Static Controller Synthesis for Peak-to-Peak Gain Minimization as an Optimization Problem

Abstract

An optimization approach to linear control systems has recently become very popular. For example, the linear feedback matrix in the classical linear-quadratic regulator problem can be viewed as a variable, and the problem can be reduced to the minimization of the performance indicator for this variable. To this end, one can apply the gradient method and obtain a justification of the convergence. This approach has been successfully applied to a number of problems, including output feedback optimization. The present paper is the first to apply this approach to the peak-to-peak gain minimization problem. A gradient method for finding a static state or output feedback is written out and justified. A number of examples are considered, including the single and double pendulums.

APPENDIX

Lemma A.1.

Let $X$ and $Y$ be solutions to the dual Lyapunov equations with the Hurwitz matrix $A$ ,

$$ {A}^{\top }X+XA+W=0$$

and

$$ AY+Y{A}^{\top }+V=0.$$

Then

$$ \mathrm{tr}\thinspace (XV)=\mathrm{tr}\thinspace (YW).$$

Proof of Lemma A.1. Indeed, by direct computation we have

$$ \begin {aligned} \mathrm{tr}\thinspace (XV)&=\mathrm{tr}\thinspace \left (X(-AY-Y{A}^{\top })\right )=-\mathrm{tr}\thinspace (XAY)-\mathrm{tr}\thinspace (XY{A}^{\top }) \\ &=-\mathrm{tr}\thinspace {(XAY)}^{\top }-\mathrm{tr}\thinspace {({A}^{\top }XY)}^{\top }=\mathrm{tr}\thinspace \left (Y(-{A}^{\top }X-XA)\right )=\mathrm{tr}\thinspace (YW). \end {aligned} $$

The proof of Lemma A.1 is complete. $\quad \blacksquare$

The next lemma contains some well-known results (see, e.g., [19]), needed in the exposition to follow.

Lemma A.2.

• 1. For matrices $A$ and $B$ of appropriate dimensions one has the relations

  $$ \begin {gathered} \begin {aligned} \|AB\|_F&\leqslant \|A\|_F\|B\|, \\[.3em] |\mathrm{tr}\thinspace AB|&\leqslant \|A\|_F\|B\|_F, \\[.3em] \|A\|&\leqslant \|A\|_F, \end {aligned} \\[.3em] AB+{B}^{\top }{A}^{\top }\leqslant \varepsilon A{A}^{\top }+\frac {1}{\varepsilon }{B}^{\top }B\quad \text {for any}\quad \varepsilon >0. \end {gathered}$$

• 2. For positive semidefinite matrices $A$ and $B$ one has the relations

  $$ 0\leqslant \lambda _{\min }(A)\lambda _{\max }(B)\leqslant \lambda _{\min }(A)\mathrm{tr}\thinspace B\leqslant \mathrm{tr}\thinspace AB\leqslant \lambda _{\max }(A)\mathrm{tr}\thinspace B\leqslant \mathrm{tr}\thinspace A\mathrm{tr}\thinspace B. $$

Lemma A.3.

For a solution $P$ of the Lyapunov equation

$$ AP+P{A}^{\top }+Q=0$$

with a Hurwitz matrix $A$ and $Q\succ 0$ one has the estimates

$$ \lambda _{\max }(P)\geqslant \frac {\lambda _{\min }(Q)}{2\sigma }, \quad \lambda _{\min }(P)\geqslant \frac {\lambda _{\min }(Q)}{2\|A\|},$$

(A.1)

where $\sigma =-\max \limits _i{\mathrm {Re}\thinspace }\lambda _i(A)$ .

If, however, $Q=D{D}^{\top }$ and the pair $(A,D)$ is controllable, then

$$ \lambda _{\max }(P)\geqslant \frac {\|u^*D\|^2}{2\sigma }>0,$$

(A.2)

where

$$ u^*A=\lambda u^*,\quad {\mathrm {Re}\thinspace }\lambda =-\sigma ,\quad \|u\|=1; $$

i.e., $u$ is the left eigenvector of $A$ corresponding to the eigenvalue $\lambda$ of $A$ with the greatest real part. The vector $u$ and the number $\lambda$ can be complex; here $u^*$ designates complex conjugation and transposition.

