Exam 6 (43–49)¶

2000 Nov¶

Instructions Do as many problems as you can. Complete solutions to five problems would be considered a good performance.

Problem 43

State the inverse function theorem. 1
Suppose \(L: ℝ^3 → ℝ^3\) is an invertible linear map and that \(g: ℝ^3 → ℝ^3\) has continuous first order partial derivatives and satisfies \(\|g(x)\| ≤ C\|x\|^2\) for some constant \(C\) and all \(x ∈ ℝ^3\). Here \(\|x\|\) denotes the usual Euclidean norm on \(ℝ^3\). Prove that \(f(x) = L(x) + g(x)\) is locally invertible near 0.

Problem 44

Let \(f\) be a differentiable real valued function on the interval \((0,1)\), and suppose the derivative of \(f\) is bounded on this interval.

Prove the existence of the limit \(L = \lim_{x → 0^+} f(x)\).

Problem 45

Let \(f\) and \(g\) be Lebesgue integrable functions on \([0,1]\), and let \(F\) and \(G\) be the integrals

\[F(x) = ∫_0^x f(t) \, dt, \quad G(x) = ∫_0^x g(t) \, dt.\]

Use Fubini’s theorem and/or Tonelli’s theorem to prove that

\[∫_0^1 F(x)g(x) \, dx = F(1) G(1) - ∫_0^1 f(x)G(x) \, dx.\]

Other approaches to this problem are possible, but credit will be given only to solutions based on these theorems.

Problem 46

Let \((X, 𝔐, μ)\) be a finite measure space and suppose \(ν\) is a finite measure on \((X, 𝔐)\) that is absolutely continuous with respect to \(μ\). Prove that the norm of the Radon-Nikodym derivative \(f = \left[\frac{dν}{dμ}\right]\) is the same in \(L_∞ (μ)\) as it is in \(L_∞(ν)\).

Problem 47

Suppose that \(\{f_n\}\) is a sequence of (Lebesgue) measurable functions on \([0,1]\) such that \(\lim\limits_{n→∞} ∫_0^1 |f_n| \, dx = 0\) and there is an integrable function \(g\) on \([0,1]\) such that \(|f_n|^2 ≤ g\), for each \(n\). Prove that \(\lim\limits_{n→∞}∫_0^1 |f_n|^2 \, dx =0\).

Problem 48

Denote by \(\mathcal{P}_e\) the family of all even polynomials. Thus a polynomial \(p\) belongs to \(\mathcal{P}_e\) if and only if \(p(x) = \frac{p(x) + p(-x)}{2}\) for all \(x\). Determine, with proof, the closure of \(\mathcal{P}_e\) in \(L_1[-1,1]\). You may use without proof the fact that continuous functions on \([-1,1]\) are dense in \(L_1[-1,1]\).

Problem 49

Suppose that \(f\) is real valued and integrable with respect to Lebesgue measure \(m\) on and that there are real numbers \(a<b\) such that

\[a ⋅ m(U) ≤ ∫_U f \, dm ≤ b ⋅ m(U),\]

for all open sets \(U\) in \(ℝ\). Prove that \(a ≤ f(x) ≤ b\) a.e.

A statement of the inverse function theorem appears in the appendix of theorems.
First note that \(L\) and \(g\) both have continuous first order partial derivatives; i.e., \(L, g ∈ C^1(ℝ^3)\). Therefore, the derivative of \(f = L + g\),

\[f'(x) \triangleq J_f(x) \triangleq \left(\frac{\partial f_i}{\partial x_j}\right)_{i,j=1}^3\]

exists. Furthermore, \(J_f(x)\) is continuous in a neighborhood of the zero vector, because this is true of the partials of \(g(x)\), and the partials of \(L(x)\) are the constant matrix \(L\). Therefore, \(f ∈ C^1(ℝ^3)\). By the IFT, then, we need only show that \(f'(0)\) is invertible. Since \(f'(x) = L + g'(x)\), we must show \(f'(0) = L + g'(0)\) is invertible. Consider the matrix \(g'(0) = J_g(0)\). We claim, \(J_g(0)=0\). Indeed, if \(x_1, x_2, x_3\) are the elementary unit vectors (also known as i, j, k), then the elements of \(J_g(0)\) are

(21)¶\[\frac{∂ g_i}{∂ x_j}(0) = \lim_{h→0} \frac{g_i(0 + h x_j) - g_i(0)}{h} = \lim_{h→0} \frac{g_i(h x_j)}{h}.\]

The second equality follows by the hypothesis that \(g\) is continuous and satisfies \(\|g(x)\| ≤ C\|x\|^2\), which implies that \(g(0) =0\). Finally, to show that (21) is zero, consider

\[|g_i(h x_j)| ≤ \|g(h x_j)\| ≤ C \|h x_j\|^2 = C|h|^2,\]

which implies

\[\frac{|g_i(h x_j)|}{|h|} ≤ C \frac{|h x_j|^2}{|h|} = C|h| → 0, \text{ as } h→0.\]

This proves that \(f'(0) = L\), which is invertible by assumption, so the inverse function theorem implies that \(f(x)\) is locally invertible near \(0\).

