1 Introduction

Let \(-A\) be the generator of a \(C_0\)-semigroup \((T_t)_{t\ge 0}\) on a complex Banach space X. For \(\alpha > 0,\) we say that \((T_t)_{t\ge 0}\) is \(\alpha\)-Kreiss bounded (or simply Kreiss bounded when \(\alpha = 1\)) if \(\sigma (A) \subset \overline{{\mathbb{C}}_+} = \{ \lambda \in {\mathbb{C}}: \, {\text{Re}}(\lambda ) \ge 0\}\) and

$$\begin{aligned} \left\| R(\lambda ,A)\right\| \le \frac{C}{(-{\text{Re}}(\lambda ))^{\alpha }}, \quad -1< {\text{Re}}(\lambda )<0. \end{aligned}$$
(1.1)

According to [7, (2.6)] (see also [12, Theorem 1.1.(1)] and [8]) , if \((T_t)_{t\ge 0}\)  is an \(\alpha\)-Kreiss bounded \(C_0\)-semigroup on a Hilbert space for \(\alpha >1,\) then \(\left\| T_t\right\| = O(t^{\alpha }).\) In the following, we do not need to consider the case  \(0< \alpha < 1.\) Indeed when \(0< \alpha < 1\) and \((T_t)_{t\ge 0}\)  is \(\alpha\)-Kreiss bounded, then it is exponentially stable (see [4, Remark 1.22 page 90]). Let us fix \(\gamma \in (0,1).\) In [5], the authors give an example (see [5, Example 4.4.]) of a Kreiss bounded \(C_0\)-semigroup for which there exists a constant \(C>0\) such that \(\left\| T_t\right\| \ge Ct^{\gamma }\) for \(t\ge 1.\) A natural question (listed as an open question in [11, Appendix B]) is whether the estimate \(\left\| T_t\right\| = O(t^{\alpha }),\) for an \(\alpha\)-Kreiss bounded \(C_0\)-semigroup \((T_t)_{t\ge 0}\)  on a Hilbert space, is sharp. In the next section we answer this question for \(\alpha =1.\)

2 Main results

Theorem 2.1

Let \(-A\) be the generator of a Kreiss bounded \(C_0\)-semigroup \((T_t)_{t\ge 0 }\) on a Hilbert space. Then

$$\begin{aligned} \left\| T_t\right\| = \underset{t \rightarrow \infty }{{\mathcal{O}}}\Big (t/\sqrt{\log (t)}\Big ). \end{aligned}$$
(2.1)

The proof of Theorem 2.1 is based on techniques given in [3]. We first state and prove two lemmas. It is known (see [6] or [13]) that a \(C_0\)-semigroup \((T_t)_{t\ge 0}\) on a Hilbert space, with negative generator A,  is bounded if and only if

$$\begin{aligned} \left\| R(-r + i\cdot ,A)x\right\| ^2_{L^2({\mathbb{R}},H)} \le \frac{C}{r}\left\| x\right\| ^2, \quad 0<r<1 \end{aligned}$$

and

$$\begin{aligned} \left\| R(-r + i\cdot ,A^*)x^*\right\| ^2_{L^2({\mathbb{R}},H)} \le \frac{C}{r}\left\| x^*\right\| ^2, \quad 0<r<1. \end{aligned}$$

The following lemma gives an analogous necessary condition for a \(C_0\)-semigroup to be \(\alpha\)-Kreiss bounded, for \(\alpha >0.\)

Lemma 2.2

Let \(\alpha >0\) and \((T_t)_{t\ge 0 }\) be a \(C_0\)-semigroup on a Hilbert space with negative generator A. Assume \((T_t)_{t\ge 0}\) is \(\alpha\)-Kreiss bounded then

$$\begin{aligned} \left\| R(-r + i\cdot ,A)x\right\| ^2_{L^2({\mathbb{R}},H)} \le \frac{C(1+r^{\alpha })^2}{r^{2\alpha }}\left\| x\right\| ^2, \quad 0<r<1 \end{aligned}$$
(2.2)

and

$$\begin{aligned} \left\| R(-r + i\cdot ,A^*)x^*\right\| ^2_{L^2({\mathbb{R}},H)} \le \frac{C(1+r^{\alpha })^2}{r^{2\alpha }}\left\| x^*\right\| , \quad 0<r<1. \end{aligned}$$
(2.3)

Proof

Let \(x \in H.\) First, by [9, formula (7.1) section 1.7], we have

$$\begin{aligned} R(-r +i\beta ,A)x = -\int _{0}^{\infty } {\mathrm{e}}^{i \beta s }{\mathrm{e}}^{-r s}T_{s}x{\mathrm{d}}s, \quad r > 0, \beta \in {\mathbb{R}}. \end{aligned}$$

