Questions & explanations
1. Give an example where pointwise convergence of measures fails to be uniform on a disjoint sequence.
Consider μ_n on ℕ defined by μ_n({k}) = 1 if k=n, else 0. Then μ_n(A) → 0 for every A (since only one point has mass, and it eventually leaves any fixed set). But for the disjoint sequence A_k = {k}, we have μ_n(A_n)=1, so convergence is not uniform. This violates the condition of the Vitali-Hahn-Saks theorem because the limit is not countably additive? Actually the limit is 0, which is countably additive, but the convergence is not uniform on disjoint sets? The theorem requires that the limit be finite and the measures be countably additive; here the limit is 0, so it holds? Wait, check: the theorem says if μ_n(A) converges for each A, then the limit is countably additive and the convergence is uniform on disjoint sequences. In this example, μ_n(A) → 0 for each A, so the limit is 0, which is countably additive. But the convergence is not uniform on the disjoint sequence {k} because sup_k |μ_n({k}) - 0| = 1 for n=k. This seems to contradict the theorem. Actually, the theorem requires that the measures be countably additive and the limit be finite. Here μ_n are countably additive, and
2. Apply Rouché to show that z^4+2z+1 has exactly one zero in the left half-plane.
Consider the contour consisting of the imaginary axis from -iR to iR and a semicircle to the left. On the semicircle |z|=R large, |z^4| dominates, so number of zeros in left half-plane equals number of zeros of z^4 there? Actually, use Rouché on a large semicircle. Let f(z)=z^4, g(z)=2z+1. On |z|=R large, |f|=R^4, |g|≤2R+1, so |f|>|g|. So f and f+g have same number of zeros inside the left half-plane contour. f has 2 zeros in left half-plane? z^4=0 has a zero of order 4 at 0, which is on the boundary? Better: use argument principle. Alternatively, note that z^4+2z+1 has 4 zeros total. On the imaginary axis, z=iy, then f(iy)=y^4+2iy+1. Real part y^4+1>0, so no zeros on axis. By Rouché on a large semicircle in left half-plane, the number of zeros in left half-plane equals the number of zeros of z^4 there, which is 2 (since z^4 has a zero of order 4 at 0, but 0 is on the boundary? Actually, 0 is on the imaginary axis? No, 0 is on the imaginary axis? The left half-plane does not include the imaginary axis. So we need to consider a contour that excludes the origin? This is tricky. Simpler
3. Give an example of a vector measure that does not have a Radon-Nikodym derivative with respect to Lebesgue measure.
Consider the vector measure ν on [0,1] defined by ν(A) = (∫_A g dλ, ∫_A h dλ) where g and h are not integrable? Actually, a counterexample: let ν(A) = (λ(A), 0) if A is Lebesgue measurable; this has derivative (1,0). A true counterexample: take ν(A) = (∫_A f dλ, ∫_A f dλ) with f non-integrable? Wait, need a vector measure with infinite variation. For instance, define ν on subsets of ℕ by ν(A) = (∑_{n∈A} 1/n, ∑_{n∈A} 1/n²). This is absolutely continuous with respect to counting measure, but the derivative is (1/n, 1/n²) which is integrable. Actually, a vector measure that is not of finite variation may fail. Example: let μ be a positive measure with infinite variation? Better: consider a vector measure that is not σ-finite? I recall that if the Banach space does not have the Radon-Nikodym property, there are counterexamples. For instance, take the vector measure with values in c₀ that is the limit of measures? A standard example: let ν(A) = (∫_A sin(2πnx) dλ)_{n} in ℓ²? This might not have a derivative. Simpler: let ν be the vector measure on [0,1] with values in L¹[0,1] defined by ν(
4. Use the open mapping theorem to show that a non-constant entire function cannot be bounded.
If f is entire (holomorphic on whole plane) and bounded, then by Liouville's theorem it is constant. But the open mapping theorem also gives a proof: if f is non-constant, its image is open. The only open subset of the complex plane that is bounded is empty? Actually, bounded open sets exist, but if f is entire and bounded, its image is bounded and open. However, the plane is not bounded, so the image cannot be the whole plane. But Liouville's theorem is standard. Alternatively, if f is non-constant and bounded, then the image is a bounded open set, but the only open subset of C that is also closed? This is not a direct proof. Better: Suppose f is non-constant and bounded. Then f(C) is open and bounded. But the complement of a bounded set is unbounded, so f(C) is not closed. However, continuous image of a connected set is connected, so f(C) is connected and open. That is possible, e.g., the unit disk is open and bounded. So this argument doesn't work. Actually, Liouville's theorem is the correct one. So maybe use open mapping to show that if f is non-constant, its image is open, but
