) ≤ ∥f∥_{L_{∗}^{p,∞}(R^{n})}^{p}/α^{p}.

This claim is clearly also true if E has zero measure. By the σ-finiteness of R^{n}, the claim is then also true for sets E of possibly infinite measure. This shows that

m({x ∈ R^{n} : |f(x)| > α}) ≤ ∥f∥_{L_{∗}^{p,∞}(R^{n})}^{p}/α^{p}

which gives the first inequality in (a).

• A2. (b) The homogeneity, non-degeneracy, and triangle inequality properties of the norm are all easily verified. To establish completeness one repeats the arguments of Theorem 6.6 of Folland; we let f_{n} be an absolutely convergent sequence in L_{∗}^{p,∞}, and then show that the partial sums ∫_{E} ∑_{n=0}^{N} |f_{n}| are uniformly bounded in N for any set E of finite measure. Thus ∫_{E} ∑_{n=0}^{∞} |f_{n}| is finite, hence ∑_{n=0}^{∞} f_{n} is absolutely convergent almost everywhere to some limit f, and then by dominated convergence one can bound ∫_{E} |f − ∑_{n=0}^{N} f_{n}| for each E and N; we omit the details.

• Q3. Let (X, B_{∗}, dμ) be a probability space (thus X is a set, B_{∗} is a σalgebra on X, and dμ is a positive measure on B_{∗} such that μ(X) = 1). Let B be a sub-σ-algebra of B_{∗}, thus (X, B, dμ) is also a probability space. For any 1 ≤ p ≤ ∞, let L^{p}(B) be the space real-valued of Bmeasurable, p^{th}-power integrable functions, thus L^{p}(B) is a subset of L^{p}(B_{∗}). (We of course identify two functions in L^{p} if they agree almost everywhere.)

• (a) Show that L^{2}(B) is a closed subspace of the real Hilbert space L^{2}(B_{∗}). In particular, we have an orthogonal projection map f ↦ E(f|B) from L^{2}(B_{∗}) to L^{2}(B). The expression E(f|B) is known as the conditional expectation of f with respect to the σ-algebra B.

• (b) Show that if f ∈ L^{2}(B_{∗}) is non-negative, then E(f|B) is also (almost everywhere) non-negative. (Hint: E(f|B) is the closest element of L^{2}(B) to f). Conclude that conditional expectation is a contraction on L^{∞}(B_{∗}) (viewed as a metric space).

• (c) Show that for any 1 ≤ p ≤ ∞, there is a unique way to extend the conditional expectation operator to be a map from L^{p}(B_{∗}) to L^{p}(B) which is still a contraction, thus ∥E(f|B)∥_{L^{p}} ≤ ∥f∥_{L^{p}} for all f ∈ L^{p}(B_{∗}). (Hint: do this for p = 1 first, and then use the complex interpolation method.)

• A3(a). From Folland we know that L^{2}(B) is already complete, and has the metric structure inherited from that in L^{2}(B_{∗}), hence is closed in the latter space.

• A3(b). Observe that if f is non-negative, then max(E(f|B), 0) will be closer pointwise (and hence in L^{2}(B_{∗}) norm) to f than E(f|B), and still lies in L^{2}(B). Thus we must have max(E(f|B), 0) = E(f|B), hence E(f|B) is also non-negative. This in turn implies that if f ≤ g then E(f|B) ≤ E(g|B). In particular, since −∥f∥_{L^{∞}(B_{∗})} ≤ f ≤ ∥f∥_{L^{∞}(B_{∗})}, and since constant functions lie in L^{2}(B) and are hence their own conditional expectation, we have −∥f∥_{L^{∞}(B_{∗})} ≤ E(f|B) ≤ ∥f∥_{L^{∞}(B_{∗})}, which gives the desired contraction in L^{∞}.