This means that for \(r>0,\)

$$\begin{aligned} R(-r +i\beta ,A)x = -{\mathcal{F}}^{-1}(s \mapsto {\mathrm{e}}^{-r s}T_sx)(\beta ), \end{aligned}$$
(2.4)

where \({\mathcal{F}} : L^2({\mathbb{R}},H) \mapsto L^2({\mathbb{R}},H)\) is the Fourier–Plancherel operator. By the Fourier–Plancherel Theorem

$$\begin{aligned} \left\| {\mathcal{F}}^{-1}(s \mapsto {\mathrm{e}}^{-r s}T_sx)(\cdot ) \right\| _{L^2({\mathbb{R}}, H)} = \frac{1}{\sqrt{2\pi }} \left\| {\mathrm{e}}^{-r \cdot }T_{(\cdot )}x\right\| _{L^2({\mathbb{R}}_+, H)}. \end{aligned}$$
(2.5)

The Gearhart–Prüss Theorem (see [1, Theorem 5.2.1.]) says that for each \(r>0,\) \(\underset{t>0}{\sup }\left\| {\mathrm{e}}^{-rt}T_t\right\| < \infty .\) This implies that there exists \(K>0\) such that

$$\begin{aligned} \underset{a>1}{\sup }\left\| {\mathrm{e}}^{-a \cdot }T_{(\cdot )}x\right\| _{L^2({\mathbb{R}}_+, H)} \le K\left\| x\right\| \end{aligned}$$

and then

$$\begin{aligned} \underset{a>1}{\sup }\left\| R(-a +i\beta ,A)x\right\| _{L^2({\mathbb{R}}, H)} \le \frac{K}{\sqrt{2\pi }}\left\| x\right\| . \end{aligned}$$
(2.6)

Now, let \(0<r<1\) and \(a = r+1>1.\) By the resolvent identity, we have, for \(\beta \in {\mathbb{R}},\)

$$\begin{aligned} R(-r +i\beta ,A)x&= \big (I+(r-a)R(-r +i\beta ,A)\big ) R(-a+i\beta ,A)x \\ &= \big (I-R(-r +i\beta ,A)\big ) R(-a+i\beta ,A)x. \end{aligned}$$

Using (1.1) and (2.6):

$$\begin{aligned} \left\| R(-r + i\cdot ,A)x\right\| ^2_{L^2({\mathbb{R}},H)}&\le \big (1+\sup _{\beta \in {\mathbb{R}}}\left\| R(-r+i\beta ,A)\right\| \big )^2 \\ &\quad \times \underset{a'>1}{\sup }\left\| R(-a' +i\cdot ,A)x\right\| ^2_{L^2({\mathbb{R}}_+, H)} \\ &\le 2\pi K^2\Big (1+\frac{C}{r^{\alpha }}\Big )^2\left\| x\right\| ^2 \\ &\le C'\frac{(1+r^{\alpha } )^2}{r^{2\alpha }}\left\| x\right\| ^2 . \end{aligned}$$

In the same way we obtain (2.3). \(\square\)

The next lemma says that if \((T_t)_{t\ge 0}\) and \((T^*_t)_{t\ge 0}\) satisfy a kind of Cesàro condition then \(\left\| T_t\right\|\) satisfies (2.1).

Lemma 2.3

Let \((T_t)_{t\ge 0}\) be a \(C_0\)-semigroup on a Banach space X. Assume that there exists a \(C>0\) such that for each \(t>1,\)

$$\begin{aligned} \int _{0}^t \left\| T_s x\right\| ^2 {\mathrm{d}}s \le Ct^{2} \left\| x\right\| ^2, \quad x \in X, \end{aligned}$$

and

$$\begin{aligned} \int _{0}^t \left\| T^*_s x^*\right\| ^{2} {\mathrm{d}}s \le Ct^{2} \left\| x^*\right\| ^{2}, \quad x^* \in X^* . \end{aligned}$$

Then, \(\left\| T_t\right\| = {\mathcal{O}}(t/\sqrt{\log (t)}).\)