☐

First consider the sequence \(\{x_n\} := \{\frac{1}{n}\}\) for \(n = 2, 3, 4, \dots\).

Claim 1. \(\{f(x_n)\}\) is a Cauchy sequence.

Proof. Observe

\[\left| f\bigl( \frac{1}{n+j} \bigr)- f\bigl( \frac{1}{n} \bigr) \right| = \left| \frac{ f( \frac{1}{n+j} ) - f( \frac{1}{n} ) } { \frac{1}{n+j} - \frac{1}{n} } \right| \left| \frac{1}{n+j} - \frac{1}{n} \right|,\]

and

\[\frac{1}{n+j} - \frac{1}{n} = \frac{n - (n + j)}{n(n+j)} = \frac{-j}{n+j}.\]

By the mean value theorem, there exists \(c ∈ \left[\frac{1}{n+j},\frac{1}{n}\right]\) such that

\[\left| f'(c)\right| = \frac{\left| f(\frac{1}{n+j})- f(\frac{1}{n})\right|}{\left| \frac{1}{n+j}- \frac{1}{n} \right|}.\]

Therefore,

\[\left| f\left(\frac{1}{n+j}\right)- f\left(\frac{1}{n}\right)\right| ≤ M \left| \frac{1}{n+j}- \frac{1}{n} \right| ≤ M/n,\]

so given \(ε > 0\) we have \(|f(1/m)- f(1/n)| < ε\) for all \(m,n ≥ N > M / ε\).

This proves that \(\{f(\frac{1}{n})\}\) is a Cauchy sequence, and it follows that \(\lim_{n→ ∞} f(\frac{1}{n}) = ℓ\) exists in \(ℝ\), since \(ℝ\) is a complete metric space.

☐

Claim 2. \(\lim\limits_{x↘0} f(x) = ℓ\).

Proof. Fix \(ε>0\). We show \(∃ δ>0\) such that \(0 < x < δ\) implies \(|f(x) - ℓ| < ε\).

Given \(0<a<b<1\) there exists \(c∈ [0,1]\) such that \(f(b) - f(a) = f'(c) (b-a)\), and \(|f'(c)|≤ M\) by assumption.

Therefore, \(|f(b) - f(a)| ≤ M |b-a|\) for all \(0<a<b<1\).

Let \(N> 0\) be chosen so large that \(N > ε/(2M)\) and \(n≥ N \, ⟹ \, |f(1/n) - ℓ| < ε/2\).

Then for all \(0< x < 1/n\) we have

\[|f(x) - ℓ| ≤ |f(x) - f(1/n)| + |f(1/n) - f(ℓ)| ≤ M/n + ε/2 < M\frac{ε}{2M} + \frac{ε}{2} = ε.\]

☐

☐

The left-hand side is

\[∫_0^1\left( ∫_0^x f(t) \, dt\right) g(x) \, dx = ∫_0^1∫_0^x f(t) g(x) \, dt\, dx.\]

Consider \(f̃(t,x):= f(t)\) and \(g̃(t,x):= g(x)\). Then,

\[f̃^{-1}(G) = f^{-1}(G) × [0,1] \; \text{ and } \; g̃^{-1}(G) = [0,1]× g^{-1}(G)\]

are measurable subsets of \([0,1] × [0,1]:= T × X\), so \(f̃\) and \(g̃\) are measurable on \(T× X\).