Proof of Lemma A.3. The estimates (A.1) are well known; see, e.g., [20]. Let us prove the estimate (A.2). The explicit solution of the Lyapunov equation for a Hurwitz matrix has the form

$$ P=\int \limits _0^{+\infty }e^{At}D{D}^{\top }e^{{A}^{\top } t}dt.$$

Post- and pre-multiplying this relation by $u$ and by $u^*$, respectively, and taking into account the fact that $u^*e^{At}=e^{\lambda t}u^*$ and $e^{{A}^{\top }t}u=e^{\lambda ^*t}u$, we obtain

$$ \lambda _{\max }(P)\geqslant u^*Pu=\int \limits _0^{+\infty }u^*e^{At}D{D}^{\top }e^{{A}^{\top }t}udt =\int \limits _0^{+\infty }e^{(\lambda +\lambda ^*)t} u^*D{D}^{\top }u dt=\frac {\|u^*D\|^2}{2\sigma },$$

with $\|u^*D\|{\thinspace >\thinspace }0$ by virtue of the controllability of the pair $(A,D)$; see, e.g., [5, Theorem D.1.5]. The proof of Lemma A.3 is complete. $\quad \blacksquare$

Proof of Lemma 1 .

(a) Equation (6) can be represented in the form

$$ \left (A+\frac {\alpha }{2}I\right )P+P{\left (A+\frac {\alpha }{2}I\right )}^{\top }=-\frac {1}{\alpha }D{D}^{\top }$$

and, according to [5, Lemma 1.2.3], has a unique solution if and only if the matrix $A+\frac {\alpha }{2}I$ is Hurwitz— ${\mathrm {Re}\thinspace }\lambda _i(A+\frac {\alpha }{2}I)<0$, i.e., for $0<\alpha <2\sigma$.

Let us estimate the quantity $f(\alpha )=\mathrm{tr}\thinspace CP(\alpha ){C}^{\top }$ using Lemma A.3 with the obvious replacements

$$ f(\alpha )=\mathrm{tr}\thinspace CP(\alpha ){C}^{\top }\geqslant \lambda _{\min }({C}^{\top }C)\lambda _{\max }\left (P(\alpha )\right )\geqslant \frac {\|u^*D\|^2\lambda _{\min }({C}^{\top }C)}{\alpha (2\sigma -\alpha )},$$

where $u$ has the same meaning as in Lemma A.3 and the quantity $\|u^*D\|^2$ is positive by virtue of the assumption about the controllability of the pair $(A,D)$ (and therewith also of the pair $(A+\frac {\alpha }{2}I,D)$).

Now let us show that the function $f(\alpha )=\mathrm{tr}\thinspace CP(\alpha ){C}^{\top }$ is strictly convex on the interval $(0,2\sigma )$. In accordance with [5, Lemma 1.2.3], the solution of Eq. (6) is representable in closed form as

$$ P(\alpha )=\int \limits _0^{+\infty }e^{(A+\frac {\alpha }{2}I)t}\frac {1}{\alpha }D{D}^{\top }e^{{(A+\frac {\alpha }{2}I)}^{\top }t}dt =\int \limits _0^{+\infty } \underbrace {\frac {e^{\alpha t}}{\alpha }}_{g(\alpha ,t)}\underbrace {e^{At}D{D}^{\top }e^{{A}^{\top }t}}_{h(t)}dt.$$

However, $g(\alpha ,t)>0$, $h(t)\succ 0$ for $\alpha >0$; therefore, on the interval $(0,2\sigma )$ we have

$$ P(\alpha )=\int \limits _0^{+\infty }g(\alpha ,t)h(t)dt\succ 0, \quad f(\alpha )=\mathrm{tr}\thinspace P(\alpha ){C}^{\top }_2C_2>0.$$

By a straightforward computation, we obtain

$$ g^{\prime {}\prime }(\alpha ,t)=\left ((\alpha t-1)^2+1\right )\frac {e^{\alpha t}}{\alpha ^3}\geqslant \frac {e^{\alpha t}}{\alpha ^3}=\frac {1}{\alpha ^2}g(\alpha ,t) $$

(here differentiation is with respect to $\alpha$); thus,

$$ f^{\prime {}\prime }(\alpha )=\int \limits _0^{+\infty }g^{\prime {}\prime }(\alpha ,t)h(t)dt\geqslant \frac {1}{\alpha ^2}f(\alpha )\geqslant \frac {1}{4\sigma ^2}f(\alpha ^*)>0. $$

Thus, the second derivative of the function $f(\alpha )$ is positive and tends to infinity at the endpoints of the interval $(0,2\sigma )$.