Proof

Let \(t >2\) and \(1<P<Q<t.\) We have

$$\begin{aligned} (Q-P)^2|\langle T_tx,x^* \rangle |^2&= \Big |\int _{P}^Q \langle T_{t-s}x,T_s^*x^* \rangle {\mathrm{d}}s \Big |^2\\ &\le \Big (\int _{P}^Q \left\| T_{t-s}x\right\| ^2{\mathrm{d}}s\Big )\Big (\int _{P}^Q \left\| T_{s}^*x^*\right\| ^{2}{\mathrm{d}}s\Big ) \\ &\le \Big (\int _{P}^Q \left\| T_{t-s}x\right\| ^2{\mathrm{d}}s\Big )\Big (\int _{0}^Q \left\| T_{s}^*x^*\right\| ^2{\mathrm{d}}s\Big ) \\ &\le CQ^{2}\left\| x^*\right\| ^2\Big (\int _{t-Q}^{t-P} \left\| T_{s}x\right\| ^2{\mathrm{d}}s\Big ). \end{aligned}$$

Finally taking the supremum over \(\{\left\| x^*\right\| =1 \}\) one obtains,

$$\begin{aligned} \frac{(Q-P)^2}{Q^2} \left\| T_tx\right\| ^2 \le C\int _{t-Q}^{t-P} \left\| T_{s}x\right\| ^2{\mathrm{d}}s. \end{aligned}$$
(2.7)

Now we set \(L:= \left\lfloor \log (t)/\log (2) \right\rfloor .\) For \(0 \le l \le L-1,\) \(Q=2^{l+1}\) and \(P= 2^{l}\) we have

$$\begin{aligned} \left\| T_tx\right\| ^2 \le 4C \int _{t-2^{l+1}}^{t-2^l} \left\| T_sx\right\| ^2{\mathrm{d}}s. \end{aligned}$$

Therefore

$$\begin{aligned} L\left\| T_tx\right\| ^2&= \sum _{l=0}^{L-1} \left\| T_tx\right\| ^2 \le \sum _{l=0}^{L-1} 4C \int _{t-2^{l+1}}^{t-2^l} \left\| T_sx\right\| ^2{\mathrm{d}}s \\ &\le 4C\int _{0}^t \left\| T_sx\right\| ^2 {\mathrm{d}}s \le 4C^2t^{2}. \end{aligned}$$

Hence for \(t>2,\)

$$\begin{aligned} \left\| T_tx\right\| \le \frac{2Ct}{\sqrt{L}}\left\| x\right\| \le \frac{C't}{\sqrt{\log (t)}}\left\| x\right\| . \end{aligned}$$

\(\square\)

We are now ready to prove Theorem 2.1.

Proof of Theorem 2.1

Let \(x\in H\) and \(r>0.\) By (2.4), (2.5) and Lemma 2.2,

$$\begin{aligned} \left\| {\mathrm{e}}^{-r \cdot }T_{(\cdot )}x\right\| _{L^2({\mathbb{R}}_+, H)}^2 = \left\| R(-r + i\cdot ,A)x\right\| ^2_{L^2({\mathbb{R}},H)} \le \frac{C(1+r)^2}{r^{2}}\left\| x\right\| ^2. \end{aligned}$$

Furthermore

$$\begin{aligned} \left\| {\mathrm{e}}^{-r \cdot }T_{(\cdot )}x\right\| ^2_{L^2({\mathbb{R}}_+, H)} = \int _{{\mathbb{R}}}{\mathrm{e}}^{-2r s}\left\| T_sx\right\| ^2{\mathrm{d}}s \ge {\mathrm{e}}^{-2r t}\int _{0}^t \left\| T_sx\right\| ^2{\mathrm{d}}s. \end{aligned}$$

Taking \(t = \frac{1}{r}\) we obtain

$$\begin{aligned} \int _{0}^t \left\| T_s x\right\| ^2 {\mathrm{d}}s \le C{\mathrm{e}}^{2}t^{2}\Big (1+\frac{1}{t}\Big )^2 \left\| x\right\| ^2. \end{aligned}$$

Hence, for \(t>1,\)

$$\begin{aligned} \int _{0}^t \left\| T_s x\right\| ^2 {\mathrm{d}}s \le C't^{2} \left\| x\right\| ^2, \end{aligned}$$

where \(C'=4{\mathrm{e}}^2C.\) Similarly, since \((T_t^*)_{t\ge 0}\) is a Kreiss bounded \(C_0\)-semigroup on a Hilbert space, for \(t>1\) and \(x^*\in H,\) we have

$$\begin{aligned} \int _{0}^t \left\| T^*_s x^*\right\| ^2 {\mathrm{d}}s \le C't^{2} \left\| x^*\right\| ^2. \end{aligned}$$

Finally, Lemma 2.3 allows us to conclude that \(\left\| T_t\right\| = {\mathcal{O}}(t/\sqrt{\log (t)}).\) \(\square\)