Therefore, \(f̃ g̃ = f g\) is measurable and we can apply Fubini’s theorem which asserts the following:

\[\text{ if } \; ∬ |f̃ g̃ |\, dt\, dx < ∞ \; \text{ or } \; ∬ |f̃ g̃ |\, dx\, dt < ∞,\]

then \(f̃ g̃ ∈ L_1([0,1] × [0,1])\) and

\[∬ f̃ g̃ \, dt\, dx = ∬ f̃ g̃ \, dx\, dt.\]

So we calculate,

\[\begin{split}∫_0^1 ∫_0^x |f̃ g̃ |\, dt\, dx &= ∫_0^1 ∫_0^x |f(t)||g(x)|\, dt\, dx\\ & ≤ ∫_0^1 ∫_0^1 |f||g| = \left(∫ |f|\right) \left(∫ |g|\right) < ∞,\end{split}\]

since \(f, g ∈ L_1\) by assumption. Therefore, the hypotheses of Fubini’s theorem are satisfied and we have

\[\begin{split}∫_0^1 ∫_0^x f(t)g(x)\, dt\, dx &= ∫_0^1 ∫_t^1 f(t)g(x)\, dx\, dt \\ &= ∫_0^1 \left(∫_t^1 g(x)\, dx\right) f(t) \, dt \\ &= ∫_0^1 [G(1) - G(t)] f(t)\, dt = F(1)G(1) - ∫_0^1 f(t)G(t) \, dt,\end{split}\]

as desired.

☐

It seems an implicit assumption here is that the measures are positive, so we make that assumption in what follows.

Define (as usual) the ∞-norm relative to \(μ\) of a (real- or complex-valued) function \(f\) on \(X\) as follows:

(22)¶\[\|f\|_∞ := \inf \{a∈ ℝ^∗ ∣ μ\{x : |f(x)| > a\} = 0\},\]

where \(ℝ^∗ = ℝ ∪ \{-∞, ∞\}\) and \(\inf ∅ = ∞\).

Since there are two measures in context, let us denote the norm defined in (22) by \(\|f\|_μ\), and let \(\|f\|_ν\) denote the ∞-norm relative to \(ν\).

The claim to prove is \(\left\|\frac{dν}{dμ}\right\|_μ = \left\|\frac{dν}{dμ}\right\|_ν\).

Here is the proof.

Let \(f := \frac{dν}{dμ}\) so that for each \(E ∈ 𝔐\) we have \(ν E = ∫_E f\, dμ\).

Observe that \(f ≥ 0\), since we assumed \(μ, ν\) positive.

For each \(a ∈ ℝ^∗\), let \(E_a := \{x ∈ X ∣ f(x) > a\}\).

Then \(μ E_a = 0\) only if \(ν E_a = 0\), since \(ν ≪ μ\).

Therefore, \(\{a ∈ ℝ^∗ ∣ μ E_a = 0\} ⊆ \{a ∈ ℝ^∗ ∣ ν E_a = 0\}\), so

(23)¶\[\|f\|_ν ≤ \|f\|_μ.\]

Suppose \(\|f\|_ν < \|f\|_μ\). Then there exists \(a ∈ ℝ\) such that \(ν E_a = 0\) and \(μ E_a > 0\), and

\[0 = ν E_a = ∫_{\{f > a\}} f\, dμ ≥ a μ E_a > 0,\]

a contradiction. Therefore, equality must hold in (23).

☐

Fix an arbitrarily small \(ε > 0\) and let \(m\) denote Lebesgue measure on \(ℝ\).

Since \(g ∈ L_1[0,1]\), there exists \(A ⊆ [0,1]\) such that \(g\) is bounded on \(A\) and \(∫_{[0,1]-A} g \, dm < ε\).

This and the assumption \(g ≥ |f_n|^2 ≥ 0\) imply that there exists \(M < ∞\) such that \(0≤ g(x) ≤ M\) for all \(x ∈ A\).

Define,

\[A_n^+ := {A ∩ \{|f_n|>1\}} \quad \text{ and } \quad A_n^- := {A ∩ \{|f_n|≤1\}}.\]

and observe that \(A = A_n^+ ∪ A_n^-\), a disjoint union. Thus, for every \(n∈ ℕ\),

\[\begin{split}∫_{[0,1]} |f_n|^2 \, dm &= ∫_A |f_n|^2 \, dm + ∫_{[0,1]-A} |f_n|^2 \, dm\\ &= ∫_{A_n^+} |f_n|^2 \, dm + ∫_{A_n^-} |f_n|^2 \, dm + ∫_{[0,1]-A} |f_n|^2 \, dm.\end{split}\]

On \(A_n^+ := A ∩ \{|f_n|>1\}\) we have \(|f_n| ≤ |f_n|^2 ≤ g ≤ M\), so

\[∫_{A_n^+} |f_n|^2 \, dm ≤ M ∫_{A_n^+} |f_n|\, dm ≤ M ∫_{[0,1]} |f_n|\, dm.\]