• A3(c). Since L^{2}(B_{∗}) is dense in L^{1}(B_{∗}), it suffices to show that

∥E(f|B)∥_{L^{1}(B_{∗})} ≤ ∥f∥_{L^{1}(B_{∗})}

for all f ∈ L^{2}(B_{∗}). From the positivity we have

|E(f|B)| ≤ E(|f||B)

and so it suffices to prove the claim when f (and hence E(f|B)) is nonnegative. But then ∥E(f|B)∥_{L^{1}} is the inner product of E(f|B) with the constant function 1, and similarly ∥f∥_{L^{1}} is the inner product of f with 1. Since 1 lies in the range L^{2}(B) of the orthogonal projection f ↦ E(f|B), we see that these two inner products are equal, and the claim follows. Having established the contraction on L^{1} and L^{∞}, the intermediate cases follow from the Riesz-Thorin theorem.

• Q4. Let μ_{1}, μ_{2}, … be a sequence of positive Radon measures on R^{n}, such that sup_{j} μ_{j}(R^{n}) < ∞ (thus the μ_{j} are uniformly finite). Show that there exists a subsequence μ_{j_{1}}, μ_{j_{2}}, … of this sequence which is weakly convergent to another positive Radon measure μ, thus lim_{k→∞} ∫ fdμ_{j_{k}} = ∫ fdμ for all f ∈ C_{0}(R^{n}). (Hint: first show that the space C_{0}(R^{n}) is separable, for instance using the Weierstrass approximation theorem. Then use an Arzela-Ascoli diagonalization argument (or some form of the Banach-Alaoglu theorem)).

• A4. Consider the algebra of rational functions of the form P (x)/Q(x), where P, Q are both polynomials with Q having higher degree than P and being non-negative. These can be seen to be a sub-algebra of C_{0}(R^{n}) which separates points, and is hence dense by Stone-Weierstrass. The subset of rational functions with rational coefficients is then a countable dense subset of C_{0}(R^{n}). (There are many other ways to do this; for instance by multiplying the space of polynomials with rational coefficients by a cutoff function one can uniformly approximate any continuous function with compact support, and hence approximate any function in C_{0}(R^{n}).). Now the space of Radon measures is identifiable with the dual of C_{0}(R^{n}); the separability of C_{0}(R^{n}) will show that this dual is metrizable, and hence every compact set is sequentially compact. By the Banach-Alaoglu theorem, the closed unit ball in C_{0}(R^{n})^{∗} is thus sequentially compact in the weak topology. Since the measures μ_{j} are uniformly bounded, they thus have a weakly convergent sequence, as desired. Since each of the μ_{j} are positive, we have ∫ fdμ_{j} ≥ 0 for all f ∈ C_{0}(R^{n}), hence the same property holds for the weak limit μ, and so the weak limit is also a positive Radon measure. (Alternatively, one can enumerate the dense subset of C_{0}(R^{n}) as f_{1}, f_{2}, …, and pass to a subsequence of μ_{j} where ∫ f_{1}dμ_{j} converges, then to a sub-subsequence where ∫ f_{2}dμ_{j} also converges, and so forth, and then by taking a diagonalization of these subsequences we obtain a sequence where all of the ∫ f_{n}dμ_{j} converge, and hence by density and uniform boundedness ∫ fdμ_{j} converges for all f ∈ C_{0}(R^{n})).

• Q5. Let λ ∈ (C_{0}^{∞}(R^{n}))^{∗} be a distribution, and suppose that all sufficiently high order derivatives of λ vanish, thus there exist an integer N > 0 such that

(\frac{∂}{∂x_{1}} )^{α_{1}} … (\frac{∂}{∂x_{n}})^{α_{n}}λ = 0

whenever α_{1}, …, α_{n} ≥ 0 are such that α_{1}+…+α_{n} ≥ N. Show that λ is a polynomial, i.e. there exists a polynomial P : R^{n} → R of n variables such that λ(f) = ∫R^{n} f(x)P (x) dx for all f ∈ C_{0}^{∞}(R^{n}). (Hint: as a warm up, you may try the case n = 1 and α_{1} = 1.)

