UDC 517.948+513.881

MATHEMATICS

S. I. POKHOZHAEV

ON THE NORMAL SOLVABILITY OF NONLINEAR EQUATIONS

(Presented by Academician I. M. Vinogradov on 21 XII 1967)

Let \(L\) be a linear bounded operator acting from a Banach space\(^*\) \(X\) into a Banach space \(Y\), and let \(L^*: X^* \to Y^*\) be the operator adjoint to \(L\).

Consider the equation

\[ Lx = y . \tag{1} \]

If, for some \(y \in Y\), equation (1) is solvable, then, as is known,

\[ y \in (\operatorname{Ker} L^*)^\perp \equiv \{y \in Y \mid \langle y^*, y\rangle = 0 \text{ for every } y^* \in \operatorname{Ker} L^*\}. \]

Here \(\langle y^*, y\rangle\) is the value of the functional \(y^* \in Y^*\) on the element \(y \in Y\), and

\[ \operatorname{Ker} L^* \equiv \{y^* \in Y^* \mid L^* y^* = 0\}. \]

The linear equation (1) is called normally solvable in the sense of Hausdorff (see (\(^{1}\))) if this necessary condition for solvability is also sufficient.

Let now \(A(x)\) be a continuous and continuously differentiable (nonlinear) operator acting from \(X\) into \(Y\). Consider the equation

\[ A(x) = y . \tag{2} \]

If equation (2) is solvable for some \(y \in Y\), i.e., there exists \(x_0 \in X\) such that \(A(x_0)=y\), then, obviously,

\[ y - A(x_0) \in \bigl(\operatorname{Ker} (A'(x_0))^*\bigr)^\perp . \]

Here \((A'(x_0))^*\) is the operator adjoint to the linear bounded operator \(A'(x_0)\), the derivative of the operator \(A(x)\) at \(x=x_0\).

Definition. Equation (2) is called normally solvable if: 1) the operator \(A(x)\) is of class \(C^1\) and is such that for every \(y \in Y\) there exists an \(x_0\), a point of minimum\(^ {**}\) of the functional \(\|A(x)-y\|\); 2) from the condition

\[ y - A(x_0) \in \bigl(\operatorname{Ker} (A'(x_0))^*\bigr)^\perp \]

it follows that \(y=A(x_0)\).

Remark. In the case of a reflexive Banach space \(Y\) and a linear bounded operator \(A: X \to Y\), this definition is equivalent to Hausdorff’s definition.

In this note a theorem is proved giving a sufficient condition for the normal solvability of equation (2). As a consequence, a theorem on the solvability of the nonlinear equation (2) is obtained. A class of operators \(A(x)\) is described for which the “orthogonality” condition takes the form

\[ y \in \bigl(\operatorname{Ker} (A'(x_0))^*\bigr)^\perp . \]

An extension of the operator \(A\) is considered, based on the introduction of additional “potentials.”

Theorem 1. Let \(Y\) be a reflexive Banach space and let \(A\) be an operator of class \(C^1\) acting from the Banach space \(X\) into \(Y\). Let \(A(X)\), the range of the operator \(A\), be weakly closed\(^ {***}\) in \(Y\).

Then the nonlinear equation (2) is normally solvable.

The proof is based on the following lemmas.

\(^*\) Here and below all Banach spaces are considered over the field of real or complex numbers.

\(^ {**}\) Everywhere in the text, by a minimum is meant an absolute (i.e., relative to all of \(X\)) minimum.

\(^ {***}\) Everywhere in the note, following S. Mazur, a weakly closed set is understood to mean a set that contains all weak limits of sequences of elements of this set.

Lemma 1. Under the conditions of Theorem 1, for any \(y \in Y\) the functional \(f(x) \equiv \|A(x)-y\|\) attains its lower bound.

Let \(x_0\) be a point of minimum of the functional \(f(x)\) for an arbitrarily fixed \(y \in Y\). Introduce the notation: \(A_0 = A'(x_0)\), \(RA_0\) is the range of the linear operator \(A_0\), and \(\overline{RA_0}\) is the closure of the set \(RA_0\), \(v_0 = A(x_0)-y\).

Lemma 2. Let \(v_0 \in \overline{RA_0}\).
Then \(v_0 = 0\).

Proof. Suppose that \(v_0 \ne 0\). Since \(v_0 \in \overline{RA_0}\), for \(\delta=\|v_0\|/2>0\) there is an element \(v_1 \in Y\) such that \(\|v_1-v_0\|<\delta\), and the equation \(A_0u=v_1\) has a solution. Take some solution \(u_0\) of this equation (if the solution of this equation is not unique), \(A_0u_0=v_1\), and fix it. Since \(\|v_1\|>0\), we have \(\|u_0\|>0\). Putting \(x-x_0=\lambda u_0\), we obtain \(A_0(x-x_0)=\lambda v_1\) (\(\lambda\) is a real number).