In a similar way, by a straightforward calculation of the fourth derivative, we obtain

$$ g^{(IV)}(\alpha ,t)=\left ((\alpha t-2)^2\alpha ^2 t^2 +2(2\alpha t-3)^2+6\right )\frac {e^{\alpha t}}{\alpha ^5}\geqslant \frac {6}{\alpha ^5}e^{\alpha t}=\frac {6}{\alpha ^4}g(\alpha ,t);$$

thus,

$$ f^{(IV)}(\alpha )\geqslant \frac {6}{\alpha ^4}f(\alpha )>0;$$

i.e., the second derivative $f^{\prime {}\prime }(\alpha )$ itself is convex and increases at the boundaries of the interval.

(b) Now let us derive a formula for the derivative of the function $f(\alpha )$. In Eq. (6), the solution $P$ is a function of $\alpha$. Let us differentiate this equation; by $P^{\prime }$ we mean the derivative with respect to $\alpha$,

$$ AP^{\prime }+P^{\prime }{A}^{\top }+\alpha P^{\prime }+ P-\frac {1}{\alpha ^2}D{D}^{\top }=0.$$

Comparing the equations for $P^{\prime }$ and $Y$ and applying Lemma A.1, we obtain the desired formula

$$ f^{\prime }(\alpha )=\mathrm{tr}\thinspace CP^{\prime }{C}^{\top }=\mathrm{tr}\thinspace Y\left (P-\frac {1}{\alpha ^2}D{D}^{\top }\right ). $$

(c) In a similar manner, we will obtain an expression for the second derivative $f(\alpha )$. Differentiating the equation for $P^{\prime }$ with respect to $\alpha$, we obtain

$$ AP^{\prime {}\prime }+P^{\prime {}\prime }{A}^{\top }+\alpha P^{\prime {}\prime }+ 2P^{\prime }+\frac {2}{\alpha ^3}D{D}^{\top }=0.$$

Applying again Lemma A.1 to this equation and to Eq. (10) (and bearing in mind that $X=P^{\prime }$), we obtain

$$ f^{\prime {}\prime }(\alpha )=\mathrm{tr}\thinspace CP^{\prime {}\prime }{C}^{\top }=2\mathrm{tr}\thinspace Y\left (X+\frac {1}{\alpha ^3}D{D}^{\top }\right ).$$

The proof of Lemma 1 is complete. $\quad \blacksquare$

Proof of Lemma 3. Consider a sequence of stabilizing controllers $\{K_j\}\subseteq \mathcal S$ such that $K_j\to K\in \partial \mathcal S$; i.e., $\sigma (A+BKC_1)=0$. This means that for each $\varepsilon >0$ there exists a number $N=N(\varepsilon )$ such that the inequality

$$ |\sigma (A+BK_jC_1)-\sigma (A+BKC_1)|=\sigma (A+BK_jC_1)<\varepsilon$$

holds for all $j\geqslant N(\varepsilon )$.

Let $P_j$ be the solution of the Lyapunov equation (15) associated with the controller $K_j$,

$$ \left (A_{K_j}+\frac {\alpha _j}{2}I\right )P_j+P_j{\left (A_{K_j}+\frac {\alpha _j}{2}I\right )}^{\top }+\frac {1}{\alpha _j}D{D}^{\top }=0,$$

and let $Y_j$ be a solution of the dual Lyapunov equation

$$ {\left (A_{K_j}+\frac {\alpha _j}{2}I\right )}^{\top }Y_j+Y_j\left (A_{K_j}+\frac {\alpha _j}{2}I\right )+C_2{C}^{\top }_2=0.$$

