Classical Lojasiewicz inequalities

Let ${f : \mathbb{R}^n \rightarrow \mathbb{R}}$ be a real analytic function with ${f(0) = 0}$. Let ${V := \{x \in \mathbb{R}^n | f(x) = 0\}}$ and ${K}$ be a compact subset in ${\mathbb{R}^n}$. Then the (classical) \L ojasiewicz inequality asserts that:

• There exist ${c > 0, \alpha > 0}$ such that

$\displaystyle |f(x)| \ge cd(x, V)^\alpha\quad \text{for}\ x \in K.\ \ \ \ \ (1)$

\noindent Let ${f : \mathbb{R}^n \rightarrow \mathbb{R}}$ be a real analytic function with ${f(0) = 0}$ and ${\nabla f(0) = 0}$. The \L ojasiewicz gradient inequality asserts that:

• There exist ${C > 0, \rho \in [0, 1)}$ and a neighbourhood ${U}$ of ${0}$ such that

$\displaystyle \|\nabla f(x)\| \ge C|f(x)|^\rho\quad \text{for}\ x \in U.\ \ \ \ \ (2)$

As a consequence, in (1), the order of zero of an analytic function is finite, and if ${f (x)}$ is close to ${0}$ then ${x}$ is close to the zero set of ${f}$. However, if ${K}$ is not compact, the latter is not always true and the inequality (1) does not always hold. The inequality (2) is similar to (1), it is not true in the case of $K$ is non-compact.

Two inequalities (1) and (2) have some special cases. For example, in the inequality (1), if ${f}$ has only isolated zero, i.e. ${V = f^{-1}(0) = \{(0, 0, \dots, 0)\}}$, this implies ${d(x, V) = \|x\|}$. Hence, we have

$\displaystyle |f(x)| \ge C\|x\|^\alpha, \text{for}\ x\in K.$

On the other hand, being different from (2), we have another inequality:

$\displaystyle \|\nabla f(x)\| \ge c\|x\|^\beta, \text{for}\ x \in U.$

There are some relations between ${\alpha, \beta}$ and ${\rho}$ in complex case and real cases…