advanced32 min studyLesson 348 in Mathematics

The direct method and weak lower semicontinuity

Prove that a minimizing sequence has an admissible limit whose energy achieves the infimum.

Before you begin

On this page

The question for today

What you’ll understand

State a sequential direct-method existence theorem with its full hypotheses.
Use coercivity and reflexivity to extract weakly convergent subsequences.
Verify weak closedness and lower semicontinuity for a loaded quadratic energy.

A differential equation tells us what a sufficiently regular minimizer must satisfy, but it does not prove there is a minimizer. The direct method begins at the other end: choose functions whose energies approach the infimum and show that a subsequence converges to an admissible function without increasing the limiting energy. Its three central tasks are boundedness, compactness, and lower semicontinuity. Each can fail independently. A useful existence proof names precisely which topology supplies convergence and why the constraints and objective survive that convergence.

An abstract existence theorem with checkable ingredients

Let $X$ be a reflexive Banach space and let $\mathcal A$ be a nonempty weakly sequentially closed subset. Let the extended-real objective have a finite infimum on that set and be finite at some admissible point. Assume its bounded-above sublevel sets on the admissible set are norm bounded, and it is weakly sequentially lower semicontinuous there. Choose a minimizing sequence with finite energies. Boundedness gives a weakly convergent subsequence by reflexivity and weak sequential compactness of bounded sets in reflexive Banach spaces.

The weakly sequentially closed constraint set contains the limit. Lower semicontinuity then says its energy is no greater than the limit inferior of the sequence energies, which equals the infimum. Because an admissible energy cannot be below the infimum, equality follows and the limit minimizes. This is an existence theorem, not a regularity or uniqueness theorem. Those additional conclusions need their own hypotheses, such as strict convexity for uniqueness and elliptic regularity for smoothness.

u_j\rightharpoonup u,\quad u_j\in\mathcal A\Rightarrow u\in\mathcal A,\qquad J[u]\le\liminf_{j\to\infty}J[u_j]=\inf_{v\in\mathcal A}J[v]

Weak admissibility and weak lower semicontinuity are separate requirements. Together with bounded minimizing sequences and reflexive compactness, they establish attainment.↔ Scroll wide equations horizontally.

Coercivity controls escape

Consider the loaded derivative energy on $H_0^1(0,1)$ with square-integrable load. Its positive quadratic term grows like the squared derivative norm, while Cauchy–Schwarz and Poincaré bound the linear load by a constant times that norm. A quadratic minus a linear expression is bounded below and tends to positive infinity as the norm grows. Thus a minimizing sequence cannot escape to infinite norm. Poincaré also makes the derivative norm equivalent to the full Sobolev norm on zero-trace functions.

Boundary conditions are important in this argument. Without a fixed trace or another condition controlling constants, the derivative seminorm alone does not control the full function. A sequence of constant functions can grow arbitrarily while its derivative energy remains zero. With a nonzero total load the objective may even tend to minus infinity along constants. Coercivity must therefore be verified on the actual admissible space, rather than inferred from seeing a positive derivative term in the integrand.

J[u]=\frac12\|u'\|_2^2-\int_0^1fu\,dx\ge\frac12\|u'\|_2^2-C_P\|f\|_2\|u'\|_2\qquad(u\in H_0^1(0,1))

Poincaré and Cauchy–Schwarz bound the linear load. This lower bound proves boundedness below and coercivity in the homogeneous Sobolev space.↔ Scroll wide equations horizontally.

The weak limit must not cost more

Weak convergence of Sobolev functions includes weak convergence of their derivatives in the square-integrable space. The squared norm is weakly lower semicontinuous, while pairing the function with a fixed square-integrable load is weakly continuous. Their combination gives the needed energy inequality. The zero-trace space is a closed linear subspace and hence weakly closed. For nonzero fixed boundary data use an affine translate of it. This completes the existence argument without solving the differential equation explicitly.