Then

$$ \begin {aligned} f(K_j)&=\mathrm{tr}\thinspace \left (C_2P_j{C}^{\top }_2\right )+\rho \|K_j\|^2_F\geqslant \mathrm{tr}\thinspace \left (P_jC_2{C}^{\top }_2\right ) =\mathrm{tr}\thinspace \left (Y_j\frac {1}{\alpha _j}D{D}^{\top }\right ) \\ &{}\geqslant \frac {1}{\alpha _j}\lambda _{\min }(Y_j)\|D\|_F^2 \geqslant \frac {1}{\alpha _j}\frac {\lambda _{\min }(C_2{C}^{\top }_2)}{2\|A+BK_jC+\frac {\alpha _j}{2}I\|}\|D\|_F^2 \\ &{}\geqslant \frac {\lambda _{\min }(C_2{C}^{\top }_2)}{4\sigma (A+BK_jC_1)\|A+BK_jC+\frac {\alpha _j}{2}I\|}\|D\|_F^2\geqslant \frac {\lambda _{\min }(C_2{C}^{\top }_2)}{4\varepsilon \big (\|A+BK_jC\|+\varepsilon \big )}\|D\|_F^2\xrightarrow [\varepsilon \to 0]{}+\infty , \end {aligned}$$

because

$$ 0<\alpha _j<2\sigma (A+BK_jC_1) \quad \text {and}\quad \sigma \left (A+BK_jC_1+\frac {\alpha _j}{2}I\right )\leqslant \sigma (A+BK_jC_1).$$

On the other hand,

$$ f(K_j)=\mathrm{tr}\thinspace (C_2P_j{C}^{\top }_2)+\rho \|K_j\|^2_F\geqslant \rho \|K_j\|^2_F\geqslant \rho \|K_j\|^2\xrightarrow [\|K_j\|\to +\infty ]{}+\infty .$$

The proof of Lemma 3 is complete. $\quad \blacksquare$

Proof of Lemma 4. System (12) closed by the feedback (13) acquires the closed-loop form

$$ \begin {aligned} \dot x &=(A+BKC_1)x+Dw,\\ z &=C_2x. \end {aligned}$$

(A.3)

Applying Theorem 1 to system (A.3), we arrive at the problem

$$ \min f(K,\alpha ),\quad f(K,\alpha )=\mathrm{tr}\thinspace C_2P{C}^{\top }_2+\rho \|K\|^2_F$$

under a constraint in the form of the Lyapunov equation for the matrix $P$ of the invariant ellipsoid,

$$ (A+BKC_1)P+P{(A+BKC_1)}^{\top }+\alpha P+\frac {1}{\alpha }D{D}^{\top }=0. $$

Differentiation with respect to $\alpha$ is performed just as above with the replacement of $A$ by $A_K=A+BKC_1$. To differentiate with respect to $K$, we give it an increment $\Delta K$ and denote the corresponding increment of $P$ by $\Delta P$,

$$ \big (A+B(K+\Delta K)C_1\big )(P+\Delta P)+(P+\Delta P){\big (A+B(K+\Delta K)C_1\big )}^{\top }+\alpha (P+\Delta P)+\frac {1}{\alpha }D{D}^{\top }=0,$$

or, after linearization and subtraction of this and preceding equations,

$$ \left (A+BKC_1+\frac {\alpha }{2}I\right )\Delta P+\Delta P{\left (A+BKC_1+\frac {\alpha }{2}I\right )}^{\top }+B\Delta KC_1P+P{(B\Delta KC_1)}^{\top }=0.$$

(A.4)

Let us calculate the increment in the functional $f(K)$ by linearizing the corresponding quantities,

$$ \Delta f(K) =\mathrm{tr}\thinspace C_2\Delta P{C}^{\top }_2+\rho \mathrm{tr}\thinspace {K}^{\top }\Delta K+\rho \mathrm{tr}\thinspace {(\Delta K)}^{\top }K=\mathrm{tr}\thinspace {C}^{\top }_2C_2\Delta P+2\rho \mathrm{tr}\thinspace {K}^{\top }\Delta K. $$