On \(A_n^-:= A ∩ \{|f_n|≤1\}\) we have \(|f_n|^2 ≤ |f_n|\), so

\[∫_{A_n^-} |f_n|^2 \, dm ≤ ∫_{A_n^-} |f_n|\, dm ≤ ∫_{[0,1]} |f_n|\, dm.\]

Thus, for every \(n∈ ℕ\),

\[\begin{split}∫_{[0,1]} |f_n|^2 \, dm &= ∫_{A_n^+} |f_n|^2 \, dm + ∫_{A_n^-} |f_n|^2 \, dm + ∫_{[0,1]-A} |f_n|^2 \, dm\\ &≤ ∫_{[0,1]}|f_n| \, dm + M ∫_{[0,1]}|f_n| \, dm + ∫_{[0,1] - A}g \, dm.\end{split}\]

By assumption \(\lim\limits_{n→∞}∫_{[0,1]}|f_n| \, dm= 0\), so the right-hand side approaches \(∫_{[0,1] - A}g \, dm\) as \(n→∞\) and we conclude that

\[\lim\limits_{n→∞}∫_0^1 |f_n|^2 \, dx ≤ ∫_{[0,1] - A}g \, dm < ε.\]

Since \(ε>0\) was arbitrary, this completes the proof.

☐

Define the operators \(E, O: L_1[-1,1] → L_1[-1,1]\) as follows: for \(f∈ L_1[-1,1]\),

\[E f(x) = \frac{f(x) + f(-x)}{2} \quad \text{ and } \quad O f(x) = \frac{f(x) - f(-x)}{2}.\]

Denote by \(Λ\) the collection of a.e.-even functions in \(L_1[-1,1]\); that is, \(g ∈ Λ\) iff \(g ∈ L_1[-1,1]\) and \(g(x) = (g(x) + g(-x))/2\) holds for almost every \(x∈ [-1,1]\).

Claim 1. \(E\) is continuous (i.e., bounded).

Proof. Evidently,

\[\begin{split}\|E f\|_1 &= ∫_{-1}^1 |E f(x)|\, dx = ∫_{-1}^1 \left| \frac{f(x) + f(-x)}{2} \right|\, dx \\ &≤ \frac{1}{2} \left( ∫_{-1}^1 |f(x)|\, dx + ∫_{-1}^1 |f(-x)|\, dx\right) = \|f\|_1.\end{split}\]

☐

Claim 2. If \(p_n ∈ \mathcal{P}_e\), then \(E p_n = p_n\). (obvious)

Since continuous functions on [-1,1] are dense in L₁[-1,1], for each \(g∈ L_1[-1,1]\) there is a sequence \(\{f_n\}⊆ \overline{C[-1,1]}\) such that \(f_n → g\) in \(L_1[-1,1]\).

Fix \(ε>0\) and let \(N\) be a natural number large enough so that \(\|f_N - g\|_1 < ε/2\).

There exists \(\{p_n\}⊆ \mathcal P\) such that \(\|p_n - f\|_∞ → 0\) as \(n→∞\), so

\[\begin{split}\|p_n - g\|_1 &= \|p_n - f_N + f_N - g\|_1\\ &≤ \|p_n - f_N\|_1 + \|f_N - g\|_1\\ &≤ \|p_n - f_N\|_∞ \, μ[-1,1] + ε/2 < ε,\end{split}\]

for large enough \(n ∈ ℕ\).

Since \(E\) is continuous, \(E p_n → E g\).

Claim 3. \(E p_n ∈ \mathcal P_e\). (obvious)

Claim 4. If \(g ∈ L_1[-1,1]\) and \(g\) is even, then \(E g = g\). (obvious)

Claim 5. If \(g ∈ L_1[-1,1]\) and \(g\) is even, then \(O g = 0\) a.e. (why?)

Claim 6. \(O(\bar{\mathcal P}_e) = 0\). (why?)

Claim 7. \(\bar{\mathcal P}_e = \{g ∈ L_1[-1,1]: g \text{ even a.e.}\} = Λ\).

Proof. The inclusion \(⊆\) follows from Claims 5 and 6.

To prove \(⊇\), fix \(g∈ Λ\) and observe that there exists \(\{p_n\} ⊆ \mathcal P_e\) such that \(p_n \xrightarrow{L_1} g\). Therefore, \(E p_n → E g\) by continuity of \(E\), and \(E g = g\) a.e. since \(g\) is even a.e.

Since \(E p_n\) is an even polynomial, we have \(g = \lim\limits_{n→ ∞} E p_n\), so \(g ∈ \bar{\mathcal P}_e\).

