Wegstein's Method for Calculating the Global Extremum

The essence of the problem of finding the global extremum is to find a solution with the minimum (maximum) value of the objective function. Currently, a lot is known and written about methods and problems of global optimization [1-5]. Nevertheless, due to their practical importance, global optimization problems continuously arise and are being investigated in many fields of science [6-13]. Thus, there is a constant need for the development of new methods and approaches to solve them, since there is no universal algorithm for solving them.

To date, all known methods of global optimization can be divided into two categories [1, 4, 5], deterministic methods [1, 2] and stochastic methods [14, 15]. And optimization tasks are usually divided into two types: static and dynamic. The static type includes tasks in which it is necessary to determine the values of the arguments that make up the extreme value of the objective function. A dynamic type is considered a task when a function called a control function is to be defined, in which the target functional takes a maximum or minimum value.

When solving static problems with a nonlinear objective function, great difficulties arise. The main difficulties are associated with multiextremality, large dimension and non-convexity of the objective function. To date, a considerable number of works devoted to overcoming the above difficulties have been published [12, 16-18].

In this paper, a new method for finding the global minimum based on the idea of auxiliary functions is proposed. At the same time, the object of research is not the objective function itself, but an auxiliary function of one variable constructed by converting the original goal function using the Lebesgue integral. It should be noted that the auxiliary function studied in this paper was considered in Ref. [19]. In it, with its help, an algorithm was proposed for dividing a segment in half to calculate the global extremum for smooth and multiextremal functions of several variables defined on convex compact sets. In this article, in contrast to the work [19], the main properties of this auxiliary function for any continuous goal function are studied in detail. An optimality criterion is established, and Wegstein's method for finding the global minimum of continuous on bounded and closed sets of functions of many variables is proposed, which is more effective for this situation.

Lemma 1. The auxiliary function (2) has the following properties:

1. $g_{m}(F, \alpha) \geq 0$;
2. $g_{m}(\mathrm{C}, \alpha)=0$ for any constant with $\mathrm{C} \geq \alpha$;
3. $g_{m}(k F, k \alpha)=k^{m} g_{m}(F, \alpha)$, for a real number $k \geq$ $0, \alpha \geq \bar{\alpha}$;
4. $g_{m}(k+F, k+\alpha)=g(F, \alpha)$, for any real $k$ and all $\alpha \geq \bar{\alpha}$;
5. $g_{m}(\alpha+F, \alpha)=g_{m}(F, 0)$, for all $\alpha \geq \bar{\alpha}$;
6. $g_{m}(\alpha F, \alpha)=\alpha^{m} g_{m}(F, 1)$, for each $\alpha \geq 0$;
7. $g_{m}(F, \alpha)$ is uniformly continuous on the interval $\alpha \geq \alpha_{0}$, where $\forall \alpha_{0}>\bar{\alpha}$;
8. $g_{m}(F, \alpha)$ is midpoint convex on the interval $\alpha \geq$ $\alpha_{0}$, where $\forall \alpha_{0}>\bar{\alpha}$.

Proof.

1) $g_{m}(F, \alpha)=\int_{E \backslash E(F, \alpha)}[|F(x)-\alpha|-F(x)+\alpha]^{m} d \mu+$$\int_{\mathrm{E}(F, \alpha)}[|F(x)-\alpha|-F(x)+\alpha]^{m} d \mu=$ $=\int_{E \backslash E(F, \alpha)}[(F(x)-\alpha)-F(x)+\alpha]^{m} d \mu$$+\int_{\mathrm{E}(F, \alpha)}[-(F(x)-\alpha)-F(x)+\alpha]^{m} d \mu=$ $=\int_{\mathrm{E}(F, \alpha)}[2(\alpha-F(x))]^{m} d \mu \geq 0$,

Since the expression under the last integral is non-negative on the set Е(F,α).

2) $g_{m}(\mathrm{C}, \alpha)=\int_{E}[|\mathrm{C}-\alpha|-\mathrm{C}+\alpha]^{m} d \mu=\int_{E}[(\mathrm{C}-\alpha)-\mathrm{C}+$ $\alpha]^{m} d \mu=0$.

