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A time-discontinuous forcing blocks classical regularity at its jump

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) for the cited integral and semigroup suppliers.

Let $X = C$ , $A = 0$ (a bounded generator), $0 < t_{0} < b$ , and $f : = 1_{[t_{0}, b]}$ , which is bounded and measurable but neither continuous nor H"older at $t_{0}$ . The mild solution of $u^{'} = A u + f = 0 \cdot u + f$ , $u (0) = 0$ , is $u (t) = \int_{0}^{t} 1_{[t_{0}, b]} (s) d s = {\begin{cases} 0, & 0 \leq t \leq t_{0}, \\ t - t_{0}, & t_{0} < t \leq b . \end{cases}$ It is continuous and Lipschitz and satisfies $u^{'} (t) = f (t)$ for $t \neq t_{0}$ , but it is not differentiable at $t_{0}$ (left derivative $0$ , right derivative $1$ ) and hence is not a classical solution on $[0, b]$ in the sense of Classical, strong and mild abstract Cauchy solutions. Thus $u \in C ([0, b], X) \cap C^{1} ((0, b] ∖ {t_{0}}, X)$ but $u \notin C^{1} ([0, b], X)$ , and the H"older-continuity hypothesis in Classical regularity for Holder-continuous forcing under initial compatibility cannot be lowered to mere boundedness. For continuous forcing with modulus $ω_{f} (σ) : = \sup_{∣ s - r ∣ \leq σ} ∥ f (s) - f (r) ∥$ , the analytic smoothing bound $∥ A T (σ) ∥ \leq C σ^{- 1}$ (Smoothing estimates for the semigroup generated by a sectorial operator) makes $\int_{0}^{b} ω_{f} (σ) σ^{- 1} d σ < \infty$ sufficient for the singular generator integral in the Duhamel cancellation estimate; H"older continuity is one way to meet this condition. This is a sufficient condition for that estimate, not a necessary condition for classicality, and the jump is outside its continuous-Dini hypothesis. The present $A = 0$ example shows directly that bounded measurable forcing alone does not suffice: the jump makes the mild solution nondifferentiable. The example isolates this failure of time regularity with a trivial initial datum.

Refuted claim. With $A = 0$ on $C$ , a bounded measurable forcing produces a classical solution of $u^{'} = A u + f$ . The jump forcing $f = 1_{[t_{0}, b]}$ satisfies $f \in L^{\infty} (0, b)$ and the mild solution exists, but its left and right derivatives at the jump $t_{0}$ disagree, so the mild solution is not even differentiable there and the classical notion fails without any additional time regularity of $f$ .

Facts & Assumptions

Given: $X = C$ , the zero operator $A = 0$ with $D (A) = X$ , the numbers $0 < t_{0} < b$ , the indicator $f = 1_{[t_{0}, b]}$ and the datum $x = 0$ .

[L1]

The infinitesimal generator of a strongly continuous semigroup $(T (t))_{t \geq 0}$ is $A x : = \lim_{t ↓ 0} (T (t) x - x) / t$ on its domain (Infinitesimal generator of a C0-semigroup).

[L2]

For a strongly continuous semigroup with generator $A$ and a Bochner integrable $f$ with $\int_{0}^{T_{0}} ∥ f ∥ < \infty$ , the formula $u (t) = T (t) x + \int_{0}^{t} T (t - s) f (s) d s$ defines the unique mild solution and the unique integral solution of $u^{'} = A u + f$ , $u (0) = x$ (Variation of constants for the inhomogeneous abstract Cauchy problem, Bochner-integrable function).

[L3]

A classical solution is a $u \in C^{1} ([0, T_{0}]; X)$ with $u (t) \in D (A)$ for every $t$ , $A u \in C ([0, T_{0}]; X)$ , $u^{'} (t) = A u (t) + f (t)$ for $0 < t < T_{0}$ and $u (0) = x$ ; endpoint equations are imposed only when $f$ extends continuously to $[0, T_{0}]$ (Classical, strong and mild abstract Cauchy solutions).

[L4]

For $A$ sectorial with semigroup bounds $c_{0}, c_{1}$ and $f \in C^{α} ([0, b], X)$ one has $∥ A v_{1} (t) ∥ \leq \frac{c_{1}}{α} [f]_{α} t^{α}$ for the Duhamel term $v_{1} (t) = \int_{0}^{t} T (t - s) (f (s) - f (t)) d s$ (Analytic Duhamel cancellation removes the generator singularity).

