6.7: Inverse and Implicit Functions. Open and Closed Maps
( \newcommand{\kernel}{\mathrm{null}\,}\)
I. "If f∈CD1 at →p, then f resembles a linear map (namely df) at →p." Pursuing this basic idea, we first make precise our notion of "f∈CD1 at →p."
A map f:E′→E is continuously differentiable, or of class CD1 (written f∈CD1), at →p iff the following statement is true:
Given any ε>0, there is δ>0 such that f is differentiable on the globe ¯G=¯G→p(δ), with ‖df(→x;⋅)−df(→p;⋅)‖<ε for all →x∈¯G.
By Problem 10 in §5, this definition agrees with Definition 1 §5, but is no longer limited to the case E′=En(Cn). See also Problems 1 and 2 below.
We now obtain the following result.
Let E′ and E be complete. If f:E′→E is of class CD1 at →p and if df(→p;⋅) is bijective (§6), then f is one-to-one on some globe ¯G=¯G→p(δ).
Thus f "locally" resembles df (→p;⋅) in this respect.
- Proof
-
Set ϕ=df(→p;⋅) and
‖ϕ−1‖=1ε
(cf. Theorem 2 of §6).
By Definition 1, fix δ>0 so that for →x∈¯G=¯G→p(δ).
‖df(→x;⋅)−ϕ‖<12ε.
Then by Note 5 in §2,
(∀→x∈¯G)(∀→u∈E′)|df(→x;→u)−ϕ(→u)|≤12ε|→u|.
Now fix any →r,→s∈¯G,→r≠→s, and set →u=→r−→s≠0. Again, by Note 5 in §2,
|→u|=|ϕ−1(ϕ(→u))|≤‖ϕ−1‖|ϕ(→u)|=1ε|ϕ(→u)|;
so
0<ε|→u|≤|ϕ(→u)|.
By convexity, ¯G⊇I=L[→s,→r], so (1) holds for →x∈I,→x=→s+t→u,0≤t≤1.
Noting this, set
h(t)=f(→s+t→u)−tϕ(→u),t∈E1.
Then for 0≤t≤1,
h′(t)=D→uf(→s+t→u)−ϕ(→u)=df(→s+t→u;→u)−ϕ(→u).
(Verify!) Thus by (1) and (2),
sup0≤t≤1|h′(t)|=sup0≤t≤1|df(→s+t→u;→u)−ϕ(→u)|≤ε2|→u|≤12|ϕ(→u)|.
(Explain!) Now, by Corollary 1 in Chapter 5, §4,
|h(1)−h(0)|≤(1−0)⋅sup0≤t≤1|h′(t)|≤12|ϕ(→u)|.
As h(0)=f(→s) and
h(1)=f(→s+→u)−ϕ(→u)=f(→r)−ϕ(→u),
we obtain (even if →r=→s)
|f(→r)−f(→s)−ϕ(→u)|≤12|ϕ(→u)|(→r,→s∈¯G,→u=→r−→s).
But by the triangle law,
|ϕ(→u)|−|f(→r)−f(→s)|≤|f(→r)−f(→s)−ϕ(→u)|.
Thus
|f(→r)−f(→s)|≥12|ϕ(→u)|≥12ε|→u|=12ε|→r−→s|
by (2).
Hence f(→r)≠f(→s) whenever →r≠→s in ¯G; so f is one-to-one on ¯G, as claimed.◻
Under the assumptions of Theorem 1, the maps f and f−1 (the inverse of f restricted to ¯G) are uniformly continuous on ¯G and f[¯G], respectively.
- Proof
-
By (3),
|f(→r)−f(→s)|≤|ϕ(→u)|+12|ϕ(→u)|≤|2ϕ(→u)|≤2‖ϕ‖|→u|=2‖ϕ‖|→r−→s|(→r,→s∈¯G).
This implies uniform continuity for f. (Why?)
Next, let g=f−1 on H=f[¯G].
