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.
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 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.
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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 \(x ∈ ℝ\).
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Footnotes
Complex Analysis Exams
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