Norm continuity alone would not suffice, because a weakly convergent sequence need not converge in norm. Conversely, an objective can be weakly lower semicontinuous without being differentiable. Convex norm-lower-semicontinuous functionals on Banach spaces provide a broad source of weakly lower-semicontinuous objectives. Nonconvex gradient energies can fail this property through oscillations, motivating relaxation in the next lesson. At the endpoint exponent one, reflexivity is also unavailable, so arguments using bounded sequences in that space need a different compactness mechanism.

Work it through

Existence first, a formula second

For $f(x)=6x$ , minimize $J[u]=\tfrac12\int_0^1(u')^2dx-\int_0^1 6xu\,dx$ on $H_0^1(0,1)$ . Give the existence argument and verify the explicit minimizer and energy.

The space is a reflexive Hilbert space, the zero function is admissible, and the coercive lower bound makes every minimizing sequence bounded. The zero-trace set is weakly closed and the energy is weakly lower semicontinuous, so a minimizer exists.
The weak equation is $-u''=6x$ with zero endpoints. Integrating gives $u_*(x)=x-x^3$ . Its derivative-square integral is $\int_0^1(1-3x^2)^2dx=4/5$ , and its load pairing is $\int_0^1 6x(x-x^3)dx=4/5$ .
Hence $J[u_*]=2/5-4/5=-2/5$ . The stationary cross term vanishes against all zero-trace differences, leaving half their derivative-square norm, which proves uniqueness.

The result

The direct method establishes attainment independently of the formula; the explicit solution x − x³ then has minimum energy −2/5.

The living notebook

Explore the model

Interactive lab

Interactive lab · heat

Strong decay is stronger than weak compactness

Reset and set both mode $n=2$ and diffusivity $\kappa=0.5$ . Predict the displayed spatial norm at time one is $\sqrt\pi e^{-2}$ ; Step twice and compare. Reset and set both $n=4,\kappa=0.5$ ; predict $\sqrt\pi e^{-8}$ at time one. This analytic heat example shows strong norm decay. It is not the minimizing sequence in the proof, and weak convergence alone would not justify such norm decay.

Current temperature modeInitial sineCurrent amplitude

Current amplitude: 1
Spatial L² norm: 1.772
Decay rate · κn²: 3.6 /s

Spatial mode n3Diffusivity κ0.4

t = 0 s

The exact periodic heat-equation mode is u(x,t) = exp(−κn²t) sin(nx), on [0,2π]. Playback advances physical time. The initial sine is shown separately, and the unnormalized L² norm is √π exp(−κn²t). This analytic instrument does not run a finite-difference solver or test discretization stability. 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 forChoose convergence claims carefully: a weakly compact minimizing sequence need not converge in norm.

Pause & reason

Check your understanding

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Read the checkpoint explanation

The weak limit supplied by reflexivity must remain admissible, and its energy must be bounded above by the sequence's limiting infimum. These are distinct from differentiability or norm continuity.

Your turn

Try it yourself

Drop every boundary condition and minimize $J[u]=\tfrac12\int_0^1(u')^2dx-\int_0^1u\,dx$ over $H^1(0,1)$ . Is the infimum finite? Identify the failed existence ingredient.

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

Give me a hint

Test constant functions before solving any differential equation.

Compare with the solution

For $u_c(x)=c$ , the derivative term is zero and $J[u_c]=-c$ . Letting $c$ tend to positive infinity makes the objective tend to minus infinity. It is neither bounded below nor coercive on this admissible space, so the stated direct-method hypotheses fail and there is no minimizer.

Keep the idea

Take it with you

Existence follows from boundedness, compactness, closed constraints, and lower semicontinuity.
Coercivity must control the norm on the actual admissible space.
Weak convergence requires weak, not merely norm-continuous, energy control.

Read further

Original sources and deeper explanations to follow at your own pace.

John Ball — Variational methods, lectures 1–8

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