If →x,→y∈H, let →r=g(→x) and →s=g(→y); so →r,→s∈¯G, with →x=f(→r) and →y=f(→s). Hence by (4),
|→x−→y|≥12ε|g(→x)−g(→y)|,
proving all for g, too.◻
Again, f resembles ϕ which is uniformly continuous, along with ϕ−1.
II. We introduce the following definition.
A map f:(S,ρ)→(T,ρ′) is closed (open) on D⊆S iff, for any X⊆D the set f[X] is closed (open) in T whenever X is so in S.
Note that continuous maps have such a property for inverse images (Problem 15 in Chapter 4, §2).
Under the assumptions of Theorem 1, f is closed on ¯G, and so the set f[¯G] is closed in E.
Similarly for the map f−1 on f[¯G].
- Proof for E′=E=En(Cn) (for the general case, see Problem 6)
-
Given any closed X⊆¯G, we must show that f[X] is closed in E.
Now, as ¯G is closed and bounded, it is compact (Theorem 4 of Chapter 4, §6).
So also is X (Theorem 1 in Chapter 4, §6), and so is f[X] (Theorem 1 of Chapter 4, §8).
By Theorem 2 in Chapter 4, §6, f[X] is closed, as required.◻
For the rest of this section, we shall set E′=E=En(Cn).
If E′=E=En(Cn) in Theorem 1, with other assumptions unchanged, then f is open on the globe G=G→p(δ), with δ sufficiently small.
- Proof
-
We first prove the following lemma.
f[G] contains a globe G→q(α) where →q=f(→p).
- Proof
-
Indeed, let
α=14εδ,
where δ and ε are as in the proof of Theorem 1. (We continue the notation and formulas of that proof.)
Fix any →c∈G→q(α); so
|→c−→q|<α=14εδ.
Set h=|f−→c| on E′. As f is uniformly continuous on ¯G, so is h.
Now, ¯G is compact in En(Cn); so Theorem 2(ii) in Chapter 4, §8, yields a point →r∈¯G such that
h(→r)=minh[¯G].
We claim that →r is in G (the interior of ¯G).
Otherwise, |→r−→p|=δ; for by (4),
2α=12εδ=12ε|→r−→p|≤|f(→r)−f(→p)|≤|f(→r)−→c|+|→c−f(→p)|=h(→r)+h(→p).
But
h(→p)=|→c−f(→p)|=|→c−→q|<α;
and so (7) yields
h(→p)<α<h(→r),
contrary to the minimality of h(→r) (see (6)). Thus |→r−→p| cannot equal δ.
We obtain |→r−→p|<δ, so →r∈G→p(δ)=G and f(→r)∈f[G]. We shall now show that →c=f(→r).
To this end, we set →v=→c−f(→r) and prove that →v=→0. Let
→u=ϕ−1(→v),
where
ϕ=df(→p;⋅),
as before. Then
→v=ϕ(→u)=df(→p;→u).
With →r as above, fix some
→s=→r+t→u(0<t<1)
with t so small that →s∈G also. Then by formula (3),
|f(→s)−f(→r)−ϕ(t→u)|≤12|t→v|;
also,
|f(→r)−→c+ϕ(t→u)|=(1−t)|→v|=(1−t)h(→r)
by our choice of →v,→u and h. Hence by the triangle law,
h(→s)=|f(→s)−→c|≤(1−12t)h(→r).
(Verify!)
As 0<t<1, this implies h(→r)=0 (otherwise, h(→s)<h(→r), violating (6)).
Thus, indeed,
|→v|=|f(→r)−→c|=0,
i.e.,
→c=f(→r)∈f[G] for →r∈G.
But →c was an arbitrary point of G→q(α). Hence
G→q(α)⊆f[G],
proving the lemma.◻
Proof of Theorem 2. The lemma shows that f(→p) is in the interior of f[G] if →p,f,df(→p;⋅), and δ are as in Theorem 1.
But Definition 1 implies that here f∈CD1 on all of G (see Problem 1).
