Editing Openai/6935580c-34e4-8001-9136-7bbdf1499790 (section)

=== Assistant: This is a very important logical question, and you are right to insist on clarity here. ===
This is a very important logical question, and you are right to insist on clarity here. The short answer is:

You do not “rule them out”; you change the problem.
By defining mild solutions in C([0,T];X)C([0,T];X)C([0,T];X), you are not claiming equivalence with all classical or distributional solutions. You are selecting a solution concept with built-in continuity and then proving that it coincides with classical solutions when those exist. Below is the precise logical structure.

==== A definition is a selection principle, not a theorem. ====

When you define:

: 

you are saying:

Only functions with this regularity are under consideration.

You are not asserting:
* that every classical solution must be continuous in XXX,
* nor that discontinuous classical solutions cannot exist abstractly.

You are instead defining a class of admissible solutions.

This is completely standard and unavoidable in PDE theory.

==== A “classical solution” means: ====
* the PDE holds pointwise (or in a strong sense),
* spatial derivatives exist,
* time derivative exists pointwise.

It does not automatically imply

t↦u(t)∈Xis continuous.t \mapsto u(t) \in X
\quad \text{is continuous.}t↦u(t)∈Xis continuous.
For example:
* a function may be smooth in space for each fixed ttt,
* yet jump in time in a strong norm.

So a “classical but XXX-discontinuous” object is logically possible.

Your definition of mild solution simply does not admit such objects.

==== Navier–Stokes is ill-posed unless you impose both ====
# an equation, and
# a function class.

Without (2), the problem has too many pathological solutions.

Examples across PDE theory:
* Weak solutions to Euler are non-unique without admissibility.
* Transport equations admit wild discontinuous solutions unless continuity or renormalization is imposed.
* Navier–Stokes admits distributional solutions with arbitrary time behavior unless energy or regularity constraints are added.

Thus:

: 

==== The correct logic is: ====

===== Let uuu be a classical solution on [0,T][0,T][0,T] such that =====

u∈C([0,T];X).u \in C([0,T];X).u∈C([0,T];X).
Then uuu satisfies the mild (Duhamel) formulation, hence is a mild solution.

This theorem is true and provable.

The converse:

: 

Also true and provable by semigroup smoothing.

✅ Together, these show equivalence on the intersection of classes — not universal equivalence.

==== If you allow discontinuities in time in strong norms: ====
* the Duhamel integral is not well-defined as a Bochner integral,
* uniqueness typically fails,
* continuous dependence on data fails,
* semigroup methods collapse.

Therefore:

: 

This is not an arbitrary restriction; it is the minimal structure needed.

==== : ====

==== If your concern is philosophical completeness, you can state explicitly: ====

: 

: 

This removes all ambiguity and is standard in serious PDE texts.

If you want, next I can:
* formalize this as a commuting diagram of solution concepts,
* show explicitly why time discontinuity destroys Duhamel equivalence,
* or discuss minimal continuity assumptions weaker than C([0,T];X)C([0,T];X)C([0,T];X) (e.g. weak continuity + energy bounds).

Just tell me.