A key element to proving Thom's gradient conjecture for parabolic evolution equations is to show that it holds for a certain quadratic on an infinite dimensional space. Orthogonal to its finite dimensional kernel, the quadratic operator is bounded below, but not above. The boundedness below allows it to dominate the higher order terms orthogonal to the kernel. The lack of boundedness above is not an issue since it is the direction of the gradient rather than its magnitude which is important for Thom's conjecture. In this note we provide a simplified proof for the quadratic case, and more generally analytic homogeneous functions, under some additional restrictions.
Before the general case had been proven in the finite dimensional case, Thom had already observed that the result holds for a homogeneous polynomial of any degree $m$. The argument is as follows.
Let $f:\mathbb{R}^n \to \mathbb{R}$ be a homogeneous polynomial of degree \(m \ge 2\). Write \(x = r u\) with \(r = \|x\| > 0\), \(u \in \mathbb{S}^{n-1}\). By homogeneity and Euler’s identity, and recalling that the derivative of a differentiable homogeneous function is homogenous of degree one less than the original function:
\[
\nabla f(r u) = r^{m-1}\nabla f(u), \qquad
\langle \nabla f(z), z \rangle = m f(z).
\]
Define the projected function \(F := f|_{\mathbb{S}^{n-1}}\). The gradient flow \(\dot{x} = -\nabla f(x)\) in these coordinates is
\[
\dot{r} = - m f(u)\, r^{m-1}, \qquad
\dot{u} = - r^{m-2}\, \nabla_{\mathbb{S}} F(u),
\]
where the second equation loses a power of r since u = x/r.
\[
\frac{d}{dt} F(u(t))
= \left\langle \nabla_{\mathbb{S}} F(u(t)), \, \dot{u}(t) \right\rangle
= -\, r^{\,m-2} \, \|\nabla_{\mathbb{S}} F(u)\|^2 \;\le\; 0.
\]
Introduce a transformed time variable
\[
s(t) := \int_{0}^{t} r(\tau)^{\,m-2}\, d\tau,
\]
Then in the transformed time, using the chain rule:
\[
\frac{d u}{d s} = - \nabla_{\mathbb{S}} F(u), \qquad u(0) = u_0.
\]
Thus, the projected flow is the gradient flow of \(F\) on the sphere. It's easy to see that the restriction to the unit sphere of an analytic function is also analytic. Then the standard argument of Lojasiewicz shows that the trajectory of the projection onto the sphere has finite length, which implies that Thom's conjecture holds in this case.
So which aspects of this proof this go through when we switch to a function $E:H \to \mathbb{R}$ on an infinite dimensional Hilbert space? It fails when we need to use the Lojasiwicz inequality, which no longer holds in general even for an analytic function. To continue, we'll have to impose some additional restrictions.
Define the zero set $Z := \{ x \in \mathbb{R}^n : f(x) = 0 \}$. Since $f$ is homogeneous, $Z$ has the "cone" property that if $u \in Z$, then $\lambda u \in Z$ for all $\lambda \in \mathbb{R}$. Analytic homogeneous functions also have the property that
\[
|f(x)| \;\ge\; C\,\mathrm{dist}(x,Z)^{\,m}
\]
This is just the Lojasiewicz inequality, except that we know the exponent must be $m$, because the distance to the zero set scales like $r$, while the left hand side scales like $r^m$.
Returning to trying to prove Thom's conjecture for analytic homogeneous functions in infinite dimensions, we know this growth condition will no longer hold in general. But let's try imposing this as a condition and see if it's sufficient to complete the proof. Because $Z$ is a cone, this is equivalent to the bound on $\mathbb{S}^{n-1}$
\[
|F(u)| \;\ge\; C\,\mathrm{dist}(u,\Sigma)^{\,m},
\qquad (u \in \mathbb{S}^{n-1}),
\tag{1}
\]
where $\Sigma := Z \cap \mathbb{S}$. That is, we get the same growth inequality for the projected flow on the sphere.
This growth inequality for the projected flow is a different but equivalent form of the Lojasiewicz inequality. Thus, we've again succeeded in showing the Lojasiewicz inequality for the projected flow.
Finally, because of the loss of compactness when moving to infinite dimensions, we'll also need to assume a Palais-Smale condition to ensure that that the flow on the sphere converges to a unique point. Thus we assume that any sequence $u_k\subset \mathbb{S}$ with $F(u_k)$ bounded and $\|\nabla_{\mathbb{S}}F(u_k)\|\to 0$ admits a convergent subsequence in $\mathbb{S}$. In cases where the zero set on the sphere $\Sigma$ is compact (this occurs when the zero set $Z$ of $f$ is finite dimensional for example), the Palais-Smale condition should follow from our distance growth condition (1). Of course, this is exactly the case when trying to prove Thom's conjecture for parabolic evolution equations, as the Hessian has a finite dimensional kernel.
It's important to note that our growth condition (1) is a lower bound only, and we need not assume any boundedness above.
