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 at the origin. 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
\[
Z(\nabla T) = \{x: \nabla T = 0\}.
\]
If the sequence $x_n$ lies on an analytic curve through the origin, then on that curve we must have $\nabla T = 0$. The analytic variety $Z(\nabla T)$ admits a Whitney stratification into a finite number of analytic manifolds at $0$. We hope that we can arrange that the sequence $x_n$ is asymptoting to $Z(\nabla T)$ faster than any given polynomial in $r$. From a previous post, we know this is not true for an arbitrary sequence.
Since $T$ and $\nabla T$ are analytic they satisfy Lojasiewicz inequalities. One form of which is
\[
\|\nabla T(x)\| \ge C\, \mathrm{dist}(x,Z(\nabla T))^\alpha,
\]
\[
\|T(x)\| \ge C\, \mathrm{dist}(x,Z(T))^\alpha.
\]
Note that $Z(T)$ contains $Z(\nabla T)$ so we can use $Z(T)$ for both. We also use suboptimal constants in exchange for the simplicity of having the same constants in both inequalities.
$\bf{Theorem:}$ A non-analytic function $f = T + \omega:\mathbb{R}^n \to \mathbb{R}$ satisfies the Lojasiewicz inequality if the flat or non-analytic part $\omega$ satisfies
\[
\lim_{x \to 0} \frac{\,|\omega(x)| + \|\nabla \omega(x)\|\,}{\operatorname{dist}(x,Z(T))^N} \;=\; 0
\]
for all positive integers $N$. The same Lojasiewicz exponent as for $T$ may be used.
It's important to realise that this condition is not satisfied by just any function with zero Taylor series at the origin.
To derive the gradient inequality for $f=T+\omega$ under this assumption, fix a small parameter $\eta>0$. By our earlier argument, we can restrict attention to a polynomial neighbourhood $\mathcal{H}$ of $z(F)$ (so that $\operatorname{dist}(x,Z(f))\le r^k$ for some $k$). Then using our growth condition on $\omega$ and $\nabla \omega$ and the Lojasiewicz inequalities for $T$ and $\nabla T$, we can achieve that for all $x\in \mathcal{H}$,
\[
|\omega(x)| \le \eta\,|T(x)|, \qquad \|\nabla\omega(x)\| \le \eta\,\|\nabla T(x)\|.
\]
On $x\in \mathcal{H}$ the triangle inequality for $f$ implies
\[
|f(x)| = |T(x)+\omega(x)| \le |T(x)| + |\omega(x)| \le (1+\eta)\,|T(x)|,
\]
and for the gradients one has
\[
\|\nabla f(x)\| = \|\nabla T(x) + \nabla\omega(x)\| \ge \|\nabla T(x)\| - \|\nabla\omega(x)\| \ge (1-\eta)\,\|\nabla T(x)\|.
\]
Using the other form of the Lojasiewicz inequality for the analytic function $T$, there exist constants $c>0$ and $\rho \in \bigl[\tfrac{1}{2},1\bigr)$ such that
\[
\|\nabla T(x)\| \ge c\,|T(x)|^{\rho}
\]
for $x$ sufficiently close to the origin. Combining this estimate with the inequalities on $\mathcal{H}$ yields
\[
\|\nabla f(x)\| \ge (1-\eta)\,c\,|T(x)|^{\rho} \ge (1-\eta)\,c\,(1+\eta)^{-\rho}\,|f(x)|^{\rho}.
\]
Hence $f$ satisfies a Lojasiewicz inequality on $\mathcal{H}$:
\[
\|\nabla f(x)\| \ge c'\,|f(x)|^{\rho},\quad x \in \mathcal{H},
\]
where the modified constant is
\[
c' = \frac{(1-\eta)\,c}{(1+\eta)^{\rho}}.
\]
Since $\eta$ may be chosen arbitrarily small, the constant $c'$ can be made as close to $c$ as desired.