☐

Suppose on the contrary that there exists a measurable set \(E\) of positive measure such that \(f(x) > b\) for all \(x ∈ E\).

Define \(E_n := \{x∈ ℝ ∣ f(x) > b + \frac{1}{n}\}\). Then \(E = ⋃_{n=1}^∞ E_n\) and there exists \(N ∈ ℕ\) such that \(m (E_N) > 0\).

Since \(f ∈ L_1(m)\), for all \(ε>0\) there exists \(δ>0\) such that

\[m(S) < δ \; ⟹ \; ∫_S |f| \, dm < ε.\]

If \(ε_0 = 1/(2N)\), and if \(δ_0>0\) is chosen appropriately, then

(24)¶\[\left|∫_S f \, dm\right| ≤ ∫_S |f| \, dm < ε_0 = \frac{1}{2N},\]

whenever and \(m(S) < δ_0\).

Now, for all \(δ_1> 0\) there exists an open set \(U\) containing \(E_N\) such that,

(25)¶\[m(U-E_N) < δ_0 \quad \text{ and } \quad m(E_N) > m(U)-δ_1.\]

From the first of these inequalities along with (24) it follows that

(26)¶\[\left|∫_{U-E_N} f\, dm\right| < \frac{1}{2N}.\]

From the second inequality in (25) follows the second inequality in the sequence of calculations below.

(27)¶\[\begin{split}∫_U f \, dm &= ∫_{U-E_N} f \, dm + ∫_{E_N} f \, dm \\ &≥ ∫_{U-E_N} f \, dm + \bigl(b + \frac{1}{N}\bigr) ⋅ m(E_N)\\ &> ∫_{U-E_N} f \, dm + \bigl(b + \frac{1}{N}\bigr) ⋅ (m(U) - δ_1).\end{split}\]

Combining (26) and (27) we have, for arbitrary \(δ_1>0\),

(28)¶\[∫_U f \, dm > \bigl(b + \frac{1}{N}\bigr) ⋅ (m(U) - δ_1) - \frac{1}{2N}.\]

We will show that an appropriate choice of \(δ_1>0\) yields

(29)¶\[\bigl(b + \frac{1}{N}\bigr)⋅(m(U) - δ_1) > b ⋅ m(U) + \frac{1}{2N}.\]

From this and (28) will follow,

\[∫_U f \, dm > b ⋅ m(U),\]

contradicting the hypothesis that \(∫_U f \, dm ≤ b⋅m(U)\) holds for every open set \(U\) in \(ℝ\).

Calculate,

\[\begin{split}\bigl(b + \frac{1}{N}\bigr)⋅(m(U) - δ_1) &= \bigl(b + \frac{1}{N}\bigr)⋅m(U) - \bigl(b + \frac{1}{N}\bigr)⋅δ_1 \\ &= b⋅m(U) + \frac{m(U)}{N} - b ⋅ δ_1 - \frac{δ_1}{N}.\end{split}\]

We want,

\[b⋅m(U) + \frac{m(U)}{N} - b ⋅ δ_1 - \frac{δ_1}{N} > b ⋅ m(U) + \frac{1}{2N},\]

equivalently, \(m(U)/N - (b - 1)⋅ δ_1/N > 1/(2N)\).

Equivalently, we must prove any one (hence all) of the following (equivalent) conditions:

\[\begin{split}\begin{array}{rrcl} & m(U) - (b ⋅ N + 1)⋅ δ_1 &>& \frac{1}{N}\\ ⟺ & m(U) - \frac{1}{N} &>& (b ⋅ N + 1)⋅ δ_1\\ ⟺ & \bigl(m(U) - \frac{1}{N}\bigr) ⋅ (|b| ⋅ N + 1)^{-1} &>& δ_1. \end{array}\end{split}\]

Thus if we choose \(δ_1\) so that it satisfies

\[δ_1 < \frac{m(E_N) - 1/N}{|b| ⋅ N + 1} < \frac{m(U) - 1/N}{|b| ⋅ N + 1},\]

then our goal (29) will be satisfied and the desired contradiction will obtain.

A similar argument can be used to show that the existence of a nonnegligible set \(E\) on which \(f < a\) will lead to a contradiction. Therefore, \(a ≤ f(x) ≤ b\) holds for almost every \(x ∈ ℝ\).

☐

Footnotes

1: Historically speaking, questions involving the inverse function theorem do not seem very popular. This is one of the few exams on which it appears.