Exam focus: Uniform convergence is what lets you swap limit with integral/derivative/continuity. Compute sup|fₙ − f| explicitly; xⁿ, nx(1−x)ⁿ, nxe^{−nx²}, x/n are the recurring families.
Lec-47 Uniform Convergence
NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)
The sup-norm test; contrast with pointwise convergence.
Lec-48 Uniform Convergence and Integration
NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)
Exactly when you may swap limit and integral.
Lec-49 Uniform Convergence and Differentiation
NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)
Why uniform convergence of fₙ says nothing about fₙ′.
Solution
sup|| = 1/n → 0, while nx) diverges for most .
Solution
(b) M-test with test with of the tail is unbounded near 1. (c) converges pointwise but not uniformly (the partial sums are unbounded near 0 in sup norm… the limit function is discontinuous at multiples of .
Solution
Max at x = 1/(n+1) equals ⁺, so not uniform on . On kills it uniformly.
Solution
By calculus the maximum of || is at with value , so || — uniform convergence on all of . This is the standard example where the sup can be computed exactly.
Solution
The pointwise limit is f(x) = 0 for x ∈ [0,1) and f(1) = 1, so (1) holds and (4) fails — the limit jumps at x = 1. (2) fails: each is continuous, so uniform convergence would force a continuous limit; concretely sup|| = 1 for every n, taking x slightly below 1. (3) On [0, 1/2] we have sup|| , which is uniform convergence.
Solution
Pointwise limit is 0 on [0,1) and 1 at x=1 — discontinuous, so convergence cannot be uniform.
Exam focus: Radius via limsup |aₙ|^{1/n} (Cauchy–Hadamard), never assume the ratio limit exists. Behaviour on the boundary circle is decided separately (Abel); differentiation/integration keep the radius.
Lec-46 Sequences and Series of Functions
NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)
Sets up function series before specialising to power series.
Lecture 3.1 - Power series
NPTEL · Complex Analysis (Pranav Haridas, KSoM)
Radius of convergence via Cauchy–Hadamard, done over ℂ but identical over ℝ.
Lecture 3.2 - Differentiation of Power series
NPTEL · Complex Analysis (Pranav Haridas, KSoM)
Termwise differentiation keeps the radius.
Solution
| takes values 1 and 3, limsup = 3, so R = 1/3. The ratio test fails here.
Solution
The alternating harmonic series converges (Leibniz), so Abel's theorem gives the sum as the limit x → 1⁻ of ln(1 + x) = ln 2.
Solution
M-test with gives (a), hence (d). (b) termwise differentiation inside. as x → 1⁻.
Solution
(1) and (4) are the definition of the radius of convergence — absolute convergence strictly inside, divergence strictly outside. (3) Term-by-term differentiation never changes the radius, since limsup | = limsup |. (2) Cannot be concluded: boundary behaviour is not determined by R — diverges at x = 2 while converges there.
Exam focus: Arzelà–Ascoli = uniformly bounded + equicontinuous ⇒ a uniformly convergent subsequence. It is the compactness criterion in C[a,b]; the failures are xⁿ (not equicontinuous) and constants n (not bounded).
Lec52 Equicontinuous family of Functions: Arzela - Ascoli Theorem
NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)
The statement and proof — this is the whole subtopic in one lecture.
Lec-51 Approximation of a Continuous Function by Polynomials: Weierstrass
NPTEL · Real Analysis (S. H. Kulkarni, IIT Madras)
Weierstrass approximation, the companion result.
Solution
(b) is 1-Lipschitz uniformly; (d) is 5-Lipschitz. and sin(nx) have derivatives blowing up with n.
Solution
(1) is Arzelà–Ascoli. (2) fails: near 0 the functions oscillate on scales ~1/n — the points 0 and get arbitrarily close while the values stay 1 apart. (3) fails at the right endpoint: drops from 1 to near 0 over an interval of length ~1/n. (4) holds: the tail shares a modulus of continuity with the (uniformly continuous) limit, and the finitely many leading terms each have one.