[L5]

The classical regularity theorem assumes $x \in D (A)$ and $f \in C^{α} ([0, b], X)$ for some $α \in (0, 1)$ and concludes classicality of the mild solution (Classical regularity for Holder-continuous forcing under initial compatibility).

[L6]

For the analytic contour semigroup generated by a sectorial operator with vertex $0$ , $T (σ) X \subseteq D (A)$ and $∥ A T (σ) ∥ \leq C σ^{- 1}$ for $σ > 0$ , with $C$ depending on the sectoriality bounds (Smoothing estimates for the semigroup generated by a sectorial operator).

Counterexample

technique · direct

1.1L1L2givenalgebra

The semigroup and the mild solution. On $X$ the operator $A = 0$ has domain $X$ and generates the identity semigroup $T (t) = I$ : for every $x$ the difference quotient $(T (t) x - x) / t = 0$ converges to $0 = A x$ , so $D (A) = X$ and the generator is $0$ ; the forcing $f = 1_{[t_{0}, b]}$ is bounded and measurable, hence Bochner integrable on $(0, b)$ with $\int_{0}^{b} ∥ f ∥ = b - t_{0} < \infty$ , and [L2] with $x = 0$ gives the unique mild solution $u (t) = \int_{0}^{t} 1_{[t_{0}, b]} (s) d s$ , that is $u (t) = 0$ for $0 \leq t \leq t_{0}$ and $u (t) = t - t_{0}$ for $t_{0} < t \leq b$ ; this $u$ is continuous, equals $0$ at the origin and is Lipschitz with constant $1$ on $[0, b]$ .

2.1step 1.1givenalgebra

The differentiability failure at the jump. For $0 < h < t_{0}$ the left difference quotient of $u$ at $t_{0}$ is $(u (t_{0}) - u (t_{0} - h)) / h = 0$ , while for $0 < h < b - t_{0}$ the right quotient is $(u (t_{0} + h) - u (t_{0})) / h = h / h = 1$ ; hence the one-sided limits differ and $u$ is not differentiable at $t_{0}$ , although on each open piece $u^{'} (t) = f (t)$ (the derivative is $0$ on $(0, t_{0})$ and $1$ on $(t_{0}, b)$ ), so $u \in C^{1} ((0, b] ∖ {t_{0}}, X)$ and $u \notin C^{1} ([0, b], X)$ .

3.1step 2.1L3L4L5L6givenalgebra∎

Why boundedness is not enough. A classical solution on $[0, b]$ must be $C^{1}$ on the closed interval with $A u$ continuous and $u^{'} (t) = A u (t) + f (t) = f (t)$ for $0 < t < b$ by [L3], so [step 2.1] shows that this mild solution is not classical even though the generator $A = 0$ is bounded, the datum is trivial and $f$ is bounded; therefore the H"older hypothesis of [L5] cannot be weakened to mere boundedness. For a continuous forcing with modulus $ω_{f} (σ) : = \sup_{∣ s - r ∣ \leq σ} ∥ f (s) - f (r) ∥$ , [L6] bounds the generator integrand in the cancellation term by $C ω_{f} (σ) σ^{- 1}$ . Thus $\int_{0}^{b} ω_{f} (σ) σ^{- 1} d σ < \infty$ is sufficient for that cancellation estimate; in the H"older case it yields $C [f]_{α} t^{α} / α$ . This sufficient estimate is not a necessary characterization of classicality. In the present example $A = 0$ , hence $A T (σ) = 0$ ; the failure follows directly from the unequal one-sided derivatives in [step 2.1], not from a singular generator kernel. All functions here are explicit, so no choice principle beyond Dependent Choice is used.

Remarks

The same witness works in any nonzero Banach space after multiplying both $f$ and $u$ by a fixed nonzero vector.

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Sources

Roland Schnaubelt, Evolution Equations, Karlsruhe Institute of Technology (2023/24 course, complete lecture notes) (standard reference, not scraped)
Gerald Teschl, Partial Differential Equations: From Classical to Modern (2026 author manuscript; complete 392-page archived text) (standard reference, not scraped)