Also, df(→x;⋅) is bijective for any →x∈G by our choice of G and Theorems 1 and 2 in §6.
Thus f maps all →x∈G onto interior points of f[G]; i.e., f maps any open set X⊆G onto an open f[X], as required.◻
Note 1. A map
f:(S,ρ)⟷ onto (T,ρ′)
is both open and closed ("clopen") iff f−1 is continuous - see Problem 15(iv)(v) in Chapter 4, §2, interchanging f and f−1.
Thus ϕ=df(→p;⋅) in Theorem 1 is "clopen" on all of E′.
Again, f locally resembles df(→p;⋅).
III. The Inverse Function Theorem. We now further pursue these ideas.
Under the assumptions of Theorem 2, let g be the inverse of fG(f restricted to G=G→p(δ)).
Then g∈CD1 on f[G] and dg(→y;⋅) is the inverse of df(→x;⋅) whenever →x=g(→y),→x∈G.
Briefly: "The differential of the inverse is the inverse of the differential."
- Proof
-
Fix any →y∈f[G] and →x=g(→y); so →y=f(→x) and →x∈G. Let U=df(→x;⋅).
As noted above, U is bijective for every →x∈G by Theorems 1 and 2 in §6; so we may set V=U−1. We must show that V=dg(→y;⋅).
To do this, give →y an arbitrary (variable) increment Δ→y, so small that →y+Δ→y stays in f[G] (an open set by Theorem 2).
As g and fG are one-to-one, Δ→y uniquely determines
Δ→x=g(→y+Δ→y)−g(→y)=→t,
and vice versa:
Δ→y=f(→x+→t)−f(→x).
Here Δ→y and →t are the mutually corresponding increments of →y=f(→x) and →x=g(→y). By continuity, →y→→0 iff \vec{t} \rightarrow \overrightarrow{0}.
As U=d f(\vec{x} ; \cdot),
\lim _{\vec{t} \rightarrow \overline{0}} \frac{1}{|\vec{t}|}|f(\vec{x}+\vec{t})-f(\vec{t})-U(\vec{t})|=0,
or
\lim _{\vec{t} \rightarrow \overrightarrow{0}} \frac{1}{|\vec{t}|}|F(\vec{t})|=0,
where
F(\vec{t})=f(\vec{x}+\vec{t})-f(\vec{t})-U(\vec{t}).
As V=U^{-1}, we have
V(U(\vec{t}))=\vec{t}=g(\vec{y}+\Delta \vec{y})-g(\vec{y}).
So from (9),
\begin{aligned} V(F(\vec{t})) &=V(\Delta \vec{y})-\vec{t} \\ &=V(\Delta \vec{y})-[g(\vec{y}+\Delta \vec{y})-g(\vec{y})]; \end{aligned}
that is,
\frac{1}{|\Delta \vec{y}|}|g(\vec{y}+\Delta \vec{y})-g(\vec{y})-V(\Delta \vec{y})|=\frac{|V(F(\vec{t}))|}{|\Delta \vec{y}|}, \quad \Delta \vec{y} \neq \overrightarrow{0}.
Now, formula (4), with \vec{r}=\vec{x}, \vec{s}=\vec{x}+\vec{t}, and \vec{u}=\vec{t}, shows that
|f(\vec{x}+\vec{t})-f(\vec{x})| \geq \frac{1}{2} \varepsilon|\vec{t}|;
i.e., |\Delta \vec{y}| \geq \frac{1}{2} \varepsilon|\vec{t}|. Hence by (8),
\frac{|V(F(\vec{t}))|}{|\Delta \vec{y}|} \leq \frac{|V(F(\vec{t}) |}{\frac{1}{2} \varepsilon|\vec{t}|}=\frac{2}{\varepsilon}\left|V\left(\frac{1}{|\vec{t}|} F(\vec{t})\right)\right| \leq \frac{2}{\varepsilon}\|V\| \frac{1}{|\vec{t}|}|F(\vec{t})| \rightarrow 0 \text { as } \vec{t} \rightarrow \overrightarrow{0}.