Consider the Lyapunov equation (20) dual to (A.4). By Lemma A.1, from Eqs. (A.4) and (20) we have

$$ \Delta f(K)=\mathrm{tr}\thinspace 2C_1PYB\Delta K+2\rho \mathrm{tr}\thinspace {K}^{\top }\Delta K=\langle 2(\rho K+{B}^{\top }YP{C}^{\top }_1),\Delta K\rangle .$$

Thus,

$$ \nabla _Kf(K,\alpha )=2(\rho K+{B}^{\top }YP{C}^{\top }_1).$$

The proof of Lemma 4 is complete. $\quad \blacksquare$

Proof of Lemma 5. Let us calculate

$$ \nabla ^2_Kf(K)[E,E]=\langle \nabla ^2_Kf(K)[E],E\rangle$$

by taking the derivative in the direction $E\in \mathbb R^{p\times l}$ of $\nabla _Kf(K)[E]=\langle \nabla _Kf(K),E\rangle$.

Linearizing the relevant quantities, we calculate the increment in the functional $\nabla _Kf(K)[E]$ in direction $E$,

$$ \begin {aligned} \Delta \nabla _Kf(K)[E]&=2\big (\rho K+\rho \delta E+{B}^{\top }(Y+\Delta Y)(P+\Delta P){C}^{\top }_1\big )-2\left (\rho K+{B}^{\top }YP{C}^{\top }_1\right ) \\ &{}=2\Big (\rho K+\rho \delta E+{B}^{\top }\big (Y+\delta Y^{\prime }(K)[E]\big )\big (P+\delta P^{\prime }(K)[E]\big ){C}^{\top }_1\Big )-2\left (\rho K+{B}^{\top }YP{C}^{\top }_1\right ) \\ &{}=2\delta \Big (\rho E+{B}^{\top }\big (YP^{\prime }(K)[E]+Y^{\prime }(K)[E]P\big ){C}^{\top }_1\Big ), \end {aligned}$$

where

$$ \begin {aligned} \Delta P&=P(K+\delta E)-P(K)=\delta P^{\prime }(K)[E], \\ \Delta Y&=Y(K+\delta E)-Y(K)=\delta Y^{\prime }(K)[E]. \end {aligned} $$

Thus, denoting $P^{\prime }=P^{\prime }(K)[E]$ and $Y^{\prime }=Y^{\prime }(K)[E]$, we have

$$ \frac 12\nabla ^2_Kf(K)[E,E]= \big \langle \rho E+{B}^{\top }(YP^{\prime }+Y^{\prime }P){C}^{\top }_1,E\big \rangle .$$

Further, $P=P(K)$ is a solution to Eq. (15); let us write it in increments in direction $E$,

$$ \left (A+B(K+\delta E)C_1\right )(P+\delta P^{\prime })+(P+\delta P^{\prime }){\left (A+B(K+\delta E)C_1\right )}^{\top }+\alpha (P+\delta P^{\prime })+\frac {1}{\alpha }D{D}^{\top }=0, $$

or

$$ \begin {aligned} (A+BKC_1)(P+\delta P^{\prime })&{}+(P+\delta P^{\prime }){(A+BKC_1)}^{\top }\\ &{}+\alpha (P+\delta P^{\prime })+\delta \left (BEC_1P+P{(BEC_1)}^{\top }\right )+\frac {1}{\alpha }D{D}^{\top }=0. \end {aligned}$$

Subtracting Eq. (6) from the resulting relation, we arrive at Eq. (22).

Further, $Y=Y(K)$ is a solution of the Lyapunov equation (20); let us write it in increments in direction $E$,

$$ {\left (A+B(K+\delta E)C_1\right )}^{\top }(Y+\delta Y^{\prime })+(Y+\delta Y^{\prime })\left (A+B(K+\delta E)C_1\right )+\alpha (Y+\delta Y^{\prime })+{C}^{\top }_2C_2=0, $$

or

$$ \begin {aligned} {(A+BKC_1)}^{\top }(Y+\delta Y^{\prime })&{}+(Y+\delta Y^{\prime })(A+BKC_1)+\\ &{}+\alpha (Y+\delta Y^{\prime })+\delta \left ({(BEC_1)}^{\top }Y+YBEC_1\right )+{C}^{\top }_2C_2=0. \end {aligned} $$

