A function $f:\mathbb{R}^n \to \mathbb{R}$ satisfies a Lojasiewicz inequality at $0$ if in a neighbourhood of $0$ we have
\[
|\nabla f| \geq c|f|^\rho,
\]
for some $c>0$ and $\rho \in [\frac{1}{2},1)$.
It is well-known that the Lojasiewicz inequality holds for analytic functions. While analyticity is sufficient for the Lojasiewicz inequality to hold, it is not necessary. Trivial examples like $f(x) = x^2 + e^{1/x}$ demonstrate this. What then is an appropriate weaker condition?
A function $f:\mathbb{R}^n \to \mathbb{R}$ is analytic at $0$ if it is locally equal to its Taylor series $T(x)$, i.e., $f(x)=T(x)$. For a non-analytic function let's write
\[
f(x) = T(x) + \omega(x),
\]
where $\omega$ has a Taylor series which is identically zero. In other words, $\omega$ is the "non-analytic" part of the function. For the Lojasiewicz inequality to hold, $\omega$ need not be zero, and it is in fact only necessary that $\omega$ is dominated by the function's Taylor series in a certain sense.
To see this, observe that if the Lojasiewicz inequality does not hold, then for any sequences $c_n \to 0$ and $\rho_n \to 1$, we can find a sequence $x_n \to 0$ such that
\[
|\nabla f(x_n)| < c_n|f(x_n)|^{\rho_n}.
\]
We can choose the sequence $x_n$ to converge to $0$ as fast as we like.
Let $\mathcal{C}$ be the set of smooth curves emanating from $0$, parameterised by arc length. Consider the sets
\[
\mathcal{C}_{a,k}^\varepsilon = \{\gamma \in \mathcal{C}; |\nabla f(\gamma(t))| \geq at^k \; \forall \; t \in [0,\varepsilon) \},
\]
\[
X_{a,k}^\varepsilon = \cup_{\gamma \in \mathcal{C}_{a,k}^\varepsilon} \gamma([0,\varepsilon)).
\]
Clearly the Lojasiewicz inequality holds inside any such set $X_{a,k}^\varepsilon$, even for a function which is not analytic. Thus the sequence $x_n$ is eventually outside $X_{a,k}^\varepsilon$ for any $k \in N$ arbitrarily large and any $a,\varepsilon$ arbitrarily small. Intuitively, we might guess that the sequence $x_n$ is (in some approriate sense) asymptoting to the analytic variety
\[
V = \{x: \nabla T = 0\}.
\]
If the sequence $x_n$ lies on an analytic curve through the origin, then on that curve we must have $\nabla T = 0$. From a previous post, we hope that we can arrange that the sequence $x_n$ is asymptoting to some analytic curve $\gamma$ faster than any given polynomial in $r$ (this is not quite true due to the existence of sequences whose projection onto the unit sphere converges slower than any polynomial in r - we'll gloss over this!). On this curve we must have $\nabla T = 0$. It's natural to conjecture the following:
$\bf{Conjecture:}$ A non-analytic function $f = T + \omega:\mathbb{R}^n \to \mathbb{R}$ satisfies the Lojasiewicz inequality if and only if it satisfies the Lojasiewicz inequality on the set $\nabla T = 0$ (in a neighbourhood of 0).
To begin to prove this conjecture, note that the analytic variety $V$ admits a Whitney stratification into a finite number of analytic manifolds at $0$. Thus, we can assume that our sequence $x_n$ is asymptoting to an analytic manifold $M$ faster than any polynomial in $r$. As for a function, a curve can be written as the sum of an analytic part and a non-analytic part. Let $\lambda(t) = \gamma(t) + \bar \gamma(t)$, where $\gamma$ is an analytic curve inside $V$ and $\bar \gamma$ is a curve with Taylor series in $t$ identically zero. By assumption, the Lojasiwicz inequality holds on $\gamma$. If we can show that the Lojasiwicz inequality also holds on all curves that deviate from $V$ by smaller than polynomial terms, we are done.
