# Implicit function

A function $f : E \rightarrow Y$ given by an equation $F ( x , y ) = z _ {0}$, where $F: X \times Y \rightarrow Z$, $x \in X$, $y \in Y$, $E \subset X$, and $X$, $Y$ and $Z$ are certain sets, i.e. a function $f$ such that $F ( x , f ( x) ) = z _ {0}$ for any $x \in E$. If $X$, $Y$ and $Z$ are topological spaces and if $F ( x _ {0} , y _ {0} ) = z _ {0}$ for some point $( x _ {0} , y _ {0} ) \in X \times Y$, then under certain conditions the equation $F ( x , y ) = z _ {0}$ is uniquely solvable in one of the variables in some neighbourhood of $( x _ {0} , y _ {0} )$. Properties of the solution of this equation are described by implicit-function theorems.

The simplest implicit-function theorem is as follows. Suppose that $X$ and $Y$ are subsets of the real line $\mathbf R$, let $x _ {0} \in X$, $y _ {0} \in Y$, and let $( x _ {0} , y _ {0} )$ be an interior point of the plane set $X \times Y$; if $F$ is continuous in some neighbourhood of $( x _ {0} , y _ {0} )$, if $F ( x _ {0} , y _ {0} ) = 0$ and if there are a $\delta > 0$ and an $\epsilon > 0$ such that $F ( x , y )$, for any fixed $x \in ( x _ {0} - \delta , x _ {0} + \delta )$, is strictly monotone on $( y _ {0} - \epsilon , y _ {0} + \epsilon )$ as a function of $y$, then there is a $\delta _ {0} > 0$ such that there is a unique function

$$f : ( x _ {0} - \delta _ {0} , x _ {0} + \delta _ {0} ) \rightarrow \ ( y _ {0} - \epsilon , y _ {0} + \epsilon )$$

for which $F ( x , f ( x) ) = 0$ for all $x \in ( x _ {0} - \delta _ {0} , x _ {0} + \delta _ {0} )$; moreover, $f$ is continuous and $f ( x _ {0} ) = y _ {0}$. Figure: i050310a

The hypotheses of this theorem are satisfied if $F$ is continuous in a neighbourhood of $( x _ {0} , y _ {0} )$, if the partial derivative $F _ {y}$ exists and is continuous at $( x _ {0} , y _ {0} )$, if $F ( x _ {0} , y _ {0} ) = 0$, and if $F _ {y} ( x _ {0} , y _ {0} ) \neq 0$. If in addition the partial derivative $F _ {x}$ exists and is continuous at $( x _ {0} , y _ {0} )$, then the implicit function $f$ is differentiable at $x _ {0}$, and

$$\frac{d f ( x _ {0} ) }{dx} = - \frac{F _ {x} ( x _ {0} , y _ {0} ) }{F _ {y} ( x _ {0} , y _ {0} ) } .$$

This theorem has been generalized to the case of a system of equations, that is, when $F$ is a vector function. Let $\mathbf R ^ {n}$ and $\mathbf R ^ {m}$ be $n$- and $m$- dimensional Euclidean spaces with fixed coordinate systems and points $x = ( x _ {1} \dots x _ {n} )$ and $y = ( y _ {1} \dots y _ {m} )$, respectively. Suppose that $F$ maps a certain neighbourhood $W$ of $( x _ {0} , y _ {0} ) \in \mathbf R ^ {n} \times \mathbf R ^ {m}$( $x _ {0} \in \mathbf R ^ {n}$, $y _ {0} \in \mathbf R ^ {m}$) into $\mathbf R ^ {m}$ and that $F _ {i}$, $i = 1 \dots m$, are the coordinate functions (of the $n + m$ variables $x _ {1} \dots x _ {n} , y _ {1} \dots y _ {m}$) of $F$, that is, $F = ( F _ {1} \dots F _ {m} )$. If $F$ is differentiable on $W$, if $F( x _ {0} , y _ {0} ) = 0$ and if the Jacobian

$$\left . \frac{\partial ( F _ {1} \dots F _ {m} ) }{\partial ( y _ {1} \dots y _ {m} ) } \right | _ {( x _ {0} , y _ {0} ) } \ \neq 0 ,$$

