Forgot password?
 Register account
View 235|Reply 2

[函数] 在原点没有极小值,但沿所有线都有极小值

[Copy link]

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

hbghlyj Posted 2022-12-6 01:35 |Read mode
mathcounterexamples
我们在这里看一个来自意大利数学家 Giuseppe Peano 的例子,一个定义在 $\Bbb R^2$ 上的实函数 $f$。 $f$ 沿所有穿过原点的直线在原点处具有极小值,但是 $f$ 作为二元函数在原点处没有极小值。 \[f(x, y) = (y-3x^2)(y-x^2)\] 特别地,在开集 $U=\{(x,y) \in \mathbb{R}^2 \ : \ x^2 < y < 3x^2\}$ 上 $f<0$,在抛物线 $y=x^2$ 和 $y=3x^2$ 上 $f=0$,其他地方 $f>0$。
考虑通过原点的直线 $D:y=λx,λ>0$。我们有\[f(x, \lambda x)= x^2(\lambda-3x)(\lambda -x).\] 对于 \(x \in (-\infty,\frac{\lambda}{3}) \setminus \{0\}\), 有 \(f(x, \lambda x) > 0\). 而 \(f( 0,0)=0\), 这证明了 \(f\) 沿 \(D\) 在原点处具有极小值。

沿着 \(x\) 轴,\(f(x,0)=3 x^ 4\) 在原点处具有极小值。

沿着 \(y\) 轴,\(f(0,y)=y^2\) 在原点处具有极小值。

沿着抛物线 \(\mathcal{P}:y = 2 x^2\) 我们有 \(f(x,2 x^2)=-x^4<0\) 对于 \(x \neq 0\)。因为 \(\mathcal{P}\) 通过原点,\(f\) 在原点的所有邻域内既取正值也取负值。这证明了 \((0,0)\) 不是 $f$ 的极小值点。

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

 Author| hbghlyj Posted 2022-12-12 17:48

在原点不可导,但存在全部的方向导数

en.wikipedia.org/wiki/Fréchet_derivative#Rela … e_Gateaux_derivative
A function $f : U ⊆ V → W$ is called Gateaux differentiable at $x ∈ U$ if $f$ has a directional derivative along all directions at $x$.
If $f$ is Fréchet differentiable at $x$, it is also Gateaux differentiable there, and $g$ is just the linear operator $A = D f ( x )$.

However, not every Gateaux differentiable function is Fréchet differentiable. This is analogous to the fact that the existence of all directional derivatives at a point does not guarantee total differentiability (or even continuity) at that point. For example, the real-valued function $f$ of two real variables defined by
\[f(x,y)=\begin{cases}{\frac {x^{3}}{x^{2}+y^{2}}}&(x,y)\neq (0,0)\\0&(x,y)=(0,0)\end{cases}\]
is continuous and Gateaux differentiable at the origin $(0,0)$, with its derivative at the origin being
\[g(a,b)=\begin{cases}{\frac {a^{3}}{a^{2}+b^{2}}}&(a,b)\neq (0,0)\\0&(a,b)=(0,0)\end{cases}\]
The function $g$ is not a linear operator, so this function is not Fréchet differentiable.

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

 Author| hbghlyj Posted 2022-12-12 17:56
WolframAlpha显示local minimum
$$\min  \left\{\frac{x^3}{x^2+y^2}\right\}=0 \text{ at }(x,y)\approx (0,1.35466)$$
刷新几次, 后面的数还会变
1.35438
1.35587

说明WolframAlpha算法里面有随机数

Mobile version|Discuz Math Forum

2025-5-31 10:40 GMT+8

Powered by Discuz!

× Quick Reply To Top Edit