# 判断函数凹凸性

## 几个凸函数/凹函数

f ( θ x + ( 1 − θ ) y ) = max ⁡ i ( θ x i + ( 1 − θ ) y i ) ⩽ θ max ⁡ i x i + ( 1 − θ ) max ⁡ i y i = θ f ( x ) + ( 1 − θ ) f ( y ) \begin{aligned} f(\theta x+(1-\theta) y) &=\max _{i}\left(\theta x_{i}+(1-\theta) y_{i}\right) \\ & \leqslant \theta \max _{i} x_{i}+(1-\theta) \max _{i} y_{i} \\ &=\theta f(x)+(1-\theta) f(y) \end{aligned} f(θx+(1θ)y)=imax(θxi+(1θ)yi)θimaxxi+(1θ)imaxyi=θf(x)+(1θ)f(y)

∇ 2 f ( x , y ) = 2 y 3 [ y 2 − x y − x y x 2 ] = 2 y 3 [ y − x ] [ y − x ] T ⪰ 0 \nabla^{2} f(x, y)=\frac{2}{y^{3}}\left[\begin{array}{cc} y^{2} & -x y \\ -x y & x^{2} \end{array}\right]=\frac{2}{y^{3}}\left[\begin{array}{c} y \\ -x \end{array}\right]\left[\begin{array}{c} y \\ -x \end{array}\right]^{T} \succeq 0 2f(x,y)=y32[y2xyxyx2]=y32[yx][yx]T0

v T ∇ 2 f ( x ) v = 1 ( 1 T z ) 2 ( ( ∑ i = 1 n z i ) ( ∑ i = 1 n v i 2 z i ) − ( ∑ i = 1 n v i z i ) 2 ) ⩾ 0 v^{T} \nabla^{2} f(x) v=\frac{1}{\left(1^{T} z\right)^{2}}\left(\left(\sum_{i=1}^{n} z_{i}\right)\left(\sum_{i=1}^{n} v_{i}^{2} z_{i}\right)-\left(\sum_{i=1}^{n} v_{i} z_{i}\right)^{2}\right) \geqslant 0 vT2f(x)v=(1Tz)21((i=1nzi)(i=1nvi2zi)(i=1nvizi)2)0

∂ 2 f ( x ) ∂ x k 2 = − ( n − 1 ) ( ∏ i = 1 n x i ) 1 / n n 2 x k 2 , ∂ 2 f ( x ) ∂ x k ∂ x l = ( ∏ i = 1 n x i ) 1 / n n 2 x k x l ∀ k ≠ l \frac{\partial^{2} f(x)}{\partial x_{k}^{2}}=-(n-1) \frac{\left(\prod_{i=1}^{n} x_{i}\right)^{1 / n}}{n^{2} x_{k}^{2}}, \quad \frac{\partial^{2} f(x)}{\partial x_{k} \partial x_{l}}=\frac{\left(\prod_{i=1}^{n} x_{i}\right)^{1 / n}}{n^{2} x_{k} x_{l}} \quad \forall k \neq l xk22f(x)=(n1)n2xk2(i=1nxi)1/n,xkxl2f(x)=n2xkxl(i=1nxi)1/nk=l

∇ 2 f ( x ) = − ∏ i = 1 n x i 1 / n n 2 ( n diag ⁡ ( 1 / x 1 2 , ⋯   , 1 / x n 2 ) − q q T ) \nabla^{2} f(x)=-\frac{\prod_{i=1}^{n} x_{i}^{1 / n}}{n^{2}}\left(n \operatorname{diag}\left(1 / x_{1}^{2}, \cdots, 1 / x_{n}^{2}\right)-q q^{T}\right) 2f(x)=n2i=1nxi1/n(ndiag(1/x12,,1/xn2)qqT)

v T ∇ 2 f ( x ) v = − ∏ i = 1 n x i 1 / n n 2 ( n ∑ i = 1 n v i 2 / x i 2 − ( ∑ i = 1 n v i / x i ) 2 ) ⩽ 0 v^{T} \nabla^{2} f(x) v=-\frac{\prod_{i=1}^{n} x_{i}^{1 / n}}{n^{2}}\left(n \sum_{i=1}^{n} v_{i}^{2} / x_{i}^{2}-\left(\sum_{i=1}^{n} v_{i} / x_{i}\right)^{2}\right) \leqslant 0 vT2f(x)v=n2i=1nxi1/n(ni=1nvi2/xi2(i=1nvi/xi)2)0

g ( t ) = log ⁡ det ⁡ ( Z + t V ) = log ⁡ det ⁡ ( Z 1 / 2 ( I + t Z − 1 / 2 V Z − 1 / 2 ) Z 1 / 2 ) = ∑ i = 1 n log ⁡ ( 1 + t λ i ) + log ⁡ det ⁡ Z \begin{aligned} g(t) &=\log \operatorname{det}(Z+t V) \\ &=\log \operatorname{det}\left(Z^{1 / 2}\left(I+t Z^{-1 / 2} V Z^{-1 / 2}\right) Z^{1 / 2}\right) \\ &=\sum_{i=1}^{n} \log \left(1+t \lambda_{i}\right)+\log \operatorname{det} Z \end{aligned} g(t)=logdet(Z+tV)=logdet(Z1/2(I+tZ1/2VZ1/2)Z1/2)=i=1nlog(1+tλi)+logdetZ

g ′ ( t ) = ∑ i = 1 n λ i 1 + t λ i , g ′ ′ ( t ) = − ∑ i = 1 n λ i 2 ( 1 + t λ i ) 2 g^{\prime}(t)=\sum_{i=1}^{n} \frac{\lambda_{i}}{1+t \lambda_{i}}, \quad g^{\prime \prime}(t)=-\sum_{i=1}^{n} \frac{\lambda_{i}^{2}}{\left(1+t \lambda_{i}\right)^{2}} g(t)=i=1n1+tλiλi,g(t)=i=1n(1+tλi)2λi2

原文作者：小羊冲呀
原文地址: https://blog.csdn.net/qq_41758867/article/details/106196157
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