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A mild solution need not be classical

Statement refuted

Assume Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain), hence Countable Choice. Let $1 \leq p < \infty$ , $X = L^{p} (R)$ , let $(T (t))_{t \geq 0}$ be the right-translation semigroup with generator $A f = f^{'}$ , $D (A) = W^{1, p} (R)$ (The right-translation semigroup on Lp has the weak derivative as generator), and let $f_{0} : = 1_{(0, 1)} \in L^{p} (R)$ . Then $f_{0} \notin D (A)$ and $u (t) : = T (t) f_{0} = 1_{(- t, 1 - t)}$ is the unique mild solution of $u^{'} = A u$ , $u (0) = f_{0}$ (Classical, strong and mild abstract Cauchy solutions, Variation of constants for the inhomogeneous abstract Cauchy problem), but $u$ is not a classical solution: for every $t \geq 0$ , $u (t) = 1_{(- t, 1 - t)}$ is a nondegenerate indicator and hence not in $W^{1, p} (R) = D (A)$ , whereas a classical solution must satisfy $u (t) \in D (A)$ for all $t \geq 0$ , in particular at $t = 0$ . The difference quotients $(T (h) f_{0} - f_{0}) / h$ have no limit in $L^{p}$ , consistently with $f_{0} \notin D (A)$ .

Refuted claim. Every mild solution of the homogeneous abstract Cauchy problem $u^{'} = A u$ , $u (0) = x$ is a classical solution. The right-translation semigroup on $L^{p} (R)$ provides a mild solution whose initial datum lies outside the generator domain, so the classical-solution condition $u (t) \in D (A)$ fails at every time.

Facts & Assumptions

Given: Dependent Choice; $1 \leq p < \infty$ , $X = L^{p} (R)$ , the right-translation semigroup $(T (t))_{t \geq 0}$ with $(T (t) g) (s) = g (s + t)$ , its generator $A f = f^{'}$ with $D (A) = W^{1, p} (R)$ (The right-translation semigroup on Lp has the weak derivative as generator), and $f_{0} : = 1_{(0, 1)}$ with $u (t) : = T (t) f_{0} = 1_{(- t, 1 - t)}$ .

[F1]

The generator of the right-translation semigroup is the weak derivative with domain $W^{1, p} (R)$ , so $D (A) = W^{1, p} (R)$ (The right-translation semigroup on Lp has the weak derivative as generator, Weak derivative of a locally integrable function, Integer-order Sobolev spaces and their norms).

[F2]

The mild solution of $u^{'} = A u$ , $u (0) = x$ is the continuous function $u (t) = T (t) x + \int_{0}^{t} T (t - s) f (s) d s$ with $f = 0$ , which is the unique mild solution and the unique integral solution (Classical, strong and mild abstract Cauchy solutions, Variation of constants for the inhomogeneous abstract Cauchy problem).

[F3]

A classical solution on $[0, \infty)$ must satisfy $u (t) \in D (A)$ for every $t$ , in particular at $t = 0$ (Classical, strong and mild abstract Cauchy solutions).

[F4]

$L^{p}$ classes are almost-everywhere classes with the norm of The space $L^{p} (μ)$ as the quotient by null functions; a jump discontinuity is the model of a function without a locally integrable weak derivative, and the one-dimensional computation used here is carried out in [step 1.2].

Counterexample

technique · direct: identify the mild solution and test membership in $W^{1,p}$ by the defining weak-derivative identity

1.1F2

$u (t) = T (t) f_{0} = 1_{(- t, 1 - t)}$ and $u$ is the unique mild solution: with $x = f_{0}$ and $f = 0$ the variation-of-constants formula gives $u (t) = T (t) f_{0}$ , which is continuous, and [F2] gives uniqueness.

1.2F1F4given

Suppose $v \in L^{p} (R)$ were a weak derivative of $f_{0} = 1_{(0, 1)}$ . Testing on each of the open intervals $(- \infty, 0)$ , $(0, 1)$ and $(1, \infty)$ gives $\int v φ = 0$ for every test supported there. The injective distribution embedding Locally integrable functions embed in distributions applies because $v$ is locally integrable (Hölder on compact intervals, or its $L^{1}$ integrability when $p = 1$ ) and Countable Choice holds. It gives $v = 0$ almost everywhere on all three intervals, hence on $R$ since ${0, 1}$ is null. Take one smooth compactly supported test $φ$ with $φ (0) = 0$ , $φ (1) = 1$ . Then $\int f_{0} φ' = φ (1) - φ (0) = 1$ , while $- \int v φ = 0$ , contradicting the weak-derivative identity. This covers $p = 1$ without an invalid shrinking $L^{\infty}$ norm estimate.

2.1F1F3step 1.2

Consequently $f_{0} \notin D (A)$ by [F1], so $u (0) \notin D (A)$ ; by [F3] $u$ is not a classical solution, and the same computation applies to every $u (t) = 1_{(- t, 1 - t)}$ , so membership in $D (A)$ fails at every $t \geq 0$ .

3.1F1step 2.1

The difference quotients $\frac{T (h) f_{0} - f_{0}}{h}$ have no limit in $L^{p}$ : if they had a limit $g$ , then $f_{0} \in D (A)$ with $A f_{0} = g$ by the definition of the generator [F1], contradicting [step 2.1].

4.1step 1.1step 2.1step 3.1∎

Hence the mild solution $u (t) = 1_{(- t, 1 - t)}$ of $u^{'} = A u$ , $u (0) = f_{0}$ is not classical, and the initial datum lies outside the generator domain; mild solutions are exactly the device that keeps such data admissible.

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Sources

Gerald Teschl, Partial Differential Equations: From Classical to Modern (2026 author manuscript; complete 392-page archived text) (standard reference, not scraped)
Mathew A. Johnson, Math 951 Lecture Notes, Chapter 6: Introduction to Semigroup Methods, University of Kansas (complete 37-page chapter) (standard reference, not scraped)
Klaus-Jochen Engel and Rainer Nagel, One-Parameter Semigroups for Linear Evolution Equations, Graduate Texts in Mathematics 194 (complete author-hosted monograph) (standard reference, not scraped)