Consider the polynomial function \(f(s,t) = 9s^3 - 6st + t^2\). Let \(P = (1,3)\).
Carefully state the conclusion of the implicit function theorem concerning the equation \(f(s,t) = 0\) when \(f\) is considered as a function from \(\mathbb{R}^2\) to \(\mathbb{R}\).
Carefully state the conclusion of the implicit function theorem concerning the equation \(f(s,t) = 0\) when \(f\) is considered as a function from \(\mathbb{C}^2\) to \(\mathbb{C}\).
Use the implicit function theorem for functions from \(\mathbb{R}^{\times}\mathbb{R}^2 \to \mathbb{R}^2\) to prove (b). (There are various approaches to this, including the definition of complex derivative, the Cauchy–Riemann equations, and consideration of total derivatives.)
The source typesets the domain in part (c) as \(\mathbb{R}^{\times}\mathbb{R}^2\), a typographical error. Viewing \(f\) on \(\mathbb{C}^2 = \mathbb{R}^2 \times \mathbb{R}^2\) with values in \(\mathbb{C} = \mathbb{R}^2\), the intended theorem is the implicit function theorem for maps \(\mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}^2\).
Solution
We have \[ f(1,3)=9-18+9=0, \] and \[ f_s(s,t)=27s^2-6t, \qquad f_t(s,t)=-6s+2t. \] Hence \[ f_s(1,3)=9, \qquad f_t(1,3)=0. \]
1. Part (a): there are open intervals\[ 1\in U\subseteq\RR, \qquad 3\in V\subseteq\RR \]and a unique smooth function\[ \phi:V\longrightarrow U \]such that\[ \phi(3)=1 \]and, for \((s,t)\in U\times V\),\[ f(s,t)=0 \quad\Longleftrightarrow\quad s=\phi(t). \]Moreover,\[ \phi'(3)=\boxed{0}. \]
The polynomial \(f:\RR^2\to\RR\) is smooth, \[ f(1,3)=0, \] and the derivative with respect to the variable to be solved for satisfies \[ f_s(1,3)=9\neq0. \] The real implicit function theorem therefore gives the stated neighborhoods and unique smooth function \(\phi\).
Differentiating \[ f(\phi(t),t)=0 \] gives \[ f_s(\phi(t),t)\phi'(t)+f_t(\phi(t),t)=0. \] At \(t=3\) this yields \[ \phi'(3) = -\frac{f_t(1,3)}{f_s(1,3)} = 0. \]
2. Parts (b) and (c): there are open neighborhoods\[ 1\in U_{\CC}\subseteq\CC, \qquad 3\in V_{\CC}\subseteq\CC \]and a unique holomorphic function\[ \Phi:V_{\CC}\longrightarrow U_{\CC} \]such that\[ \Phi(3)=1 \]and, for \((s,t)\in U_{\CC}\times V_{\CC}\),\[ f(s,t)=0 \quad\Longleftrightarrow\quad s=\Phi(t). \]Furthermore,\[ \Phi'(3)=0. \]
2.1. Realify the complex polynomial as\[ \mathcal F:\RR^2_s\times\RR^2_t\longrightarrow\RR^2, \qquad \mathcal F(s,t) = \bigl(\Re f(s,t),\Im f(s,t)\bigr). \]At\[ (s,t)=(1,3), \]the partial real derivative\[ D_s\mathcal F(1,3):\RR^2\longrightarrow\RR^2 \]is multiplication by the complex number \(9\), hence is the invertible matrix\[ 9I_2. \]
For fixed \(t\), the polynomial \(s\mapsto f(s,t)\) is holomorphic with complex derivative \[ f_s(s,t)=27s^2-6t. \] The real derivative of a holomorphic map \(\CC\to\CC\) is the real-linear map given by multiplication by its complex derivative. At \((1,3)\) that complex number is \(9\), so the corresponding real matrix is \(9I_2\).
2.2. The real implicit function theorem gives real neighborhoods\[ 1\in U\subseteq\RR^2_s, \qquad 3\in V\subseteq\RR^2_t \]and a unique \(C^1\) map\[ \Phi_{\RR}:V\longrightarrow U \]whose graph is exactly the zero set of \(\mathcal F\) in \(U\times V\).
Step 1 gives the invertibility hypothesis for the real implicit function theorem applied to \[ \mathcal F:\RR^2_s\times\RR^2_t\to\RR^2. \] Thus the theorem gives the stated map and local graph description.
2.3. After shrinking \(U\) and \(V\) if necessary,\[ D\Phi_{\RR}(t) = -\bigl(D_s\mathcal F(\Phi_{\RR}(t),t)\bigr)^{-1} D_t\mathcal F(\Phi_{\RR}(t),t) \]is complex-linear for every \(t\in V\).
Since \[ f_s(1,3)=9\neq0, \] continuity allows the neighborhoods to be chosen so that \[ f_s(s,t)\neq0 \] throughout \(U\times V\). Differentiating \[ \mathcal F(\Phi_{\RR}(t),t)=0 \] gives the displayed formula.
Because \(f\) is a polynomial in the complex variables \(s\) and \(t\), the real maps \[ D_s\mathcal F \qquad\text{and}\qquad D_t\mathcal F \] are multiplication by the complex numbers \(f_s\) and \(f_t\), respectively. They are therefore complex-linear. The inverse of the nonzero multiplication map \(D_s\mathcal F\) is also complex-linear, so the displayed composition is complex-linear.
2.4. Under the identifications \(\RR^2\cong\CC\), the map \(\Phi_{\RR}\) isholomorphic.
Fix \(t_0\in V\). Since \(\Phi_{\RR}\) is real differentiable at \(t_0\) and \(D\Phi_{\RR}(t_0)\) is complex-linear by step 3, there is \(\lambda\in\CC\) such that \[ \Phi_{\RR}(t_0+h)-\Phi_{\RR}(t_0) = \lambda h+r(h), \qquad \frac{\abs{r(h)}}{\abs h}\longrightarrow0. \] For nonzero complex \(h\), \[ \frac{\Phi_{\RR}(t_0+h)-\Phi_{\RR}(t_0)}h = \lambda+\frac{r(h)}h \longrightarrow \lambda. \] Thus the complex derivative exists at every \(t_0\in V\), so \(\Phi_{\RR}\) is holomorphic. Denote the resulting complex map by \(\Phi\) and the neighborhoods by \(U_{\CC},V_{\CC}\).
2.5. The holomorphic map \(\Phi\) has the graph and uniqueness stated instep 2, and\[ \Phi'(3)=0. \]
The graph and uniqueness are exactly those supplied by the real implicit function theorem in step 2, merely rewritten under \(\RR^2\cong\CC\).
Differentiating \[ f(\Phi(t),t)=0 \] complexly gives \[ f_s(\Phi(t),t)\Phi'(t)+f_t(\Phi(t),t)=0. \] At \(t=3\), \[ \Phi'(3) = -\frac{f_t(1,3)}{f_s(1,3)} = 0. \]
2.6. Q.E.D.
Steps <2>1–<2>5 derive the complex implicit-function conclusion in step 2 entirely from the real implicit function theorem, as required in part (c).
3. Q.E.D.
Step 1 gives the real conclusion in part (a), while step 2 and its substeps give the complex conclusion in part (b) and the requested real-IFT proof in part (c).