3) $g_{m}(k F, k \alpha)=\int_{E}[|k F(x)-k \alpha|-k F(x)+k \alpha]^{m} d \mu=$$k^{m} \int_{E}[|F(x)-\alpha|-F(x)+\alpha]^{m} d \mu=k^{m} g_{m}(F, \alpha)$.

4) $g_{m}(k+F, k+\alpha)=\int_{E}[|k+F(x)-k-\alpha|-k-$ $F(x)+k+\alpha]^{m} d \mu=\int_{E}[|F(x)-\alpha|--F(x)+\alpha]^{m} d \mu=$ $g_{m}(F, \alpha) .$

5) $g_{m}(\alpha+F, \alpha)=\int_{E}[|\alpha+F(x)-\alpha|-\alpha-F(x)+\alpha]^{m} d \mu=$ $=\int_{E}[|F(x)-0|-F(x)+0]^{m} d \mu=g_{m}(F, 0)$.

6) $g_{m}(\alpha F, \alpha)=\int_{E}[|\alpha F(x)-\alpha|-\alpha F(x)+k \alpha]^{m} d \mu=$ $\alpha^{m} \int_{E}[|F(x)-1|-F(x)+1]^{m} d \mu==\alpha^{m} g_{m}(F, 1) .$

7) Let's choose arbitrary $\alpha_{1}, \alpha_{2}$ such that $\alpha_{1}>\alpha_{2} \geq \alpha_{0}$. Let 's estimate the difference modulus:

$\left|g_{m}\left(F, \alpha_{1}\right)-g_{m}\left(F, \alpha_{2}\right)\right|==\left|\int_{E}\left[\left|F(x)-\alpha_{1}\right|-F(x)+\alpha_{1}\right]^{m} d \mu-\int_{E}\left[\left|F(x)-\alpha_{2}\right|-F(x)+\alpha_{2}\right]^{m} d \mu\right|=$$=\left|\int_{E\left(F, \alpha_{1}\right)}\left[2\left(\alpha_{1}-F(x)\right)\right]^{m} d \mu-\int_{E\left(F, \alpha_{2}\right)}\left[2\left(\alpha_{2}-F(x)\right)\right]^{m} d \mu\right|=$$=\int_{\mathrm{E}\left(F, \alpha_{2}\right)}\left[2\left(\alpha_{1}-F(x)\right)\right]^{m} d \mu+\int_{\mathrm{E}\left(F, \alpha_{1}\right) \backslash \mathrm{E}\left(F, \alpha_{2}\right)}\left[2\left(\alpha_{1}-F(x)\right)\right]^{m} d \mu-\int_{\mathrm{E}\left(F, \alpha_{2}\right)}\left[2\left(\alpha_{2}-F(x)\right)\right]^{m} d \mu \mid \leq$$\leq \int_{\mathrm{E}\left(F, \alpha_{2}\right)}\left[2\left(\alpha_{1}-\alpha_{2}\right)\left(\left(2\left(\alpha_{1}-F(x)\right)\right)^{m-1}+\left(2\left(\alpha_{1}-F(x)\right)\right)^{m-2}\left(2\left(\alpha_{2}-F(x)\right)\right)+\cdots+\right.\right.$$\left.\left.\left(\left(2\left(\alpha_{1}-F(x)\right)\right)\right)\left(2\left(\alpha_{2}-F(x)\right)\right)^{m-2}+\left(2\left(\alpha_{2}-F(x)\right)\right)^{m-1}\right)\right] d \mu\left|++\int_{E\left(F, \alpha_{1}\right) \backslash E\left(F, \alpha_{2}\right)}\left[2\left(\alpha_{1}-F(x)\right)\right]^{m} d \mu\right|=J_{1}+J_{2}$.

Note that the following inequalities are met. On the set $\mathrm{E}\left(F, \alpha_{2}\right):\left(2\left(\alpha_{1}-F(x)\right)\right)^{m-1} \leq\left(2\left(\alpha_{1}-\alpha_{0}\right)\right)^{m-1}=K$, $\left(2\left(\alpha_{2}-F(x)\right)\right)^{m-1} \leq\left(2\left(\alpha_{1}-\alpha_{0}\right)\right)^{m-1}=K$.