Since \vec{t} \rightarrow \overrightarrow{0} as \Delta \vec{y} \rightarrow \overrightarrow{0} (change of variables!), the expression (10) tends to 0 as \Delta \vec{y} \rightarrow \overrightarrow{0}.
By definition, then, g is differentiable at \vec{y}, with d g(\vec{y};)=V=U^{-1}.
Moreover, Corollary 3 in §6, applies here. Thus
\left(\forall \delta^{\prime}>0\right)\left(\exists \delta^{\prime \prime}>0\right) \quad\|U-W\|<\delta^{\prime \prime} \Rightarrow\left\|U^{-1}-W^{-1}\right\|<\delta^{\prime}.
Taking here U^{-1}=d g(\vec{y}) and W^{-1}=d g(\vec{y}+\Delta \vec{y}), we see that g \in C D^{1} near \vec{y}. This completes the proof.\quad \square
Note 2. If E^{\prime}=E=E^{n}\left(C^{n}\right), the bijectivity of \phi=d f(\vec{p} ; \cdot) is equivalent to
\operatorname{det}[\phi]=\operatorname{det}\left[f^{\prime}(\vec{p})\right] \neq 0
(Theorem 1 of §6).
In this case, the fact that f is one-to-one on G=G_{\vec{p}}(\delta) means, componentwise (see Note 3 in §6), that the system of n equations
f_{i}(\vec{x})=f\left(x_{1}, \ldots, x_{n}\right)=y_{i}, \quad i=1, \ldots, n,
has a unique solution for the n unknowns x_{k} as long as
\left(y_{1}, \ldots, y_{n}\right)=\vec{y} \in f[G].
Theorem 3 shows that this solution has the form
x_{k}=g_{k}(\vec{y}), \quad k=1, \ldots, n,
where the g_{k} are of class C D^{1} on f[G] provided the f_{i} are of class C D^{1} near \vec{p} and det \left[f^{\prime}(\vec{p})\right] \neq 0. Here
\operatorname{det}\left[f^{\prime}(\vec{p})\right]=J_{f}(\vec{p}),
as in §6.
Thus again f "locally" resembles a linear map, \phi=d f(\vec{p} ; \cdot).
IV. The Implicit Function Theorem. Generalizing, we now ask, what about solving n equations in n+m unknowns x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{m}? Say, we want to solve
f_{k}\left(x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{m}\right)=0, \quad k=1,2, \ldots, n,
for the first n unknowns (or variables) x_{k}, thus expressing them as
x_{k}=H_{k}\left(y_{1}, \ldots, y_{m}\right), \quad k=1, \ldots, n,
with H_{k} : E^{m} \rightarrow E^{1} or H_{k} : C^{m} \rightarrow C.
Let us set \vec{x}=\left(x_{1}, \ldots, x_{n}\right), \vec{y}=\left(y_{1}, \ldots, y_{m}\right), and
(\vec{x}, \vec{y})=\left(x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{m}\right)
so that (\vec{x}, \vec{y}) \in E^{n+m}\left(C^{n+m}\right).
Thus the system of equations (11) simplifies to
f_{k}(\vec{x}, \vec{y})=0, \quad k=1, \ldots, n
or
f(\vec{x}, \vec{y})=\overrightarrow{0},
where f=\left(f_{1}, \ldots, f_{n}\right) is a map of E^{n+m}\left(C^{n+m}\right) into E^{n}\left(C^{n}\right) ; f is a function of n+m variables, but it has n components f_{k}; i.e.,
f(\vec{x}, \vec{y})=f\left(x_{1}, \ldots, x_{n}, y_{1}, \ldots, y_{m}\right)
is a vector in E^{n}\left(C^{n}\right).
