Talylor’s theorem with remainder to order 2

Lemma(Taloy’s theorem with remainder to order 2). Let \(f: \mathbb{R}^2 \rightarrow \mathbb{R}^1\) is \(C^\infty\), then there exist \(C^\infty\) functions \(g_{11}, g_{12},g_{22}\) on \(\mathbb{R}^2\) such that \[ \begin{align*} f(x,y) = f(0,0) & + \frac{\partial f}{\partial x}(0,0) x + \frac{\partial f}{\partial y}(0,0) y \\ & + x^2g_{11}(x,y) + 2xyg_{12}(x,y) + y^2 g_{22}(x,y) \end{align*} \]

Proof. Trivially, \((p,m) = (0,0)\) is star-shaped in \(\mathbb{R}^2\) and the line segement \(p+t(x-p), 0<t<1\) and \(m+j(y-m), 0<j<1\) lies in \(\mathbb{R}^2\). So that \(f(p+t(x-p), m+j(y-m))\) is defined.

By the chain rule, \[ \begin{align*} &\frac{d}{dt} ~ \frac{d}{dj} f(p+t(x-p), m+j(y-m)) \\ & = \sum (x^i-p^i) \frac{\partial f}{\partial x^i}(p+t(x-p), m+j(y-m)) \\ & + \sum (y^i-m^i) \frac{\partial f}{\partial y^i}(p+t(x-p),m+j(y-m) \end{align*} \]

If we double integrate both sides with respect to \(t\) and \(j\) from \(0\) to \(1\), we get \[ \begin{align*} & [[f(p+t(x-p), m+j(y-m))]_{t=0}^{t=1}]_{j=0}^{j=1}\\ & =\sum(x^i-p^i) \int_{0}^{1} \int_{0}^{1} \frac{\partial f}{\partial x^i}(p+t(x-p), m+j(y-m)) dtdj \\ & + \sum (y^i-m^i) \int_{0}^{1} \int_{0}^{1} \frac{\partial f}{\partial y^i}(p+t(x-p), m+j(y-m) dtdj ~~ (1)\\ \end{align*} \]

Let \[ \begin{align*} & g_1(x,y)=\int_{0}^{1} \int_{0}^{1} \frac{\partial f}{\partial x^i}(p+t(x-p),m+j(y-m)) dtdj\\ & g_2(x,y) = \int_{0}^{1} \int_{0}^{1}\frac{\partial f}{\partial y^i}(p+t(x-p), m+j(y-m)) dtdj\\ \end{align*} \]

Then both \(g_1(x,y), g_2(x,y)\) are \(C^{\infty}\) and (1) becomes \[ f(x,y)-f(p,m) = \sum(x^i-p^i)g_1(x,y) + \sum(y^i-m^i)g_2(x,y) ~~(2) \]

In case \(p=0, m=0\), and \(n=1\), (2) becomes \[ f(x,y) = f(0,0)+ xg_1(x,y)+yg_2(x,y) ~~ (3) \]

Applying this lemma second time, that is to double differentiate \(f(p+t(x-p),m+j(y-m))\) with repect to \(t\) and \(j\) \[ \begin{align*} & \frac{d^2}{dt^2} \frac{d^2}{dj^2}f(p+t(x-p),m+j(y-m)) \\ & = \sum(x^i-p^i)^2 \big[\frac{\partial^2}{\partial x^{2i}} f(p+t(x-p),m+j(y-m)) \big]+ (y^i-m^i)(x^i-p^i) \big[\frac{\partial^2}{\partial x \partial y}f(p+t(x-p), m+j(y-m))\big]\\ & + \sum(y^i-p^i)^2\big[\frac{\partial^2}{\partial y^{2i}}f(p+t(x-p),m+j(y-m)) \big]+ (y^i-m^i)(x^i-p^i)\big[\frac{\partial^2 }{\partial x \partial y} f(p+t(x-p), m+j(y-m)) \big] ~~(4) \end{align*} \]

Similarly we can integrate (4) and obtain \[ \begin{align*} & g_{11}(x,y) = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{\partial^2 }{\partial x^{2i}} f(p+t(x-p), m+j(y-m))dtdjdtdj \\ & g_{12}(x,y) = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{\partial^2}{\partial x \partial y} f(p+t(x-p), m+j(y-m))dtdjdtdj\\ & g_{22}(x,y) = \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{\partial^2}{\partial y^{2i}} f(p+t(x-p), m+j(y-m)) dtdjdtdj\\ \end{align*} \] for some \(g_{11}, g_{12},g_{22}\) are \(C^{\infty}\).

We can finally obtain the equaiton in the case of \(n=2\) \[ \begin{align*} f(x,y)& =f(0,0) + (x^2g_{11}(x,y) + xy(g_{12})(x,y)) + (xyg_{12}(x,y)+y^2g_{22}(x,y))\\ & = f(0,0)+ x^2 g_{11}(x,y) + 2xy (g_{12}(x,y)) + y^2 g_{22}(x,y)~~ \square \end{align*} \]