And on the set $\mathrm{E}\left(F, \alpha_{1}\right) \backslash \mathrm{E}\left(F, \alpha_{2}\right)$ :

$\left(2\left(\alpha_{1}-F(x)\right)\right)^{m}=\left(2\left(\alpha_{1}-F(x)\right)\right)^{m-1} 2\left(\alpha_{1}-F(x)\right) \leq 2 K\left|\alpha_{1}-\alpha_{2}\right|$.

Using them it is easy to get estimates:

$J_{1} \leq 2 m K\left|\alpha_{1}-\alpha_{2}\right| \mu\left(\mathrm{E}\left(F, \alpha_{2}\right)\right) \leq 2 m K\left|\alpha_{1}-\alpha_{2}\right| \mu(\mathrm{E})=B_{1}\left|\alpha_{1}-\alpha_{2}\right|$где $B_{1}=2 m K \mu(\mathrm{E})$;$J_{2} \leq 2 K\left|\alpha_{1}-\alpha_{2}\right|\left(\mu\left(\mathrm{E}\left(F, \alpha_{1}\right)\right)-\mu\left(\mathrm{E}\left(F, \alpha_{2}\right)\right)\right)=$$2 K\left|\alpha_{1}-\alpha_{2}\right| \mu\left(\mathrm{E}\left(F, \alpha_{1}\right)\right) \leq 2 K \mu(\mathrm{E}) \leq B_{2}\left|\alpha_{1}-\alpha_{2}\right|$, где $B_{2}=2 K \mu(\mathrm{E})$.

Thus, for any $\varepsilon>0$, it is sufficient to take the number $\delta=$ $\varepsilon /\left(B_{1}+B_{2}\right)$, that for arbitrary $\alpha_{1}, \alpha_{2}$ for which $\left|\alpha_{1}-\alpha_{2}\right|<$ $\delta$ there is an inequality:

$\left|g_{m}\left(F, \alpha_{1}\right)-g_{m}\left(F, \alpha_{2}\right)\right| \leq J_{1}+J_{2}<\left(B_{1}+B_{2}\right)\left|\alpha_{1}-\alpha_{2}\right|$$<\frac{\left(B_{1}+B_{2}\right) \varepsilon}{\left(B_{1}+B_{2}\right)}=\varepsilon$

This proves that (2) is uniformly continuous.

8) Let's choose arbitrary $\alpha_{1}, \alpha_{2}$ such that $\alpha_{1}>\alpha_{2} \geq \alpha_{0}$. Using the Cauchy inequality $\left(u_{1}+u_{2}\right)^{m} \leq 2^{m-1}\left(u_{1}^{m}+u_{2}^{m}\right)$, for any $u_{1}, u_{2} \geq 0$ and $m \in \boldsymbol{N}$. Let's show the fairness of inequality:

$g_{m}\left(F, \frac{\alpha_{1}+\alpha_{2}}{2}\right) \leq \frac{1}{2}\left(g_{m}\left(F, \alpha_{1}\right)+g_{m}\left(F, \alpha_{2}\right)\right)$.

$g_{m}\left(F, \frac{\alpha_{1}+\alpha_{2}}{2}\right)=\int_{E}\left[\left|F(x)-\frac{\alpha_{1}+\alpha_{2}}{2}\right|-F(x)+\frac{\alpha_{1}+\alpha_{2}}{2}\right]^{m} d \mu=$$=\frac{1}{2^{m}} \int_{E}\left[\left|2 F(x)-\left(\alpha_{1}+\alpha_{2}\right)\right|-2 F(x)+\left(\alpha_{1}+\alpha_{2}\right)\right]^{m} d \mu \leq$$\frac{1}{2^{m}} \int_{E}\left[\left|F(x)-\alpha_{1}+F(x)-\alpha_{2}\right|-\left(F(x)-\alpha_{1}\right)-\left(F(x)-\alpha_{2}\right)\right]^{m} d \mu \leq$$\leq \frac{1}{2^{m}} \int_{E}\left[\left(\left|F(x)-\alpha_{1}\right|-F(x)+\alpha_{1}\right)+\left(\left|F(x)-\alpha_{2}\right|-F(x)+\alpha_{2}\right)\right]^{m} d \mu=I$.