We now consider the differentiability condition of the operator \(A(x)\):
\[ A(x)=A(x_0)+A_0(x-x_0)+\omega(x_0,x-x_0), \]
where
\[ \|\omega(x_0,x-x_0)\|/\|x-x_0\|\to 0 \quad \text{as } \|x-x_0\|\to 0. \]

Since \(f(x)\equiv \|A(x)-y\|\ge \|A(x_0)-y\|\), for any \(x\in X\) we have
\[ \|v_0\|\le \|v_0+A_0(x-x_0)+\omega(x_0,x-x_0)\|. \]

In particular, for \(x-x_0=\lambda u_0\) we obtain
\[ \|v_0\|\le \|v_0+\lambda v_1+\omega(x_0,\lambda u_0)\|= \]
\[ =\|v_0+\lambda v_0+\lambda(v_1-v_0)+\omega(x_0,\lambda u_0)\|\le \]
\[ \le |1+\lambda|\cdot \|v_0\|+|\lambda|\cdot \|v_1-v_0\|+\|\omega(x_0,\lambda u_0)\|. \]

Since \(\|v_1-v_0\|<\|v_0\|/2\) and
\[ \|\omega(x_0,\lambda u_0)\|/|\lambda|\cdot \|u_0\|\to 0 \quad \text{as } \lambda\to 0, \]
we have
\[ \|v_0\|< |1+\lambda|\cdot \|v_0\|+\tfrac12|\lambda|\cdot \|v_0\|+\varepsilon|\lambda|\cdot \|u_0\|, \]
where \(\varepsilon\to 0\) as \(\lambda\to 0\).

The elements \(v_0\) and \(u_0\) do not depend on \(\lambda\), and by assumption \(\|v_0\|>0\). Put \(\|u_0\|/\|v_0\|=C_0\). Then we have
\[ \|v_0\|< |1+\lambda|\cdot \|v_0\|+\tfrac12|\lambda|\cdot \|v_0\|+\varepsilon C_0|\lambda|\cdot \|v_0\| \]
or \((\|v_0\|>0)\)
\[ 1< |1+\lambda|+\tfrac12|\lambda|+\varepsilon C_0\cdot |\lambda|, \quad \text{where } \varepsilon\to 0 \text{ as } \lambda\to 0. \]

The inequality obtained is contradictory for negative sufficiently small \(\lambda\). Lemma 2 is proved.

Proof of Theorem 1. Let \(y\) be an element of \(Y\). By Lemma 1 there exists in \(X\) an element \(x_0\), a point of minimum of the functional
\[ f(x)\equiv \|A(x)-y\|. \]
Let now
\[ y-A(x_0)=-v_0 \in (\operatorname{Ker} A_0^*)^\perp. \]
By a known lemma from the theory of bounded linear operators (see (2), p. 516), we have
\[ (\operatorname{Ker} A_0^*)^\perp=\overline{RA_0}. \]
Then by Lemma 2 we obtain \(y=A(x_0)\).

Theorem 1 can also be formulated as follows:

Theorem \(1'\). Let the conditions of Theorem 1 be fulfilled.
Then, for solvability of equation (2), it is necessary and sufficient that
\[ y-A(x_0)\in (\operatorname{Ker}(A'(x_0))^*)^\perp, \]
where \(x_0\) is a point of (absolute) minimum of the functional \(f(x)\equiv \|A(x)-y\|\).

From Theorem 1 there follows directly a theorem on the solvability of equation (2).

Theorem 2. Let \(Y\) be a reflexive Banach space and let \(A\), an operator of class \(C^1\), act from a Banach space \(X\) into \(Y\). Let \(A(X)\), the range of the operator \(A\), be weakly closed in \(Y\).

Then, if $\operatorname{Ker}(A'(x))^*=\{0\}$ for every $x$ in $X$, equation (2) is solvable for every $y$ in $Y$.

Remark. This theorem (in the case of a reflexive Banach space $Y$) generalizes the well-known theorem on the solvability of a linear equation (1). In the case of a linear operator $L$, the weakly closed range of the operator $L$ coincides with the closed (strong) range of the operator $L$.

The “orthogonality” condition $y-A(x)\in(\operatorname{Ker}(A'(x))^*)^\perp$ has a form different from the corresponding condition for the case of a linear operator.

Let us define a class of operators $A(x)$ for which the “orthogonality” condition takes the form $y\in(\operatorname{Ker}(A'(x))^*)^\perp$, i.e., for which $A(x)\in(\operatorname{Ker}(A'(x))^*)^\perp$.

Theorem 3. In order that $A(x)\in(\operatorname{Ker}(A'(x))^*)^\perp$, it is sufficient and, if the linear operator $A'(x)$ (for fixed $x$) has a closed range, necessary that there exist a mapping $\Phi(x)$ from $X$ into $X$ such that the operator $A(x)$ of class $C^1$ has the form

\[ A(x)=A'(x)\cdot \Phi(x). \tag{3} \]

Remark. The last relation for an operator $A(x)$ of class $C^1$ is equivalent to the following:

\[ A(x)=\left.\frac{d}{dt}A(x+t\Phi(x))\right|_{t=0}, \tag{4} \]

where $t$ is a real number.

