最小二乘目标函数为凸函数

证明：

$${\Vert \mathbf{M}\left(\frac{{\mathit{s}}_1+{\mathit{s}}_2}{2}\right)-\mathit{c}\Vert }_2^2\le \frac{\parallel \mathbf{M}{\mathit{s}}_1-\mathit{c}{\parallel }_2^2+\parallel \mathbf{M}{\mathit{s}}_2-\mathit{c}{\parallel }_2^2}{2}.$$

注意到

$$\begin{array}{rl}{\Vert \mathbf{M}\left(\frac{{\mathit{s}}_1+{\mathit{s}}_2}{2}\right)-\mathit{c}\Vert }_2^2& ={(\mathbf{M}\left(\frac{{\mathit{s}}_1+{\mathit{s}}_2}{2}\right)-\mathit{c})}^{\mathrm{T}}(\mathbf{M}\left(\frac{{\mathit{s}}_1+{\mathit{s}}_2}{2}\right)-\mathit{c})\\ & ={(\frac{\mathbf{M}{\mathit{s}}_1+\mathbf{M}{\mathit{s}}_2}{2}-\mathit{c})}^{\mathrm{T}}(\frac{\mathbf{M}{\mathit{s}}_1+\mathbf{M}{\mathit{s}}_2}{2}-\mathit{c})\\ & =\frac{1}{4}{\mathit{s}}_1{\mathbf{M}}^{\mathrm{T}}\mathbf{M}{\mathit{s}}_1+\frac{1}{4}{\mathit{s}}_2{\mathbf{M}}^{\mathrm{T}}\mathbf{M}{\mathit{s}}_2+\frac{1}{2}{\mathit{s}}_1{\mathbf{M}}^{\mathrm{T}}\mathbf{M}{\mathit{s}}_2-({\mathit{s}}_1+{\mathit{s}}_2{)}^{\mathrm{T}}{\mathbf{M}}^{\mathrm{T}}\mathit{c}+{\mathit{c}}^{\mathrm{T}}\mathit{c}\end{array}$$

而

$$\begin{array}{rl}\parallel \mathbf{M}{\mathit{s}}_1-\mathit{c}{\parallel }_2^2+\parallel \mathbf{M}{\mathit{s}}_2-\mathit{c}{\parallel }_2^2& =(\mathbf{M}{\mathit{s}}_1-\mathit{c}{)}^{\mathrm{T}}(\mathbf{M}{\mathit{s}}_1-\mathit{c})+(\mathbf{M}{\mathit{s}}_2-\mathit{c}{)}^{\mathrm{T}}(\mathbf{M}{\mathit{s}}_2-\mathit{c})\\ & ={\mathit{s}}_1{\mathbf{M}}^{\mathrm{T}}\mathbf{M}{\mathit{s}}_1+{\mathit{s}}_2{\mathbf{M}}^{\mathrm{T}}\mathbf{M}{\mathit{s}}_2-2{\mathit{s}}_1^{\mathrm{T}}{\mathbf{M}}^{\mathrm{T}}\mathit{c}-2{\mathit{s}}_2^{\mathrm{T}}{\mathbf{M}}^{\mathrm{T}}\mathit{c}+2{\mathit{c}}^{\mathrm{T}}\mathit{c}.\end{array}$$

因此只需证明

$$\begin{array}{rl}& \frac{1}{4}{\mathit{s}}_1{\mathbf{M}}^{\mathrm{T}}\mathbf{M}{\mathit{s}}_1+\frac{1}{4}{\mathit{s}}_2{\mathbf{M}}^{\mathrm{T}}\mathbf{M}{\mathit{s}}_2+\frac{1}{2}{\mathit{s}}_1{\mathbf{M}}^{\mathrm{T}}\mathbf{M}{\mathit{s}}_2-({\mathit{s}}_1+{\mathit{s}}_2{)}^{\mathrm{T}}{\mathbf{M}}^{\mathrm{T}}\mathit{c}+{\mathit{c}}^{\mathrm{T}}\mathit{c}\\ \le & \frac{1}{2}{\mathit{s}}_1{\mathbf{M}}^{\mathrm{T}}\mathbf{M}{\mathit{s}}_1+\frac{1}{2}{\mathit{s}}_2{\mathbf{M}}^{\mathrm{T}}\mathbf{M}{\mathit{s}}_2-{\mathit{s}}_1^{\mathrm{T}}{\mathbf{M}}^{\mathrm{T}}\mathit{c}-{\mathit{s}}_2^{\mathrm{T}}{\mathbf{M}}^{\mathrm{T}}\mathit{c}+{\mathit{c}}^{\mathrm{T}}\mathit{c}.\end{array}$$

消除公共项，只需证明

$$\begin{array}{r}\frac{1}{4}{\mathit{s}}_1{\mathbf{M}}^{\mathrm{T}}\mathbf{M}{\mathit{s}}_1+\frac{1}{4}{\mathit{s}}_2{\mathbf{M}}^{\mathrm{T}}\mathbf{M}{\mathit{s}}_2-\frac{1}{2}{\mathit{s}}_1{\mathbf{M}}^{\mathrm{T}}\mathbf{M}{\mathit{s}}_2\ge 0.\end{array}$$

上面的不等式可以写成

$$({\mathit{s}}_1-{\mathit{s}}_2{)}^{\mathrm{T}}{\mathbf{M}}^{\mathrm{T}}\mathbf{M}({\mathit{s}}_1-{\mathit{s}}_2)\ge 0.$$

也即

$${(\mathbf{M}({\mathit{s}}_1-{\mathit{s}}_2))}^{\mathrm{T}}(\mathbf{M}({\mathit{s}}_1-{\mathit{s}}_2))\ge 0.$$

得证。