Let E^{\prime}=E^{n+m}\left(C^{n+m}\right), E=E^{n}\left(C^{n}\right), and let f : E^{\prime} \rightarrow E be of class C D^{1} near
(\vec{p}, \vec{q})=\left(p_{1}, \ldots, p_{n}, q_{1}, \ldots, q_{m}\right), \quad \vec{p} \in E^{n}\left(C^{n}\right), \vec{q} \in E^{m}\left(C^{m}\right).
Let [\phi] be the n \times n matrix
\left(D_{j} f_{k}(\vec{p}, \vec{q})\right), \quad j, k=1, \ldots, n.
If \operatorname{det}[\phi] \neq 0 and if f(\vec{p}, \vec{q})=\overrightarrow{0}, then there are open sets
P \subseteq E^{n}\left(C^{n}\right) \text { and } Q \subseteq E^{m}\left(C^{m}\right),
with \vec{p} \in P and \vec{q} \in Q, for which there is a unique map
H : Q \rightarrow P
with
f(H(\vec{y}), \vec{y})=\overrightarrow{0}
for all \vec{y} \in Q; furthermore, H \in C D^{1} on Q.
Thus \vec{x}=H(\vec{y}) is a solution of (11) in vector form.
- Proof
-
With the above notation, set
F(\vec{x}, \vec{y})=(f(\vec{x}, \vec{y}), \vec{y}), \quad F : E^{\prime} \rightarrow E^{\prime}.
Then
F(\vec{p}, \vec{q})=(f(\vec{p}, \vec{q}), \vec{q})=(\overrightarrow{0}, \vec{q}),
since f(\vec{p}, \vec{q})=\overrightarrow{0}.
As f \in C D^{1} near (\vec{p}, \vec{q}), so is F (verify componentwise via Problem 9(ii) in §3 and Definition 1 of §5).
By Theorem 4, §3, \operatorname{det}\left[F^{\prime}(\vec{p}, \vec{q})\right]=\operatorname{det}[\phi] \neq 0 (explain!).
Thus Theorem 1 above shows that F is one-to-one on some globe G about (\vec{p}, \vec{q}).
Clearly G contains an open interval about (\vec{p}, \vec{q}). We denote it by P \times Q where \vec{p} \in P, \vec{q} \in Q ; P is open in E^{n}\left(C^{n}\right) and Q is open in E^{m}\left(C^{m}\right).
By Theorem 3, F_{P \times Q} (F restricted to P \times Q) has an inverse
g : A \underset{\text { onto }}{\longleftrightarrow} P \times Q,
where A=F[P \times Q] is open in E^{\prime} (Theorem 2), and g \in C D^{1} on A. Let the map u=\left(g_{1}, \ldots, g_{n}\right) comprise the first n components of g (exactly as f comprises the first n components of F ).
Then
g(\vec{x}, \vec{y})=(u(\vec{x}, \vec{y}), \vec{y})
exactly as F(\vec{x}, \vec{y})=(f(\vec{x}, \vec{y}), \vec{y}). Also, u : A \rightarrow P is of class C D^{1} on A, as g is (explain!).
Now set
H(\vec{y})=u(\overrightarrow{0}, \vec{y});
here \vec{y} \in Q, while
(\overrightarrow{0}, \vec{y}) \in A=F[P \times Q],
for F preserves \vec{y} (the last m coordinates). Also set
\alpha(\vec{x}, \vec{y})=\vec{x}.
Then f=\alpha \circ F (why?), and
f(H(\vec{y}), \vec{y})=f(u(\overrightarrow{0}, \vec{y}), \vec{y})=f(g(\overrightarrow{0}, \vec{y}))=\alpha(F(g(\overrightarrow{0}, \vec{y}))=\alpha(\overrightarrow{0}, \vec{y})=\overrightarrow{0}
by our choice of \alpha and g (inverse to F). Thus
f(H(\vec{y}), \vec{y})=\overrightarrow{0}, \quad \vec{y} \in Q,
as desired.
Moreover, as H(\vec{y})=u(\overrightarrow{0}, \vec{y}), we have
\frac{\partial}{\partial y_{i}} H(\vec{y})=\frac{\partial}{\partial y_{i}} u(\overrightarrow{0}, \vec{y}), \quad \vec{y} \in Q, i \leq m.
