advanced32 min studyLesson 785 in Mathematics

Dual spaces and weak topologies

Identify which functionals test convergence and distinguish weak-star limits from genuinely weak limits on a dual space.

On this page

What you’ll understand

  • Define weak and weak-star topology using their respective testing spaces.
  • Use the classical sequence-space dual identifications with correct endpoints.
  • Exhibit a bounded weak-star convergent sequence that is not weakly convergent.

Weak convergence asks every bounded scalar functional to agree on a limit, while norm convergence controls all unit-norm functionals uniformly. On a dual space there is another topology, weak-star, testing only evaluations coming from the original space. The difference depends on the chosen predual and can survive even for bounded sequences. Sequence spaces make these ideas concrete, because coordinate tests, summable tests and all dual tests can be written explicitly. They also show why coordinatewise convergence alone does not identify the weak topology without a boundedness and duality argument.

The dual is complete even before the original space is

The bounded linear functionals on a normed space X form its dual X*, with operator norm given by the largest absolute evaluation on the unit ball. A norm-Cauchy sequence of functionals is pointwise Cauchy in the scalar field; its pointwise limit is linear and satisfies the inherited uniform bound. This proves X* Banach even if X is incomplete. Hahn–Banach makes the canonical evaluation map J:X→X** an isometry. Whether this map reaches every element of the bidual is the later question of reflexivity.

For sequences, the dual of ℓᵖ is ℓᑫ when 1≤p<∞ and 1/p+1/q=1, with q=∞ at p=1. The dual of c₀ is ℓ¹. Pairing means summing a bounded linear coefficient sequence against the input, with a consistent complex-linear convention rather than silently conjugating the wrong argument. The dual of ℓ∞ is larger than ℓ¹; identifying it with summable coefficients would erase important functionals. These endpoint differences are structural, not merely an inconvenient exception in notation.

∥f∥=sup⁡∥x∥≤1∣f(x)∣,Jx(f)=f(x),∥Jx∥=∥x∥,(ℓp)∗=ℓq (1≤p<∞),c0∗=ℓ1\|f\|=\sup_{\|x\|\leq1}|f(x)|,\qquad Jx(f)=f(x),\quad\|Jx\|=\|x\|,\qquad(\ell^p)^*=\ell^q\ (1\leq p<\infty),\quad c_0^*=\ell^1
The sequence dualities use the usual coefficient pairing. The excluded endpoint ℓ∞ has additional bounded functionals beyond summable coefficient tests.↔ Scroll wide equations horizontally.

Weak and weak-star choose different tests

The weak topology σ(X,X*) is the smallest topology making every bounded functional on X continuous. A net converges weakly exactly when all its scalar evaluations converge. The weak-star topology σ(X*,X) on the dual tests f at each fixed x in X. The weak topology on that same dual instead tests every element of X**. Since JX sits inside X**, weak convergence on X* implies weak-star convergence, but the reverse can fail. Norm convergence implies weak convergence in either setting.

A weakly convergent sequence is norm bounded by uniform boundedness. In a sequence space, boundedness plus convergence against finitely supported tests can sometimes be upgraded to all dual tests by approximating the testing coefficients in their own norm. This succeeds for the weak topology of c₀, whose dual coefficients are summable, but not for ℓ¹, whose dual coefficients can be arbitrary bounded sequences. The failure is easy to see: coordinate unit vectors have every fixed coordinate tending to zero, while the constant-one ℓ∞ functional evaluates each at one.

xα⇀x  ⟺  f(xα)→f(x) ∀f∈X∗,fα⇀∗f  ⟺  fα(x)→f(x) ∀x∈Xx_\alpha\rightharpoonup x\iff f(x_\alpha)\to f(x)\ \forall f\in X^*,\qquad f_\alpha\overset{*}{\rightharpoonup}f\iff f_\alpha(x)\to f(x)\ \forall x\in X
The testing family defines the topology. On X*, weak convergence tests X**, while weak-star convergence tests only the embedded predual X.↔ Scroll wide equations horizontally.

