Functionals and admissible function spaces
Objective: Why must the norm or topology on X be declared?
A functional assigns a number to an entire curve or field; its domain must encode smoothness and boundary data before “nearby” functions or extrema have meaning.
Read the admissible family before differentiating: A functional J:X→R maps a function u in an admissible set X to a scalar; a typical integral functional is J[u]=∫_a^bL(x,u,u′)dx with fixed, free, or constrained traces. The result to establish is: If X is convex and L(x,y,p) is convex in (y,p), then J is convex on X; if strict convexity acts on every nonzero admissible difference, a minimizer is unique. Keep the topology, endpoint traces, constraints, regularity, and allowed signs of each variation attached to every step.
The worked question is: For J[u]=∫₀¹(u′)²dx, compare u=x and u=x² under endpoints u(0)=0,u(1)=1. Begin with “Both functions satisfy the endpoint conditions.” and finish with “For u=x², J=∫₀¹4x²dx=4/3.” Retain every integration-by-parts boundary term, then decide whether the route proves only stationarity or an actual minimum.
If X is convex and L(x,y,p) is convex in (y,p), then J is convex on X; if strict convexity acts on every nonzero admissible difference, a minimizer is unique.
For u,v∈X and 0≤t≤1, convexity of X makes tu+(1−t)v admissible.
Apply pointwise convexity of L to the values and derivatives of this combination.
Integrate the pointwise inequality; strict inequality for nonidentical functions gives uniqueness.
For J[u]=∫₀¹(u′)²dx, compare u=x and u=x² under endpoints u(0)=0,u(1)=1.
- Both functions satisfy the endpoint conditions.
- For u=x, J=∫₀¹1dx=1.
- For u=x², J=∫₀¹4x²dx=4/3.
Result: Within these two admissible candidates, u=x has smaller energy 1 than 4/3; this comparison alone does not yet prove global minimality.