According to the statement 1) of Lemma 1, two expressions in parentheses under the last integral have non-negative values, so applying the above Cauchy inequality we get the following estimate.

$I \leq \frac{1}{2^{m}} \int_{E} 2^{m-1}\left[\left(\left|F(x)-\alpha_{1}\right|-F(x)+\alpha_{1}\right)^{m}+\left(\left|F(x)-\alpha_{2}\right|-F(x)+\alpha_{2}\right)^{m}\right] d \mu=$$=\frac{1}{2}\left(g_{m}\left(F, \alpha_{1}\right)+g_{m}\left(F, \alpha_{2}\right)\right) .$.

The lemma is proved.

Lemma 2. The auxiliary function (2) is differentiable at each point $\alpha>\bar{\alpha}$, where $\bar{\alpha}=\min _{E} F(x)$.

Proof. Let $\alpha+h>\alpha>\bar{\alpha}$, consider the increment of the auxiliary function.

$g_{m}(F, \alpha+h)-g_{m}(F, \alpha)=$$=\int_{E}[|F(x)-(\alpha+h)|-F(x)+(\alpha+h)]^{m} d \mu-\int_{E}[|F(x)-\alpha|-F(x)+\alpha]^{m} d \mu=$$=\int_{\mathrm{E}(F, \alpha+\mathrm{h})}[2(\alpha+h-F(x))]^{m} d \mu-\int_{\mathrm{E}(F, \alpha)}[2(\alpha-F(x))]^{m} d \mu=$$=\int_{\mathrm{E}(F, \alpha)}[2(\alpha+h-F(x))]^{m} d \mu+\int_{\mathrm{E}(F, \alpha+h) \backslash \mathrm{E}(F, \alpha)}[2(\alpha+h-F(x))]^{m} d \mu-\int_{\mathrm{E}(F, \alpha)}[2(\alpha-F(x))]^{m} d \mu=$$=\int_{\mathrm{E}(F, \alpha)}\left[2 h\left((\alpha+h-F(x))^{m-1}+(\alpha+h-F(x))^{m-2}(\alpha-F(x))+\ldots\right.\right.$$+(\alpha-F(x))^{m-2}(\alpha+h-F(x))$$\left.\left.+(\alpha-F(x))^{m-1}\right)\right] d \mu+$$+\int_{\mathrm{E}(F, \alpha+\mathrm{h}) \backslash \mathrm{E}(F, \alpha)}[2(\alpha+h-F(x))]^{m} d \mu=I_{1} h+I_{2}$.

It is easy to check that

$\begin{aligned} \lim _{h \rightarrow 0} I_{1} &=2 \int_{\mathrm{E}(F, \alpha)} m[2(\alpha-F(x))]^{m-1} d \mu \\ &=2 m g_{m-1}((F, \alpha)) . \end{aligned}$

It is not difficult to see that $I_{2}=\int_{\mathrm{E}(F, \alpha+\mathrm{h}) \backslash \mathrm{E}(F, \alpha)}[2(\alpha+h-$ $F(x))]^{m} d \mu$ is infinitesimal for $\mathrm{h} \rightarrow 0$, since the expression under the integral in square brackets does not exceed $2 h$, and due to the continuity of the measure $\mu$ takes place:

$0 \leq I_{2} \leq(2 h)^{m} \mu(\mathrm{E}(F, \alpha+\mathrm{h})-\mu(\mathrm{E}(F, \alpha))) \rightarrow 0$

Thus,

$\frac{d g_{m}}{d \alpha}=2 m g_{m-1}((F, \alpha))$.

Lemma 2 is proved.

Corollary 1. When m=1

$\frac{d g_{1}}{d \alpha}=2 \mu(E(F, \alpha))$.

Corollary 2. For any given natural number m, the equality holds,

$\frac{d^{m} g_{m}}{d \alpha^{m}}=(2)^{m} m ! \mu(E(F, \alpha))$.