then there are neighbourhoods $U$ and $V$ of $x _ {0} \in \mathbf R ^ {n}$ and $y _ {0} \in \mathbf R ^ {m}$, respectively, $U \times V \subset W$, and a unique mapping $f : U \rightarrow V$ such that $F ( x , f ( x) ) = 0 \in \mathbf R ^ {m}$ for all $x \in U$. Here $f ( x _ {0} ) = y _ {0}$, $f$ is differentiable on $U$, and if $f = ( f _ {1} \dots f _ {m} )$, then the explicit expression for the partial derivatives $\partial f _ {j} / \partial x _ {i}$, $i = 1 \dots n$, $j = 1 \dots m$, can be found from the system of $m$ linear equations in these derivatives:

$$\frac{\partial F _ {k} }{\partial x _ {i} } + \sum _ { j= } 1 ^ { m } \frac{\partial F _ {k} }{\partial y _ {j} } \frac{\partial f _ {j} }{\partial x _ {i} } = 0 ,$$

$k = 1 \dots m$, $i$ is fixed $( i = 1 \dots n )$. Sometimes the main assertion of the theorem is stated as follows: There are neighbourhoods $U$ of $x _ {0}$ in $\mathbf R ^ {n}$ and $W _ {0}$ of $( x _ {0} , y _ {0} )$ in $\mathbf R ^ {n} \times \mathbf R ^ {m}$, $W _ {0} \subset W$, and a unique mapping $f : U \rightarrow \mathbf R ^ {m}$ such that $( x , f ( x) ) \in W _ {0}$ and $F ( x , f ( x) ) = 0$ for all $x \in U$. In other words, the conditions

$$( x , y ) \in W _ {0} ,\ \ F ( x , y ) = 0$$

are equivalent to $x \in U$, $y = f ( x)$. In this case one says that the equation $F ( x , y ) = 0$ is uniquely solvable in the neighbourhood $W _ {0}$ of $( x _ {0} , y _ {0} )$.

The classical implicit-function theorem thus stated generalizes to the case of more general spaces in the following manner. Let $X$ be a topological space, let $Y$ and $Z$ be affine normed spaces over the field of real or complex numbers, that is, affine spaces over the relevant field to which are associated normed vector spaces $\mathbf Y$ and $\mathbf Z$, $\mathbf Y$ being complete, let ${\mathcal L} ( \mathbf Y , \mathbf Z )$ be the set of continuous linear mappings from $\mathbf Y$ into $\mathbf Z$, and let $W$ be an open set in the product space $X \times Y$, $( x _ {0} , y _ {0} ) \in W$, $x _ {0} \in X$, $y _ {0} \in Y$.

Let $F : W \rightarrow Z$ be a continuous mapping and $F ( x _ {0} , y _ {0} ) = z _ {0}$. If for every fixed $x$ and $( x , y ) \in W$ the mapping $F$ has a partial Fréchet derivative $F _ {y} \in {\mathcal L} ( \mathbf Y , \mathbf Z )$, if $F _ {y} ( x , y ) : W \rightarrow {\mathcal L} ( \mathbf Y , \mathbf Z )$ is a continuous mapping and if the linear mapping $F _ {y} ( x _ {0} , y _ {0} ) : \mathbf Y \rightarrow \mathbf Z$ has a continuous inverse linear mapping (that is, it is an invertible element of ${\mathcal L} ( \mathbf Y , \mathbf Z )$), then there exist open sets $U \subset X$ and $V \subset Y$, $x _ {0} \in U$, $y _ {0} \in V$, such that for any $x \in U$ there is a unique element $y \in V$, denoted by $y = f ( x)$, satisfying the equations

$$f ( x) \in V \ \ \textrm{ and } \ F ( x , f ( x) ) = z _ {0} .$$

The function $y = f ( x)$ thus defined is a continuous mapping from $U$ into $V$, and $y _ {0} = f ( x _ {0} )$.