Proof. Sufficiency. We have

\[ \langle y^*, A(x)\rangle=\langle y^*, A'(x)\cdot \Phi(x)\rangle =\langle (A'(x))^*y^*,\ \Phi(x)\rangle . \]

Therefore, if $y^*\in\operatorname{Ker}(A'(x))^*$ ($x$ fixed), we obtain
$\langle y^*,A(x)\rangle=0$, i.e. $A(x)\in(\operatorname{Ker}(A'(x))^*)^\perp$.

Necessity. Let $A(x)\in(\operatorname{Ker}(A'(x))^*)^\perp$. Since the range of the linear operator $A'(x)$ ($x$ fixed) is closed and the element $A(x)$ belongs to $(\operatorname{Ker}(A'(x))^*)^\perp$, by Banach’s theorem (see ($^3$), p. 284) the equation $A'(x)\cdot h=A(x)$ is solvable with respect to $h$. Denoting the solution $h$, depending on $x$, by $\Phi(x)$, we obtain the second assertion of the theorem.

Remark. It is clear from the proof that the mapping $\Phi(x)$ is, generally speaking, not uniquely determined.

Let us note that a positively homogeneous operator $A(x)$, $A(kx)=k^aA(x)$ ($k>0$, $a>0$), of class $C^1$ satisfies the condition of Theorem 3. Indeed, it is enough in relation (4) to put $\Phi(x)=x/a$.

From Theorems 1 and 3 we obtain the following theorem.

Theorem 4. Suppose that the conditions of Theorem 1 are fulfilled and the operator $A(x)$ admits the representation (3).

Then, for the solvability of equation (2) it is necessary and sufficient that
$y\in(\operatorname{Ker}(A'(x_0))^*)^\perp$, where $x_0$ is a point of (absolute) minimum of the functional $f(x)=\|A(x)-y\|$.

In those cases when the range of the operator $A$ does not coincide with the space $Y$, the operator $A$ can be extended by additionally introducing a Banach space $Z$—a space of “potentials.” In the theory of boundary-value problems such an extension corresponds to the introduction of additional potentials (see ($^4$)).

Let $A$ be an operator of class $C^1$ acting from a Banach space $X$ into a Banach space $Y$, and let $B$ be an operator of class $C^1$ acting from a Banach space $Z$ into $Y$, with $B(0)=0$. Consider the Banach space $X\times Z$, for example, with norm $\|(x,z)\|=\|x\|+\|z\|$, and define the extension $\mathfrak{A}:X\times Z\to Y$ of the operator $A$ by the formula $\mathfrak{A}(x,z)=\mathfrak{A}(x,0)+\mathfrak{A}(0,z)$, where $\mathfrak{A}(x,0)=A(x)$, $\mathfrak{A}(0,z)=B(z)$. For the derivative $\mathfrak{A}'(x,z)$ we have
$\mathfrak{A}'(x,z)\cdot(u,w)=A'(x)\cdot u+B'(z)\cdot w$. The linear operator

\((\mathfrak A'(x,z))^*\) (for fixed \(x,z\)) has kernel \(\operatorname{Ker}(\mathfrak A'(x,z))^*\), determined by the relation

\[ \operatorname{Ker}(\mathfrak A'(x,z))^*=\operatorname{Ker}(A'(x))^*\cap \operatorname{Ker}(B'(z))^* . \]

Applying Theorem 2, we obtain

Theorem 5. Let \(Y\) be a reflexive Banach space. Let
\(\mathfrak A:X\times Z\to Y\) be an operator of class \(C^1\), defined by the relation \(\mathfrak A(x,z)=A(x)+B(z)\), and suppose that it has in \(Y\) a weakly closed range.

Then, if \(\operatorname{Ker}(A'(x))^*\cap \operatorname{Ker}(B'(z))^*=\{0\}\) for all \((x,z)\) in \(X\times Z\), the equation \(A(x)+B(z)=y\) is solvable for every \(y\) in \(Y\).

A sufficient condition for the weak closedness of the range of the operator \(A\) is contained in the following lemma.

Lemma 3. Let \(X\) be a reflexive Banach space. Let the operator \(A:X\to Y\) be defined on all of \(X\) and be weakly closed (i.e., from \(x_n\to x_0\) weakly in \(X\) and \(A(x_n)\to y_0\) weakly in \(Y\) as \(n\to\infty\) it follows that \(y_0=A(x_0)\)), and such that for every bounded set \(G\subseteq A(X)\) there exists a bounded set \(\Omega\subset X\) such that \(A(\Omega)=G\).

Then \(A(X)\), the range of the operator \(A\), is weakly closed in \(Y\).

In conclusion the author expresses gratitude to Prof. M. I. Vishik for discussion of the work.

Moscow
Power Engineering Institute

Received
18 XII 1967

REFERENCES

¹ F. Hausdorff, Set Theory, Supplement, Moscow, 1937.
² N. Dunford, J. T. Schwartz, Linear Operators. General Theory, Moscow, 1962.
³ B. Iosida, Functional Analysis, Moscow, 1967.
⁴ M. I. Vishik, G. I. Eskin, UMN, 22, no. 1 (133), 15 (1967).