Questions: Weak Convergence

Question 1

In L²([0,1]), consider the sequence fₙ(t) = sin(nπt). Which statement correctly describes its convergence?

• A. It converges strongly (in norm) to zero, since the oscillations become infinitely rapid.
• B. It converges weakly to zero, but its norm does not converge to zero — strong convergence fails.
• C. It converges neither weakly nor strongly, since its norm stays constant.
• D. It converges weakly and strongly to zero, since every integral ∫fₙg → 0.

Answer: (B)

This is the canonical example of the weak/strong distinction. By the Riemann-Lebesgue lemma, ∫sin(nπt)g(t)dt → 0 for every g ∈ L², so fₙ converges weakly to zero. But ||fₙ||₂ = 1/√2 for all n, so strong (norm) convergence fails completely. The sequence oscillates wildly but its inner products with every test function converge — 'indistinguishable by measurement' but not geometrically close. Option A is the common misconception: rapid oscillation is exactly what produces weak convergence (oscillations cancel in integrals) while preventing strong convergence.

Question 2

Why is weak convergence indispensable in variational problems and PDE theory, where strong convergence often cannot be guaranteed?

• A. Weak convergence implies strong convergence in the spaces where variational problems live, so it is a convenient shorthand.
• B. Variational problems only require that the solution minimize a functional, not that it satisfy the PDE pointwise, so strong convergence is irrelevant.
• C. In reflexive Banach spaces, every bounded sequence has a weakly convergent subsequence, restoring the compactness needed to extract limit points of minimizing sequences.
• D. Weak convergence is easier to verify numerically, making it the standard in computational PDE.

Answer: (C)

The key application: in infinite-dimensional spaces, the unit ball is not compact under the norm, so bounded minimizing sequences need not have norm-convergent subsequences. In a reflexive Banach space, the Banach-Alaoglu theorem provides a substitute: every bounded sequence has a weakly convergent subsequence. This is the infinite-dimensional analogue of Bolzano-Weierstrass. To find a minimizer of a functional, you extract a weakly convergent subsequence from a minimizing sequence, then show the weak limit is itself a minimizer — typically by showing the functional is weakly lower semicontinuous.

Question 3

In finite-dimensional vector spaces, weak convergence and norm convergence are equivalent.

• T. True
• F. False

Answer: True

In finite dimensions, every basis element gives a bounded linear functional, and convergence in every coordinate is equivalent to norm convergence. The interesting distinction between weak and strong convergence is an infinite-dimensional phenomenon. In ℝⁿ, the norm is equivalent to coordinate-wise convergence, which is exactly what bounded linear functionals test. This is why the theory of weak convergence is developed in infinite-dimensional spaces — in finite dimensions, there's nothing new to say.

Question 4

If a sequence (xₙ) converges weakly to x in a Banach space, then ||xₙ - x|| → 0.

• T. True
• F. False

Answer: False

This is the central misconception the topic addresses. Weak convergence is strictly weaker than norm convergence in infinite-dimensional spaces — it does not imply norm convergence. The sequence sin(nπt) in L² converges weakly to 0 while ||sin(nπt)||₂ = 1/√2 for all n. The converse is true: norm convergence implies weak convergence (by linearity and boundedness of every functional). A sequence that converges weakly may oscillate wildly in space while still having its 'correlation' with every test function converge.

Question 5

Explain why a sequence can converge weakly to zero while its norm remains bounded away from zero. What is the sequence 'doing' geometrically?

Model answer: Weak convergence to zero means every bounded linear functional applied to the sequence gives values tending to zero — the sequence 'looks like zero' to every observer. But this requires only that inner products (or more generally, functional evaluations) converge, not that the vectors themselves approach zero geometrically. In L², a sequence like sin(nπt) achieves this through rapid oscillation: the oscillations cancel out in every integral against a fixed function g, driving ∫fₙg → 0, but the energy ||fₙ||² = ∫|fₙ|² stays constant because the integrand is always nonnegative. The vectors are spreading their energy over rapidly oscillating regions, invisible to any fixed test function, but still present in the norm.

The distinction is between 'indistinguishable by measurement' (weak) and 'geometrically close' (strong). Weak convergence tests only what bounded linear functionals can detect — essentially, correlations with test functions. Strong convergence requires the vectors to actually approach their limit in distance. In infinite dimensions, a sequence can evade all fixed measurements through wild oscillation while still having substantial norm — this is impossible in finite dimensions, which is why the distinction only matters in infinite-dimensional spaces.