As u \in C D^{1}, all \partial u / \partial y_{i} are continuous (Definition 1 in §5); hence so are the \partial H / \partial y_{i}. Thus by Theorem 3 in §3, H \in C D^{1} on Q.
Finally, H is unique for the given P, Q; for
\begin{aligned} f(\vec{x}, \vec{y})=\overrightarrow{0} & \Longrightarrow(f(\vec{x}, \vec{y}), \vec{y})=(\overrightarrow{0}, \vec{y}) \\ & \Longrightarrow F(\vec{x}, \vec{y})=(\overrightarrow{0}, \vec{y}) \\ & \Longrightarrow g(F(\vec{x}, \vec{y}))=g(\overrightarrow{0}, \vec{y}) \\ & \Longrightarrow(\vec{x}, \vec{y})=g(\overrightarrow{0}, \vec{y})=(u(\overrightarrow{0}, \vec{y}), \vec{y}) \\ & \Longrightarrow \vec{x}=u(\overrightarrow{0}, \vec{y})=H(\vec{y}). \end{aligned}
Thus f(\vec{x}, \vec{y})=\overrightarrow{0} implies \vec{x}=H(\vec{y}); so H(\vec{y}) is the only solution for \vec{x}. \quad \square
Note 3. H is said to be implicitly defined by the equation f(\vec{x}, \vec{y})=\overrightarrow{0}. In this sense we say that H(\vec{y}) is an implicit function, given by f(\vec{x}, \vec{y})=\overrightarrow{0}.
Similarly, under suitable assumptions, f(\vec{x}, \vec{y})=\overrightarrow{0} defines \vec{y} as a function of \vec{x}.
Note 4. While H is unique for a given neighborhood P \times Q of (\vec{p}, \vec{q}), another implicit function may result if P \times Q or (\vec{p}, \vec{q}) is changed.
For example, let
f(x, y)=x^{2}+y^{2}-25
(a polynomial; hence f \in C D^{1} on all of E^{2}). Geometrically, x^{2}+y^{2}-25=0 describes a circle.
Solving for x, we get x=\pm \sqrt{25-y^{2}}. Thus we have two functions:
H_{1}(y)=+\sqrt{25-y^{2}}
and
H_{2}(y)=-\sqrt{25-y^{2}}.
If P \times Q is in the upper part of the circle, the resulting function is H_{1}. Otherwise, it is H_{2}. See Figure 28.
V. Implicit Differentiation. Theorem 4 only states the existence (and uniqueness) of a solution, but does not show how to find it, in general.
The knowledge itself that H \in C D^{1} exists, however, enables us to use its derivative or partials and compute it by implicit differentiation, known from calculus.
(a) Let f(x, y)=x^{2}+y^{2}-25=0, as above.
This time treating y as an implicit function of x, y=H(x), and writing y^{\prime} for H^{\prime}(x), we differentiate both sides of (x^{2}+y^{2}-25=0\) with respect to x, using the chain rule for the term y^{2}=[H(x)]^{2}.
This yields 2 x+2 y y^{\prime}=0, whence y^{\prime}=-x / y.
Actually (see Note 4), two functions are involved: y=\pm \sqrt{25-x^{2}}; but both satisfy x^{2}+y^{2}-25=0; so the result y^{\prime}=-x / y applies to both.
Of course, this method is possible only if the derivative y^{\prime} is known to exist. This is why Theorem 4 is important.
(b) Let
f(x, y, z)=x^{2}+y^{2}+z^{2}-1=0, \quad x, y, z \in E^{1}.
Again f satisfies Theorem 4 for suitable x, y, and z.
Setting z=H(x, y), differentiate the equation f(x, y, z)=0 partially with respect to x and y. From the resulting two equations, obtain \frac{\partial z}{\partial x} and \frac{\partial z}{\partial y}.