Lemma 3. Let $\left\{\alpha_{i}\right\}$ be a decreasing sequence whose limit for $i \rightarrow \infty$ is $\alpha_{0}$ and $g_{m}\left(F, \alpha_{0}\right)>0$. Then $\alpha_{0}>$ $\bar{\alpha}=\min _{E} F(x)$.

Proof. Suppose the opposite, let $\alpha_{0} \leq \bar{\alpha}$. Since $g_{m}\left(F, \alpha_{0}\right)>0$, the set $E\left(F, \alpha_{0}\right)=\left\{x \in \mathrm{E} \mid F(x)<\alpha_{0}\right\}$ is not empty. Therefore, the inequality $F(x)<\bar{\alpha}$ holds on this set $E\left(F, \alpha_{0}\right)$. And this contradicts the fact that $\bar{\alpha}$ is the value of the global minimum. Hence, $\alpha_{0}>\bar{\alpha}$.

Lemma 3 is proved.

Despite the fact that the Wegstein’s method [20] converges slower than some other methods, it requires almost twice as many calculations of the values of the auxiliary function (2). And calculating the values of the auxiliary function leads to the numerical calculation of multiple integrals with a large number of variables, which is associated with great difficulties. Therefore, Wegstein's method turns out to be more effective for this situation. Besides, it always converges. Next, we will give a brief description of this method. According to (3) we will solve the equation $g_{m}(F, \alpha)=0$ with respect to $\alpha$. Below we will need:

$\Psi(\alpha)=\alpha-A g_{m}(F, \alpha),$

$A=\frac{1}{2^{m+p} m B}, B=\int_{E}|F(x)|^{m-1} d \mu.$

In accordance with the study [20], we can choose the coefficient A at our discretion. Therefore, here we choose the smallest positive integer p in such a way that the following inequalities are satisfied:

$0<1-2 m A g_{m-1}(F, \tau)<1$.         (6)

The iterative procedure, according to Wegstein [20], consists of the following two formulas:

$\alpha_{n+1}=\Psi\left(\tilde{\alpha}_{n}\right)$;              (7)

$\tilde{\alpha}_{n+1}=\alpha_{n+1}-\frac{\left[\alpha_{n+1}-\alpha_{n}\right]\left[\alpha_{n+1}-\tilde{\alpha}_{n}\right]}{\alpha_{n+1}-\alpha_{n}-\tilde{\alpha}_{n}+\tilde{\alpha}_{n-1}}$        (8)

After choosing a suitable initial approximation $x_{0} \in E, \varepsilon>$ 0 and accepting $\alpha_{0}=F\left(x_{0}\right), \alpha_{1}=\Psi\left(\alpha_{0}\right), \tilde{\alpha}_{0}:=\alpha_{0}, \tilde{\alpha}_{1}:=$ $\alpha_{1}$, we apply formulas (7) and (8). Then we check the condition $\frac{\left|\widetilde{\alpha}_{n}-\widetilde{\alpha}_{n+1}\right|}{\left|\widetilde{\alpha}_{n}\right|} \geq \varepsilon$. If it is executed, then we repeat the process; otherwise, the iteration ends and we take the root equal to $\tilde{\alpha}_{n+1}$.

To determine the coordinates of the global minimum points, we apply the method of statistical tests [21]. Suppose that $l$ Wegstein iterations have been performed, that is, $\tilde{\alpha}_{n+1}=\tilde{\alpha}_{l}$.

Following [21], we randomly select $s$ different $n$ dimensional points $x_{j}=\left(x_{j 1}, \ldots, x_{j n}\right)$ from the set E. Calculate the values of the objective function at each of these points, denote $F_{j}=F\left(x_{j}\right), j=1, \ldots, s$. Next, define the sets $W_{k}=\left\{x_{j} \in \mathrm{E}: F\left(x_{j}\right) \leq \tilde{\alpha}_{k}, k=0,1, \ldots, l\right\}$, denote $\xi_{k}$ the number of elements in $W_{k}$. For each $W_{k}$, we define the values of $\mathrm{n}$ functions:

$\bar{x}_{i}\left(\tilde{\alpha}_{k}\right)=\frac{\sum_{j=1}^{\xi_{k}} x_{j i}\left[2\left(\tilde{\alpha}_{k}-F_{j}\right)\right]^{m}}{\sum_{j=1}^{\xi_{k}}\left[2\left(\tilde{\alpha}_{k}-F_{j}\right)\right]^{m}}, i=1, \ldots, n$            (9)

For each of the functions (9) we construct an approximating parabola:

$x_{i}(\tilde{\alpha})=a_{i} \tilde{\alpha}^{2}+b_{i} \tilde{\alpha}+c_{i}, i=1, \ldots, n .$           (10)

The coefficients of the quadratic trinomial (10) can be calculated by solving the following systems of linear equations.

$\left\{\begin{array}{c}M_{4} a_{i}+M_{3} b_{i}+M_{2} c_{i}=M_{2,1}, \\ M_{3} a_{i}+M_{2} b_{i}+M_{1} c_{i}=M_{1,1}, i=1, \ldots, n. \\ M_{2} a_{i}+M_{1} b_{i}+M_{0} c_{i}=M_{0,1};\end{array}\right.$

Here the coefficients and free terms of the system are determined by the following formulas [21].

$M_{0}=1, M_{1}=\frac{1}{l} \sum_{k=1}^{l} \tilde{\alpha}_{k}, M_{2}=\frac{1}{l} \sum_{k=1}^{l} \tilde{\alpha}_{k}^{2}, M_{3}=\frac{1}{l} \sum_{k=1}^{l} \tilde{\alpha}_{k}^{3}$,$M_{4}=\frac{1}{l} \sum_{k=1}^{l} \tilde{\alpha}_{k}^{4}$

$M_{0,1}=\frac{1}{l} \sum_{k=1}^{l} \bar{x}_{i}\left(\tilde{\alpha}_{k}\right), M_{1,1}=\frac{1}{l} \sum_{k=1}^{l} \tilde{\alpha}_{k} \cdot \bar{x}_{i}\left(\tilde{\alpha}_{k}\right), M_{2,1}=$ $\frac{1}{l} \sum_{k=1}^{l} \tilde{\alpha}_{k}^{2} \cdot \bar{x}_{i}\left(\tilde{\alpha}_{k}\right) .$

Substituting the value $\tilde{\alpha}_{l}$ instead of $\tilde{\alpha}$ in (10), we obtain approximate coordinates of the global minimum $x^{*}=$ $\left(x_{1}\left(\tilde{\alpha}_{l}\right), \ldots, x_{n}\left(\tilde{\alpha}_{l}\right)\right)$. It should be noted that if the objective function is multiextremal, then the set $W_{k}$ at the $k$ - th iteration can be decomposed into the union of several sets $W_{k}=$ $\bigcup_{\eta=1}^{r_{k}} W_{k}^{(\eta)}$. After that, the above procedure for determining the coordinates of the global minimum is performed for each partial set $W_{k}^{(\eta)}$.

Next, we formulate and prove a theorem on the convergence rate of this algorithm.

Theorem 2. Let the conditions of Theorem 1 be fulfilled. Then the iterative sequence $\left\{\tilde{\alpha}_{n}\right\}$  constructed according to the Wegstein’s algorithm (8) converges to the global minimum of the objective function $F(x)$ with linear convergence rate, that is, there are $\beta \in(0,1)$ and for $N \in \mathbb{N}$ and such that $\forall n>N$  is fulfilled $\left|\tilde{\alpha}_{n+1}-\bar{\alpha}\right|<\beta\left|\tilde{\alpha}_{n}-\bar{\alpha}\right|$.