If $X$ is also an affine normed space, then under certain conditions the implicit function $f : x \mapsto y$ which satisfies the equation

$$\tag{1 } F ( x , y ) = z _ {0}$$

is also differentiable. Namely, let $X$, $Y$ and $Z$ be affine normed spaces, let $W$ be an open set in $X \times Y$, let $F : W \rightarrow Z$, $F ( x _ {0} , y _ {0} ) = z _ {0}$, $x _ {0} \in Y$, and let $f$ be the implicit mapping given by (1), taking a certain neighbourhood $U$ of $x _ {0}$ into an open subset $V$ of $Y$, $U \times V \subset W$. Thus, for all $x \in U$,

$$\tag{2 } f ( x) \in V ,\ \ F ( x , f ( x) ) = z _ {0} .$$

Suppose also that $f$ is continuous at $x _ {0}$ and that $f ( x _ {0} ) = y _ {0}$. If $F$ is differentiable at $( x _ {0} , y _ {0} )$, if its partial Fréchet derivatives $F _ {x} ( x _ {0} , y _ {0} )$ and $F _ {y} ( x _ {0} , y _ {0} )$ are continuous linear operators taking the vector spaces $\mathbf X$ and $\mathbf Y$ associated with $X$ and $Y$ into the vector space $\mathbf Z$ associated with $Z$, and if the operator $F _ {y} ( x _ {0} , y _ {0} )$ is an invertible element of ${\mathcal L} ( \mathbf Y , \mathbf Z )$, then $f$ is differentiable at $x _ {0}$ and its Fréchet derivative is given by

$$f ^ { \prime } ( x _ {0} ) = \ - F _ {y} ^ { - 1 } ( x _ {0} , y _ {0} ) \circ F _ {x} ( x _ {0} , y _ {0} ) .$$

This is obtained as a result of formally differentiating (2):

$$F _ {x} ( x _ {0} , y _ {0} ) + F _ {y} ( x _ {0} , y _ {0} ) \circ f ^ { \prime } ( x _ {0} ) = \ 0 \in {\mathcal L} ( \mathbf X , \mathbf Y )$$

and multiplying this equality on the left by $F _ {y} ^ { - 1 } ( x _ {0} , y _ {0} )$.

If in addition the mapping $F : W \rightarrow Z$ is continuously differentiable on $W$, if the implicit function $f : U \rightarrow V$ is continuous on $U$, $U \times X \subset W$, and if for any $x \in U$ the partial Fréchet derivative $F _ {y} ( x , f ( x) )$ is an invertible element of ${\mathcal L} ( \mathbf Y , \mathbf Z )$, then $f$ is a continuously-differentiable mapping of $U$ into $V$.

In the general case one can also indicate conditions for the existence and the uniqueness of the implicit function in terms of the continuity of the Fréchet derivative: If $Z$ is complete, if the mapping $F : W \rightarrow Z$ is continuously differentiable on $W$, if $F ( x _ {0} , y _ {0} ) = z _ {0}$, and if the partial Fréchet derivative $F _ {y} ( x _ {0} , y _ {0} )$ is an invertible element of ${\mathcal L} ( \mathbf Y , \mathbf Z )$, then (1) is uniquely solvable in a sufficiently small neighbourhood of $( x _ {0} , y _ {0} )$, i.e. there exist neighbourhoods $U$ of $x _ {0}$ in $X$ and $V$ of $y _ {0}$ in $Y$, $U \times V \subset W$, and a unique implicit function $f : U \rightarrow V$ satisfying (2). Here $f$ is also continuously differentiable on $U$. In this form the implicit-function theorem for normed spaces is a direct generalization of the corresponding classic implicit-function theorem for a single scalar equation in two variables.

Furthermore, if $F : W \rightarrow Z$ is a $k$- times continuously-differentiable mapping in a neighbourhood $W$ of $( x _ {0} , y _ {0} )$, $k = 1 , 2 \dots$ then the implicit function $f : U \rightarrow V$ is also $k$ times continuously differentiable.

Far-reaching generalizations of the classic implicit-function theorem to differential operators were given by J. Nash (see Nash theorems (in differential geometry)).

How to Cite This Entry:
Implicit function. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Implicit_function&oldid=47320
This article was adapted from an original article by L.D. Kudryavtsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article