A predual limit can fail a bidual test

View ℓ∞ as the dual of ℓ¹. Let u⁽ⁿ⁾ be one in the first n coordinates and zero afterward. Against each summable sequence, the evaluations converge to its total sum, so u⁽ⁿ⁾ converges weak-star to the constant-one sequence. To test weak convergence in ℓ∞, however, one may use any element of its full dual. Extend the ordinary limit functional from the subspace c of convergent sequences to ℓ∞ by Hahn–Banach. This bounded extension vanishes on c₀ and takes value one on the constant-one sequence.

Every u⁽ⁿ⁾ belongs to c₀, so that extended functional evaluates all of them at zero. Its value at their proposed weak limit is one, ruling out weak convergence to that limit. The example needs no special translation-invariant Banach limit: an arbitrary norm-preserving extension of the ordinary limit suffices. It demonstrates why a weak-star compactness argument cannot be upgraded to a weak compactness argument just by observing bounded norms. The topology and its available scalar tests must remain explicit at every step.

A summable test sees a limit that another dual test rejects

For u⁽ⁿ⁾ in ℓ∞ as above, prove weak-star convergence to the constant-one sequence and failure of weak convergence there. Quantify the error for the ℓ¹ test aₖ=2⁻ᵏ.

  1. For any a in ℓ¹, the difference between the evaluations at u⁽ⁿ⁾ and at the constant-one sequence is the tail sum beyond n, whose absolute value is bounded by the absolute summable tail.

  2. For aₖ=2⁻ᵏ that tail is exactly 2⁻ⁿ. Thus this fixed predual test converges even though the supremum-norm difference between u⁽ⁿ⁾ and one remains one.

  3. Extend the limit functional on convergent sequences to a bounded functional L on ℓ∞. Then L(u⁽ⁿ⁾)=0 for every n, while L(1)=1, so convergence is not weak in ℓ∞.

The result

The sequence is bounded and weak-star convergent, but it fails a legitimate weak test from the larger dual of ℓ∞.

Explore the model

Interactive lab

Interactive lab · vectors

A single projection misses a persistent norm

Pause and Reset, then set angle=0°,magnitude=1. Predict horizontal evaluation one and vector norm one. For a fresh comparison Reset and set angle=90°,magnitude=1; predict horizontal evaluation zero while the vector norm stays one. This illustrates why one scalar test is insufficient. In a finite-dimensional plane, convergence against all linear tests is actually equivalent to norm convergence; the diagram does not claim to exhibit the infinite-dimensional weak-star counterexample.

A single projection misses a persistent normPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forWeak convergence requires every test in the specified dual family, not just a convenient displayed coordinate.

Check your understanding

On X*, what is tested by weak convergence that need not be tested by weak-star convergence?

Try it yourself

Compare the coordinate vectors eₙ in c₀ and in ℓ¹. Are they weakly convergent to zero in each space? Explain why identical coordinate values give different answers.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use c₀*=ℓ¹ and (ℓ¹)*=ℓ∞; ask whether the testing coefficient sequence must tend to zero.

Compare with the solution

In c₀, a dual functional has summable coefficients a, so its value on eₙ is aₙ→0. Thus eₙ converges weakly to zero there, while its norm stays one. In ℓ¹, the constant-one bounded coefficient sequence is a dual functional and evaluates every eₙ at one. Hence eₙ is not weakly convergent to zero in ℓ¹. Its coordinatewise convergence is the same in both spaces, but weak convergence depends on the full dual testing family.

Take it with you

  • The dual is Banach and the canonical bidual embedding is isometric.
  • Weak-star topology depends on the chosen predual and tests fewer functionals than weak topology.
  • Coordinate convergence can fail to imply weak convergence at sequence-space endpoints.

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