Proof of Theorem 2. The convergence of the iterative sequence $\left\{\widetilde{\alpha}_{n}\right\}$ to the global minimum $\bar{\alpha}$ follows from Lemma 3 and Theorem 1. Next, we show that for all $n \geq 3$ there is $\alpha_{n+1} \leq \tilde{\alpha}_{n}$. It follows from Wegstein's algorithm that $\alpha_{n+1}=$ $\Psi\left(\tilde{\alpha}_{n}\right)$. It is equivalent to the equality $\alpha_{n+1}-\tilde{\alpha}_{n}=$ $-A \int_{E}\left[\left|F(x)-\tilde{\alpha}_{n}\right|-F(x)+\tilde{\alpha}_{n}\right]^{m} d \mu$. The right part of the last equality is non-positive by virtue of statement 1 of Lemma 1 , therefore $\alpha_{n+1}-\tilde{\alpha}_{n} \leq 0$. Thus, for all $n \geq 3$ it is fulfilled:

$\alpha_{n+1} \leq \tilde{\alpha}_{n}$,       (11)

Since Wegstein's algorithm is applied starting from $n=3$. Now we note that based on the basic idea of this method [20], it follows that for any $n \geq 3: \tilde{\alpha}_{n}$ always lies in the interval between $\alpha_{n+1}$ and $\alpha_{n}$. Therefore, taking into account (11), the inequalities $\alpha_{n+1} \leq \tilde{\alpha}_{n} \leq \alpha_{n}$ are satisfied for all $n \geq 3$. Therefore,

$\alpha_{n+2} \leq \tilde{\alpha}_{n+1} \leq \alpha_{n+1}$.            (12)

Further, by (12) we get the following estimate of the absolute value of the difference:

$\left|\tilde{\alpha}_{n+1}-\bar{\alpha}\right| \leq\left|\tilde{\alpha}_{n+1}-\bar{\alpha}\right|=\left|\Psi\left(\tilde{\alpha}_{n}\right)-\Psi(\bar{\alpha})\right|$

$=\left|\left(\tilde{\alpha}_{n}-\bar{\alpha}\right) \Psi^{\prime}(\tau)\right|$       (13)

where, $\tau$ lies between $\tilde{\alpha}_{n}$ and $\bar{\alpha}$. Note that from the convergence of the sequence $\left\{\widetilde{\alpha}_{n}\right\}$ it follows that there will be $N \in \mathbb{N}$ such that $\forall \mathrm{n}>N$ there will be an estimate $\mid \tau-$ $F(x)|\leq| F(x) \mid$. Now we estimate the derivative $\Psi^{\prime}(\tau)$ on the set $E$, taking into account inequalities (6).

$0 \leq\left|\Psi^{\prime}(\tau)\right|=\left|\left(\alpha-A g_{m}(F, \alpha)\right)_{\alpha=\tau}^{\prime}\right|$$=\left|1-2 m A g_{m-1}((F, \tau))\right|=$$=\left|1-\frac{2 m \int_{E}[|F(x)-\tau|-F(x)+\tau]^{m-1} d \mu}{2^{m+p} m \int_{E}|F(x)|^{m-1} d \mu.}\right|$$=\left|1-\frac{2 m \int_{\mathrm{E}(F, \tau)}[2(\tau-F(x))]^{m-1} d \mu}{2^{m+p} m \int_{E}|F(x)|^{m-1} d \mu .}\right| \leq$$\leq\left|1-\frac{2^{m} m \int_{\mathrm{E}(F, \tau)}|F(x)|^{m-1} d \mu}{2^{m+p} m \int_{E}|F(x)|^{m-1} d \mu}\right|=\beta<1$.

Thus, based on (13), we obtain an estimate:

$\left|\tilde{\alpha}_{n+1}-\bar{\alpha}\right| \leq\left|\Psi^{\prime}(\tau)\right|\left|\tilde{\alpha}_{n}-\bar{\alpha}\right|<\beta\left|\tilde{\alpha}_{n}-\bar{\alpha}\right| .$

Theorem 2 is proved.

From the proof of this theorem, we can notice the following fact. Given that for any $\tau$ inequality $\int_{\mathrm{E}(F, \tau)}|F(x)|^{m-1} d \mu \leq$ $\int_{E}|F(x)|^{m-1} d \mu$, we get the following estimate of the rate of convergence:

$\beta=\left|1-\frac{2^{m} m \int_{E(F, \tau)}|F(x)|^{m-1} d \mu}{2^{m+p} m \int_{E}|F(x)|^{m-1} d \mu}\right| \leq\left|1-\frac{1}{2^{p}}\right| .$