Subtracting Eq. (8) from the resulting relation, we have

$$ {\left (A+BKC_1+\frac {\alpha }{2}I\right )}^{\top }Y^{\prime }+Y^{\prime }\left (A+BKC_1+\frac {\alpha }{2}I\right )+{(BEC_1)}^{\top }Y+YBEC_1=0.$$

(A.5)

From (22) and (A.5) we have the relation

$$ \mathrm{tr}\thinspace P^{\prime }YBEC_1=\mathrm{tr}\thinspace Y^{\prime }BEC_1P;$$

thus,

$$ \frac 12\nabla ^2_Kf(K)[E,E]= \rho \langle E,E\rangle +\big \langle {B}^{\top }(YP^{\prime }+Y^{\prime }P){C}^{\top }_1,E\big \rangle = \rho \langle E,E\rangle +2\langle {B}^{\top }YP^{\prime }{C}^{\top }_1,E\rangle .$$

The proof of Lemma 5 is complete. $\quad \blacksquare$

Corollary A.1.

The action of the Hessian of the function $f(K)$ on a matrix $E\in \mathbb R^{p\times l}$ such that $\|E\|_F=1$ satisfies the estimate

$$ \frac 12\sup _{\|E\|_F=1}|\nabla ^2_Kf(K)[E,E]|\leqslant \rho +2\|P^{\prime }\|_F\|Y\|\|B\|_F\|C_1\|.$$

Proof of Corollary A.1. According to (21),

$$ \begin {aligned} \frac 12\sup _{\|E\|_F=1}\big |\nabla ^2_Kf(K)[E,E]\big |&\leqslant \sup _{\|E\|_F=1}\rho \langle E,E\rangle +2\sup _{\|E\|_F=1}\big |\langle {B}^{\top }YP^{\prime }{C}^{\top }_1,E\rangle \big | \\ &{}=\rho \sup _{\|E\|_F=1}\|E\|^2_F+2\sup _{\|E\|_F=1}\big |\langle P^{\prime },YBEC_1\rangle \big |\\ &{}\leqslant \rho +2\|P^{\prime }\|_F\sup _{\|E\|_F=1}\|YBEC_1\|_F \leqslant \rho +2\|P^{\prime }\|_F\|Y\|\|B\|_F\|C_1\|, \end {aligned} $$

because, considering Lemma A.2, we have

$$ \|YBEC_1\|_F\leqslant \|Y\|\|B\|_F\|E\|_F\|C_1\|. $$

The proof of Corollary A.1 is complete. $\quad \blacksquare$

Proof of Lemma 6. According to Corollary A.1, it suffices to estimate from above the quantity

$$ \rho +2\|P^{\prime }\|_F\|Y\|\|B\|_F\|C_1\|.$$

We have the following estimate for $\|Y\|$:

$$ \begin {aligned} \frac {1}{\alpha }\lambda _{\min }(D{D}^{\top })\|Y\|&\leqslant \frac {1}{\alpha }\lambda _{\min }(D{D}^{\top })\mathrm{tr}\thinspace Y\leqslant \frac {1}{\alpha }\mathrm{tr}\thinspace YD{D}^{\top }=\mathrm{tr}\thinspace Y\frac {1}{\alpha }D{D}^{\top } \\ &{}=\mathrm{tr}\thinspace P{C}^{\top }_2C_2 =\mathrm{tr}\thinspace C_2P{C}^{\top }_2=f(K)-\rho \|K\|_F^2\leqslant f(K)\leqslant f(K_0); \end {aligned}$$

hence

$$ \|Y\|\leqslant \frac {\alpha }{\lambda _{\min }(D{D}^{\top })}f(K_0).$$

(A.6)

The estimate for $\alpha$ is established as follows:

$$ \begin {aligned} \alpha &<2\sigma (A+BKC_1)\leqslant 2\|A+BKC_1\| \\ &\leqslant 2\big (\|A\|+\|B\|\|K\|\|C_1\|\big )\leqslant 2\big (\|A\|+\|B\|\|K\|_F\|C_1\|\big ) \\ &\leqslant 2\big (\|A\|+\|B\|\sqrt {f(K)}\|C_1\|\big )\leqslant 2\big (\|A\|+\sqrt {f(K_0)}\|B\|\|C_1\|\big ); \end {aligned}$$

thus,

$$ \|Y\|\leqslant 2\frac {\|A\|+\sqrt {f(K_0)}\|B\|\|C_1\|}{\lambda _{\min }(D{D}^{\top })}f(K_0). $$