• A5. Let us first verify the case when N = 1. Let V ⊂ C_{0}^{∞}(R^{n}) be the subspace of C_{0}^{∞} generated by derivatives \frac{∂f}{∂x_{j}} for any f ∈ C_{0}^{∞}, 1 ≤ j ≤ n. Then λ annihilates V. We now show that V has codimension at most one inside C_{0}^{∞}, which shows that the space of λ which annihilate V is at most one-dimensional; since the constants already do this, this means that λ is constant.

Let f ∈ C_{0}^{∞}(R^{n}), let ψ ∈ C_{0}^{∞}(R) be a standard bump function with ∫ ψ = 1, and consider the function

g(x_{1}, …, x_{n−1}) := ∫R f(x_{1}, …, x_{n−1}, x_{n}) dx_{n}.

Then one checks that g ∈ C_{0}^{∞}(R^{n−1}) and f(x_{1}, …, x_{n})−g(x_{1}, …, x_{n−1})ψ(x_{n}) has mean zero in every line oriented in the x_{n} direction, and so it is the x_{n}-derivative of a test function and thus lies in V. Thus, modulo V, f is equal to a function g(x_{1}, …, x_{n−1} ∈ C_{0}^{∞}, tensored with ψ. Iterating this argument n times we see that modulo V, f is equal to a constant c, times ψ(x_{1}) … ψ(x_{n}) and hence C_{0}^{∞}(R^{n})/V is at most one dimensional, as claimed.

• Q6. Let λ ∈ (C^{∞}(R))^{∗} be a compactly supported distribution, supported in some open interval I. Show that λ agrees with a high order (distributional) derivative of a continuous function on I. More precisely, there exists a continuous function f ∈ C(R) and integers α ≥ 0 such that the distribution

λ − (\frac{d}{dx})^{α}f

is supported outside of I, where of course we use the distributional derivative. (Hint: first show that there exists an integer k ≥ 0 such that λ extends to be a continuous linear functional on C^{k}(R). Use this and the Riesz representation theorem to show that λ is a high order derivative of a Radon measure. Then show that this measure is in turn the distributional derivative of a continuous function.)

• A6. Let K be a compact interval containing the closure I in its interior, and let ψ ∈ C_{0}^{∞} be a bump function that equals one on I and is supported on K. We define a functional λ̃ : C^{∞}(K) → C by setting λ̃(f) := λ(fψ), then one easily verifies that λ̃ is continuous. Since C^{∞}(K) is a Frechet space generated by the C^{k}(K) norms, which are nested, we thus have λ̃ continuous in the C^{k}(K) topology for some k ≥ 0.

Now let x_{0} be the left endpoint of K, and define the integration operator S : C^{j}(K) → C^{j+1}(K) by

Sf(x) := ∫_{x_{0}}^{x} f(y) dy.

Then we see that the map λ̃ ∘ S^{k} is continuous from C(K) to C, and hence by the Riesz representation theorem there exists a Radon measure μ supported on K such that

λ̃ ∘ S^{k}(f) = ∫ f dμ

for any f ∈ C(K). In particular, if f ∈ C_{0}^{∞} is supported on I, then S(f^{(k)}) = f by the fundamental theorem of calculus, hence

λ(f) = λ̃(f) = ∫ f^{(k)} dμ.

If we let g(x) := ∫_{x_{0}}^{x} μ, then an integration by parts shows that

λ(f) = ∫ f^{(k)} dμ = − ∫ f^{(k+1)}g dx = ∫ f^{(k+2)}Sg dx

and hence

λ(f) = (−\frac{d}{dx})^{k+2}Sg(f).

In other words, λ(f) − (−\frac{d}{dx})^{k+2}Sg is supported outside of I. in the sense of distributions. Since g is bounded variation, Sg is continuous, and the claim follows.