5. Give an example of solving a differential equation using the Dirac delta.
Consider the equation y''(x) = δ(x) with boundary conditions y(-∞)=0, y(∞)=0. The solution is the Green's function for the second derivative. Integrating once gives y'(x) = H(x) + C, where H is the Heaviside step. Integrating again gives y(x) = x H(x) + Cx + D. Using boundary conditions, for x<0, y=0 so D=0, and for x>0, y=x + Cx, so to have y(∞)=0, we need C=-1, so y(x)= x H(x) - x = -x for x>0? Actually careful: The solution is y(x) = -|x|/2? Let's correct: The standard Green's function for d^2/dx^2 is G(x)=|x|/2? Actually, the solution to y''=δ is y(x)= (x)_+ = x for x>0, 0 for x<0, but that diverges. Usually with zero boundary conditions at infinity, the solution is y(x)= -|x|/2? Let's compute: For x>0, y''=0 so y=Ax+B, for x<0, y=Cx+D. Continuity at 0 gives B=D. Jump in derivative: y'(0+)-y'(0-)=1, so A-C=1. Boundary conditions: y(-∞)=0 implies C=0? Actually if C=0, then y constant for x<0, but then D=0 from boundary? Let's set y(-∞)=0 implies D=0, so B=0. Then y(x)=Ax for x>0, and y(x)=0 for x<0. Then A-C=1 gives A=1. So y(x)=x for x>0, 0 for x<0. That's the solution, but it di
6. Give an example of a rational function whose Julia set is the whole Riemann sphere.
Consider f(z)=1/(z^2). The Julia set is the whole sphere because the map is expanding everywhere except at 0 and ∞, which are superattracting? Actually, for f(z)=z^2, Julia set is the unit circle, not whole sphere. For f(z)=2z (linear), Julia set is empty? Typically, for a rational function of degree at least 2, if the function is everywhere expanding (like a Lattès map), the Julia set can be the whole sphere. For example, the map f(z)=z^2 - 2 has Julia set the interval [-2,2] on the real line, not whole sphere. A concrete example: f(z)=z^2 + i? Not whole sphere. Actually, the map f(z)=z^2 + c with c in the Mandelbrot set has connected Julia set, but not whole sphere. A rational function with Julia set the whole sphere is f(z)=z^2 + 0.5? No. Consider f(z)=1/(z^2) has Julia set the whole sphere? Let's check: for f(z)=1/z^2, the critical points are 0 and ∞, both map to ∞, so the Julia set is the whole sphere? I think yes, because the Fatou set is empty. So f(z)=1/z^2 is an example.
7. Give an example of a Banach space where every weakly convergent sequence is also norm convergent.
The space ℓ^1 of absolutely summable sequences has the Schur property: if a sequence in ℓ^1 converges weakly, it also converges in norm. For instance, the standard basis vectors e_n converge weakly to 0 but not in norm, so ℓ^1 does not have this property. Actually, ℓ^1 does have the Schur property: every weakly convergent sequence in ℓ^1 is norm convergent. A concrete example: the sequence (0,0,...,1,0,...) with 1 in the nth position converges weakly to 0 but not in norm, so ℓ^1 does not have the Schur property. Wait, correction: ℓ^1 does have the Schur property. The standard basis does not converge weakly to 0 because the functional that picks the nth coordinate gives 1. So it's not a counterexample. A correct example: in ℓ^1, weak convergence implies norm convergence. For instance, the sequence (1/n, 1/n, ...) is not in ℓ^1. Actually, a simple example is the zero sequence. To show the property, consider any weakly null sequence in ℓ^1; it must be norm null.