Now let us estimate $\|P\|$ from above,

$$ \lambda _{\min }({C}^{\top }_2C_2)\|P\|\leqslant \mathrm{tr}\thinspace (C_2P{C}^{\top }_2)=f(K)-\rho \|K\|^2_F\leqslant f(K)\leqslant f(K_0);$$

hence

$$ \|P\|\leqslant \frac {f(K_0)}{\lambda _{\min }({C}^{\top }_2C_2)}. $$

Finally, let us estimate $\|P^{\prime }\|_F$ from above. Considering Lemma A.2, note that

$$ \begin {aligned} &{}\lambda _{\max }\left (BEC_1P+P{(BEC_1)}^{\top }\right )=\big \|BEC_1P+P{(BEC_1)}^{\top }\big \|\leqslant \big \|P^2+BEC_1{(BEC_1)}^{\top }\big \| \\ &\qquad \qquad {}\leqslant \|P\|^2+\|B\|^2\|C_1\|^2\|E\|_F^2\leqslant \frac {f^2(K_0)}{\lambda ^2_{\min }({C}^{\top }_2C_2)}+\|B\|^2\|C_1\|^2=\xi \frac {1}{\alpha }\lambda _{\min }(D{D}^{\top }) \end {aligned} $$

for

$$ \xi =\frac {\alpha }{\lambda _{\min }(D{D}^{\top })}\left (\frac {f^2(K_0)}{\lambda ^2_{\min }({C}^{\top }_2C_2)}+\|B\|^2\|C_1\|^2\right ). $$

Therefore, for the solution $P^{\prime }$ of the Lyapunov equation (22) one has the estimate

$$ \begin {aligned} P^{\prime }&{}\preccurlyeq \xi P\preccurlyeq \frac {\alpha }{\lambda _{\min }(D{D}^{\top })}\left (\frac {f^2(K_0)}{\lambda ^2_{\min }({C}^{\top }_2C_2)}+\|B\|^2\|C_1\|^2\right )\frac {f(K_0)}{\lambda _{\min }({C}^{\top }_2C_2)}I \\ &{}\preccurlyeq 2f(K_0)\frac {\|A\|+\sqrt {f(K_0)}\|B\|\|C_1\|}{\lambda _{\min }(D{D}^{\top })\lambda _{\min }({C}^{\top }_2C_2)}\left (\frac {f^2(K_0)}{\lambda ^2_{\min }({C}^{\top }_2C_2)}+\|B\|^2\|C_1\|^2\right )I, \end {aligned} $$

and hence

$$ \|P^{\prime }\|_F\leqslant 2\sqrt {n} f(K_0) \frac {\|A\| + \sqrt {f(K_0)}\|B\|\|C_1\|} {\lambda _{\min }(D{D}^{\top })\lambda _{\min }({C}^{\top }_2C_2) } \left ( \frac {f^2(K_0)}{\lambda ^2_{\min }({C}^{\top }_2C_2) } + \|B\|^2\|C_1\|^2 \right ) .$$

(A.7)

Considering estimates (A.6) and (A.7), we obtain the quantity (23). The proof of Lemma 6 is complete. $\quad \blacksquare$

Proof of Theorem 3. First of all, Algorithm 1 is well defined at the starting point, since $K_0$ is a stabilizing controller by assumption. Further, for sufficiently small $\gamma _j$, $f(K)$ is monotone decreasing in the algorithm (motion along the antigradient); i.e., the $K_j$ remain in the domain $\mathcal S_0$, and thus we can apply the results in Lemma 6 about the Lipschitz property of the gradient.

Thus, the results on the convergence of the gradient method for unconditional minimization in [18] are applicable. In particular, condition (b) at step 3 in Algorithm 1 will be satisfied after a finite number of partitions, and convergence along the gradient will occur linearly in the gradient method. The proof of Theorem 3 is complete. $\quad \blacksquare$