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A first resolvent estimate does not ensure the prescribed semigroup bound

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). Let $X = C^{2}$ over the complex field (The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ ) have the maximum product norm $∥ (z_{1}, z_{2}) ∥_{\infty} = \max (∣ z_{1} ∣, ∣ z_{2} ∣)$ (The standard product norms on a finite product of normed spaces, Real and complex scalar conventions for normed spaces); this finite-dimensional normed space is Banach (Every finite-dimensional normed space is Banach). Let $A \in B (X)$ (A bounded linear operator between normed spaces) be the matrix $A = (\begin{array}{cc} - 1 & 4 \\ 0 & - 1 \end{array}), D (A) = X .$ For $M = 25 / 16$ and $ω = 0$ , every $λ > 0$ lies in $ρ (A)$ and the first resolvent estimate $∥ R (λ, A) ∥_{\infty} \leq \frac{M}{λ}$ holds (Resolvent and spectrum of a closed operator on a Banach space, The operator norm as the least bound and as the unit-sphere or unit-ball supremum). But the $n = 2$ Hille--Yosida estimate required by Hille-Yosida generation theorem fails at $λ = 3$ : $∥ R (3, A)^{2} ∥_{\infty} = \frac{3}{16} > \frac{25}{144} = \frac{M}{3^{2}} .$ The bounded-operator exponential semigroup generated by $A$ (The exponential series of a bounded operator) is $e^{t A} = e^{- t} (\begin{array}{cc} 1 & 4 t \\ 0 & 1 \end{array}),$ and $∥ e^{A / 2} ∥_{\infty} = 3 e^{- 1 / 2} > \frac{25}{16} .$ Thus a first-power resolvent estimate alone does not ensure the semigroup bound with the same $M > 1$ and $ω = 0$ .

Refuted claim. For every $M \geq 1$ , a closed densely defined operator whose positive real resolvent set contains $(0, \infty)$ and satisfies only $∥ R (λ, A) ∥ \leq M / λ$ for all $λ > 0$ necessarily generates a strongly continuous semigroup bounded by $M$ .

Facts & Assumptions

Given: Dependent Choice; $X = C^{2}$ with the maximum product norm, $A = (\begin{array}{cc} - 1 & 4 \\ 0 & - 1 \end{array})$ on $D (A) = X$ , and $M = 25 / 16$ , $ω = 0$ .

[F1]

The maximum product norm on $C^{2}$ is $∥ (z_{1}, z_{2}) ∥_{\infty} = \max (∣ z_{1} ∣, ∣ z_{2} ∣)$ , with complex scalar homogeneity read using the complex modulus (The standard product norms on a finite product of normed spaces, Real and complex scalar conventions for normed spaces). The complex scalar field is $C$ (The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ ).

[F2]

Every finite-dimensional normed space over $C$ is Banach (Every finite-dimensional normed space is Banach).

[F3]

The operator norm is the least constant $C$ such that $∥ B z ∥_{\infty} \leq C ∥ z ∥_{\infty}$ for every $z \in X$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum). For a matrix with nonnegative entries, the upper bound by its maximum row sum is attained on $z = (1, 1)$ whenever the largest row sum is positive.

[F4]

For $λ \in ρ (A)$ , the resolvent is $R (λ, A) = (λ I - A)^{- 1}$ (Resolvent and spectrum of a closed operator on a Banach space).

[F5]

If a closed densely defined operator generates a semigroup with $∥ T (t) ∥ \leq M e^{ω t}$ , then every $λ > ω$ satisfies $∥ R (λ, A)^{n} ∥ \leq M (λ - ω)^{- n}$ for every integer $n \geq 1$ (Hille-Yosida generation theorem).

[F6]

A bounded operator $A$ on a Banach space generates the strongly continuous exponential semigroup $e^{t A}$ (The exponential series of a bounded operator).

[F7]

The real exponential constant satisfies $e < 3$ (The elementary numerical bound $2 < e < 3$ ).

[F8]

Matrix multiplication is given by the finite coordinate sums of Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes.

Proof

technique · direct matrix calculations

1.1F1F2F3given

Let $N = (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ . Then $N^{2} = 0$ and $A = - I + 4 N$ by matrix multiplication (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes). The space $X$ is Banach by [F1, F2]. The maximum-row-sum estimate gives $∥ A z ∥_{\infty} \leq 5 ∥ z ∥_{\infty}$ , so $A$ is bounded; its domain is all of $X$ , hence dense, and continuity shows its graph is closed.

1.2F3F6F7F8algebra

Since $A = - I + 4 N$ , the two summands commute and $N^{2} = 0$ . The exponential series gives $e^{t A} = e^{- t} e^{4 t N} = e^{- t} (I + 4 t N) = e^{- t} (\begin{array}{cc} 1 & 4 t \\ 0 & 1 \end{array}) .$ For $t \geq 0$ , its maximum row sum is $(1 + 4 t) e^{- t}$ . At $t = 1 / 2$ this is $3 e^{- 1 / 2} = 3 / \sqrt{e}$ . By [F7], $e < 3$ , hence $3 / \sqrt{e} > 3 / \sqrt{3} = \sqrt{3}$ ; moreover $\sqrt{3} > 25 / 16$ because $3 > (25 / 16)^{2}$ . Thus $∥ e^{A / 2} ∥_{\infty} > 25 / 16 = M$ .

2.1F3F4F8step 1.1algebra

For every $λ > 0$ , $λ I - A = (\begin{array}{cc} λ + 1 & - 4 \\ 0 & λ + 1 \end{array}), R (λ, A) = (\begin{array}{cc} (λ + 1)^{- 1} & 4 (λ + 1)^{- 2} \\ 0 & (λ + 1)^{- 1} \end{array})$ Their product is $I$ , so every such $λ$ is in $ρ (A)$ . For a vector $z$ with $∥ z ∥_{\infty} \leq 1$ , the first row of $R z$ has modulus at most $(λ + 1)^{- 1} + 4 (λ + 1)^{- 2}$ and the second at most $(λ + 1)^{- 1}$ ; the vector $(1, 1)$ attains the first row sum. Thus $∥ R (λ, A) ∥_{\infty} = \frac{1}{λ + 1} + \frac{4}{(λ + 1)^{2}} .$ Writing $x = λ / (λ + 1) \in (0, 1)$ gives $λ ∥ R (λ, A) ∥_{\infty} = x + 4 x (1 - x) = 5 x - 4 x^{2} = \frac{25}{16} - 4 {(x - \frac{5}{8})}^{2} \leq \frac{25}{16} .$ This proves the first-power estimate for every $λ > 0$ .

3.1F3F5F8step 2.1algebra

At $λ = 3$ , $R (3, A) = (\begin{array}{cc} 1 / 4 & 1 / 4 \\ 0 & 1 / 4 \end{array}), R (3, A)^{2} = (\begin{array}{cc} 1 / 16 & 1 / 8 \\ 0 & 1 / 16 \end{array}) .$ The maximum row sum of the square is $3 / 16$ , attained on $(1, 1)$ , so $∥ R (3, A)^{2} ∥_{\infty} = \frac{3}{16} = \frac{27}{144} > \frac{25}{144} = \frac{M}{3^{2}} .$ Therefore the required second-power estimate fails.

4.1F5step 3.1step 1.2∎

The direct second-power failure in step 3.1 also rules out a semigroup with generator $A$ and the prescribed bound by [F5]. The example therefore shows that the first-power resolvent estimate alone is insufficient for a general bound with $M > 1$ .

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Sources

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)