State the most general (real) version of the implicit theorem you know and outline how it can be proved from the corresponding version of the (real) inverse function theorem.
Solution
Let \[ U\subseteq\RR^n\times\RR^m \] be open, let \[ F:U\longrightarrow\RR^m \] be \(C^r\) for some \(1\leq r\leq\infty\), and let \[ (a,b)\in U. \] Put \[ c=F(a,b). \]
1. If the partial derivative\[ D_yF(a,b):\RR^m\longrightarrow\RR^m \]is invertible, then there are neighborhoods\[ a\in A\subseteq\RR^n, \qquad b\in B\subseteq\RR^m \]and a unique \(C^r\) map\[ \phi:A\longrightarrow B \]with \(\phi(a)=b\) such that, for \((x,y)\in A\times B\),\[ \boxed{ F(x,y)=c \quad\Longleftrightarrow\quad y=\phi(x). } \]Moreover,\[ D\phi(x) = -\bigl(D_yF(x,\phi(x))\bigr)^{-1} D_xF(x,\phi(x)) \]after the neighborhoods are chosen sufficiently small.
1.1. Define\[ H:U\longrightarrow\RR^n\times\RR^m, \qquad H(x,y)=(x,F(x,y)). \]Then\[ DH(a,b) = \begin{pmatrix} I_n&0\\ D_xF(a,b)&D_yF(a,b) \end{pmatrix} \]is invertible.
The displayed derivative follows directly from the two components of \(H\). It is block lower triangular. Its diagonal blocks are \(I_n\) and \(D_yF(a,b)\), both invertible, so \(DH(a,b)\) is invertible.
1.2. The inverse function theorem gives neighborhoods\[ (a,b)\in W\subseteq U, \qquad (a,c)\in Z\subseteq\RR^n\times\RR^m \]such that\[ H|_W:W\longrightarrow Z \]is a \(C^r\) diffeomorphism.
Step 1 is exactly the invertibility hypothesis of the inverse function theorem at \((a,b)\).
1.3. After shrinking \(Z\) to a product neighborhood if necessary, the localinverse has the form\[ H^{-1}(x,z)=(x,G(x,z)) \]for a \(C^r\) map \(G\).
Because \(Z\) is open and contains \((a,c)\), it contains a product \[ A_0\times C \] of neighborhoods of \(a\) and \(c\). Restrict the inverse to this product.
If \[ H(u,v)=(x,z), \] then the first component of the definition of \(H\) gives \(u=x\). Therefore the first component of \(H^{-1}(x,z)\) is always \(x\), so the inverse has the displayed form. Since \(H^{-1}\) is \(C^r\), so is \(G\).
1.4. Define\[ \phi(x)=G(x,c). \]After shrinking \(A_0\) to a neighborhood \(A\) of \(a\) and choosing aneighborhood \(B\) of \(b\), one has\[ \phi:A\to B \]and, for \((x,y)\in A\times B\),\[ F(x,y)=c \quad\Longleftrightarrow\quad y=\phi(x). \]
Since \[ H^{-1}(a,c)=(a,b), \] we have \(G(a,c)=b\). Since \(W\) is open, choose neighborhoods \(A_1\) of \(a\) and \(B\) of \(b\) such that \[ A_1\times B\subseteq W. \] By continuity of \(x\mapsto G(x,c)\) at \(a\), shrink to a neighborhood \[ A\subseteq A_0\cap A_1 \] such that \(\phi(A)\subseteq B\). Then \(A\times B\subseteq W\) as required.
For \(x\in A\), \[ H(x,\phi(x)) = H(H^{-1}(x,c)) = (x,c), \] so \[ F(x,\phi(x))=c. \]
Conversely, if \((x,y)\in A\times B\) and \(F(x,y)=c\), then \[ H(x,y)=(x,c). \] Since \(H|_W\) is injective and \[ H(x,\phi(x))=(x,c), \] we get \(y=\phi(x)\). This proves the local graph description and uniqueness.
1.5. The derivative of the implicit function is\[ D\phi(x) = -\bigl(D_yF(x,\phi(x))\bigr)^{-1} D_xF(x,\phi(x)). \]
By continuity of \(D_yF\) and invertibility at \((a,b)\), the neighborhoods may be shrunk so that \[ D_yF(x,\phi(x)) \] is invertible for every \(x\in A\). Differentiate the identity \[ F(x,\phi(x))=c. \] The chain rule gives \[ D_xF(x,\phi(x)) + D_yF(x,\phi(x))D\phi(x) = 0. \] Multiplying by the inverse of \(D_yF(x,\phi(x))\) gives the formula.
1.6. Q.E.D.
Steps <2>1–<2>5 derive every assertion of step 1 from the inverse function theorem.
2. Q.E.D.
Step 1 states the theorem for all finite \(n,m\) and every order \(1\leq r\leq\infty\), and its substeps derive it from the inverse function theorem.