8. Give an example where the inverse function theorem fails because the derivative is zero.
Take f(z)=z^2 at z0=0. Here f'(0)=0, so the theorem does not apply. Indeed, f is not locally invertible near 0: any neighborhood of 0 contains points z and -z that map to the same value, so f is not one-to-one. The inverse would be multi-valued (square root). The theorem requires non-zero derivative to guarantee a local holomorphic inverse. If f'(z0)=0, f may still be locally invertible in some cases (e.g., f(z)=z^3 at 0? Actually f'(0)=0, and f is one-to-one near 0? No, because z^3 is one-to-one on ℂ? Actually z^3 is injective on ℂ? No, because ω^3=1 has three roots, so not injective globally; near 0, it is injective? For z^3, if z1^3=z2^3 then z1=z2 or z1=ω z2, but near 0, only z1=z2 possible? Actually, consider z1 and z2 small, if z1^3=z2^3 then (z1/z2)^3=1, so z1/z2 is a cube root of unity. For small nonzero z, z and ωz are distinct but both small, so f is not injective in any neighborhood of 0. So no local inverse exists.
9. Give an example where the monotone convergence theorem applies but the dominated convergence theorem does not.
Consider f_n(x) = n on [0, 1/n] and 0 elsewhere, with Lebesgue measure. f_n increases pointwise to f=0 almost everywhere? Actually f_n does not increase; it decreases. A correct example: f_n(x) = 1_{[n,∞)}(x) on ℝ. f_n decreases to 0, but monotone convergence requires increasing. Dominated convergence fails because no integrable dominating function exists (sup f_n = 1, not integrable on ℝ). So monotone convergence does not apply directly; you need a decreasing version. Better: f_n(x) = 1_{[0,n]}(x) on ℝ. f_n increases to 1_{[0,∞)}. Monotone convergence applies, dominated convergence also applies with dominating function 1_{[0,∞)} which is not integrable? Actually 1_{[0,∞)} is not integrable on ℝ. So dominated convergence fails because no integrable dominating function exists. So monotone convergence works, dominated does not.
10. Give an example of a Radon measure on a locally compact space that is not σ-finite.
On ℝ, consider the measure μ that assigns infinite measure to any nonempty open set, but finite measure to compact sets? Actually, a Radon measure must be locally finite, so on ℝ, the measure that is infinite on every open set would not be locally finite. A better example: on an uncountable discrete space with the counting measure, it is locally finite only if each point has finite measure, but counting measure on an uncountable set is not σ-finite but is Radon if the space is discrete? Actually, on a discrete space, each singleton is open and compact, and counting measure gives each point measure 1, so it is locally finite and inner regular, but it is not σ-finite if the space is uncountable. So counting measure on an uncountable discrete space is Radon but not σ-finite.
11. Give an example where Kummer's test works but Raabe's test is inconclusive.
Consider the series ∑ 1/(n log n). Raabe's test: n a_n/a_{n+1} - (n+1) = n ( (n+1) log(n+1) / (n log n) ) - (n+1) = (n+1) log(n+1)/log n - (n+1) = (n+1)( log(n+1)/log n - 1). As n→∞, this tends to 1? Actually compute: log(n+1)/log n ≈ 1 + 1/(n log n), so expression ≈ (n+1)/(n log n) → 0. So Raabe's test gives limit 0, inconclusive. Use Kummer's test with c_n = n log n. Then c_n a_n/a_{n+1} - c_{n+1} = n log n * ( (n+1) log(n+1) / (n log n) ) - (n+1) log(n+1) = (n+1) log(n+1) - (n+1) log(n+1) = 0. So K=0, not >0. Actually need another c_n. For this series, the test with c_n = n log n gives 0, so no conclusion. A better example: ∑ 1/(n (log n)^2). Use c_n = n log n gives limit >0? Let's skip. Actually Kummer's test is not always conclusive; it's a general framework.
12. Explain why Rouché's theorem fails if the inequality is not strict.
If |f-g| = |g| on some point of the contour, then the condition is not strict, and the conclusion may not hold. For example, take f(z)=z, g(z)=z on |z|=1, then |f-g|=0 < |g|=1 holds, but actually f=g so they have same zeros. But if f(z)=z+1, g(z)=z on |z|=2, then on |z|=2, |f-g|=1, |g|=2, so 1<2, but f has zero at -1 inside, g has zero at 0 inside, both one zero. However, consider f(z)=z^2, g(z)=z^2+ε on a contour where equality occurs; if |f-g| = |g| at some point, the argument may change. A counterexample: f(z)=z, g(z)=z+1 on |z|=1/2? Actually, a known case: f(z)=z, g(z)=z-1 on |z|=1, then |f-g|=1, |g|=|z-1|, at z=1, |g|=0, so inequality fails. Rouché's theorem requires strict inequality everywhere to ensure the winding numbers match.