Remark 2.4

Let \(\alpha >1\) and \((T_t)_{t\ge 0}\)  be an \(\alpha\)-Kreiss bounded \(C_0\)-semigroup. The computations which lead to (2.7) with \(Q=2\) and \(P=1\) give

$$\begin{aligned} \left\| T_tx\right\| ^2 \le C2^{\alpha }\int _{t-2}^{t-1}\left\| T_sx\right\| ^2{\mathrm{d}}s \le C'\int _{0}^{t}\left\| T_sx\right\| ^2{\mathrm{d}}s \le C''t^{2\alpha }\left\| x\right\| ^2. \end{aligned}$$

Then using Lemma 2.2 and following the proof of Theorem 2.1 we can state \(\left\| T_t\right\| = O(t^{\alpha }).\) This gives another proof of [12, Theorem 1.1.(1) with \(g(s) = s^{\alpha }\) for \(\alpha >1\)].

Question 2.5

In view of [2], we can naturally ask whether the estimate (1.1) can be improved. In the aforementioned paper, the authors show that if \((T_t)_{t\ge 0}\) is a positive Kreiss bounded \(C_0\)-semigroup on a \(L^1\)-space, then there exists \(0<\epsilon <1\) such that \(\left\| T_t\right\| = {\mathcal{O}}(t^{1-\epsilon }).\) Do we have same estimate on Hilbert spaces?

Let us give an application of the Theorem 2.1. Let \(W^{2,s}(\mathbb {T}^2)\) be the second order Sobolev space equipped with the standard norm. Let A be the operator with domain \(D(A) = W^{2,2}(\mathbb {T}^2) \times W^{2,1}(\mathbb {T}^2)\) defined by

$$\begin{aligned} A = \left( \begin{array}{cc} 0 &{} -1 \\ -\Delta - M\frac{\partial }{\partial x} &{} 0 \end{array}\right) , \end{aligned}$$

where \(\Delta\) is the Laplacian with \(D(\Delta ) = W^{2,2}(\mathbb {T}^2)\) and \(M :L^2(\mathbb {T}^2) \rightarrow L^2(\mathbb {T}^2)\) is the multiplication operator given by \(M(h)(x,y) = {\mathrm{e}}^{iy}h(x,y).\) Then, it is known (see [12, section 4] or [10] for details) that \(-A\) generates a \(C_0\)-group \((T_t)_{t\in {\mathbb{R}}}\) on the Hilbert space \(H = W^{1,2}(\mathbb {T}^2)\times L^2(\mathbb {T}^2),\) satisfying \(\left\| T_t\right\| = \underset{t \rightarrow \pm \infty }{{\mathcal{O}}}(|t|{\mathrm{e}}^{|t|/2}).\) With Theorem 2.1, we are able to improve this estimate:

Proposition 2.6

Let \((T_t)_{t \in {\mathbb{R}}}\) be the \(C_0\)-group on H generated by \(-A.\) Then

$$\begin{aligned} \left\| T_t\right\| = \underset{t \rightarrow \pm \infty }{{\mathcal{O}}}\Bigg (\frac{|t|}{\sqrt{\log (|t|)}}{\mathrm{e}}^{|t|/2}\Bigg ). \end{aligned}$$
(2.8)

Proof

According to [12, Lemma 4.4.], for \(\lambda \in {\mathbb{C}}_+,\) with \({\text{Re}}(\lambda ) \in (0,1),\)

$$\begin{aligned} \left\| R\Big (\lambda , A + \frac{1}{2}\Big ) \right\| \le \frac{C}{-{\text{Re}}(\lambda )} \quad \text{and} \quad \left\| R\Big (\lambda , -A + \frac{1}{2}\Big ) \right\| \le \frac{C}{-{\text{Re}}(\lambda )}. \end{aligned}$$

Then, according to Remark 2.4(2), \(({\mathrm{e}}^{-\frac{t}{2}}T_t)_{t\ge 0}\) (resp. \(({\mathrm{e}}^{-\frac{t}{2}}T_{-t})_{t\ge 0}\) ) which is generated by \(-(A+\frac{1}{2})\) (resp. \(A+\frac{1}{2}\)), satisfies \(\left\| {\mathrm{e}}^{-1/2t}T_t\right\| = {\mathcal{O}}(t/\sqrt{\log (t)})\) (resp. \(\left\| {\mathrm{e}}^{-1/2t}T_{-t}\right\| = {\mathcal{O}}(t/\sqrt{\log (t)})\)). This yields the estimate (2.8). \(\square\)