# Is the space of maps which satisfy this vanishing condition finite-dimensional?

Let $$\mathbb{D}^n \subseteq \mathbb{R}^n$$ be the closed $$n$$-dimensional unit ball. Let $$h:\mathbb{D}^n \to \mathbb{R}^{k}$$ be smooth, and suppose that $$h(x) \neq 0$$ a.e. on $$\mathbb{D}^n$$. Set $$V_h=\{ \,\,f \in C^{\infty}(\mathbb{D}^n;\mathbb{R}^{k}) \, \,\,| \, \, (df_x)^T\big(h(x)\big)=0 \, \text{ for every }\, x \in \mathbb{D}^n \, \}$$

$$V_h$$ is a real vector-space. Is it always finite-dimensional? Can it be infinite-dimensional for some $$h$$?

Edit:

Pozz showed nicely that when $$k=1$$, $$V_h$$ always coincides with the space of constant functions, and that for $$k>1$$, $$V_h$$ might be infinite-dimensional (e.g. if $$h$$ is a constant function).

Is there ever a case where $$V_h$$ is finite-dimensional when $$k>1$$? I suspect that the answer is negative, but I don't know how to prove this.

Let us write the condition $$(df_x)^T(h(x))=0$$ more explicitly. We can write $$(df_x)^T=\bigg(\nabla f^1(x)\,\bigg|\,...\,\bigg|\,\nabla f^k(x)\bigg),$$ where $$\nabla f^i(x)$$ is the column vector given by the Euclidean gradient of $$f^i$$, where the $$f^i$$'s are the components of $$f$$ for $$i=1,...,k$$. Hence the condition defining $$V_h$$ becomes $$(df_x)^T(h(x))=0\quad\forall x \qquad\Leftrightarrow\qquad \langle \partial_j f(x), h(x) \rangle =0\,\,\quad\forall j=1,...,n\quad \forall x$$ where $$\langle\cdot,\cdot\rangle$$ denotes the Euclidean product.

If $$k=1$$, we can then prove that $$V_h$$ is the $$1$$-dimensional vector space of constant functions. Indeed, if $$f\in V_h$$ then $$h(x)\partial_j f(x)=0$$ for any $$j=1,...,n$$ and any $$x$$. Since $$h(x)\neq0$$ almost everywhere, then $$\partial_j f(x)=0$$ for any $$j=1,...,n$$ and at almost every $$x$$. Since $$f$$ is smooth, then $$\nabla f$$ is actually identically zero, and thus $$f$$ is constant.

If $$k>1$$ we can find an example of $$h$$ such that $$V_h$$ is infinite-dimensional. Consider in fact $$h(x)=(1,0,...,0)$$, that is smooth and non-zero. In this case, if $$f\in V_h$$ then $$\langle\partial_j f(x),h(x)\rangle=\partial_jf^1(x)=0 \,\,\quad \forall j=1,...,n\quad\forall x.$$ This implies that any function $$f=(0,f^2,...,f^k)$$ belongs to $$V_h$$ for any choice of $$f^2,...,f^k$$ smooth. And thus $$V_h$$ is infinite-dimensional.

• Thank you! This is a nice solution. I also thought about try "testing" against constant functions $h$, but only after I posted the question... I am now interested to know if $V_h$ can be finite-dimensional when $k>1$. Do you have any idea about this? – Asaf Shachar Aug 19 '19 at 9:07
• @AsafShachar At the moment I don't have an answer to this second question! I suspect that it is not completely trivial. – Pozz Aug 24 '19 at 8:13

TL;DR: Yes, it can be finite dimensional. I think that this is possible only due to "global obstructions".

Let's consider the case $$n = 2, k = 2$$. Writing $$f = (f^1,f^2)$$ and $$h = (h^1,h^2)$$, we get the system

$$f^1_x h^1 + f^2_x h^2 = 0, \\ f^1_y h^1 + f^2_y h^2 = 0.$$

Differentiating the first equation with respect to $$y$$ and the second to $$x$$, we also get $$f^1_{yx} h^1 + f^1_x h^1_y + f^2_{yx} h^2 + f^2_x h^2_y = 0, \\ f^1_{xy} h^1 + f^1_y h^1_x + f^2_{xy} h^2 + f^2_y h^2_x = 0.$$

Comparing both equations and using the equality of mixed derivatives, we get the equation $$f_x^1 h^1_y + f^2_x h^2_y = f^1_y h^1_x + f^2_y h^2_x.$$

This gives us three linear equations for $$(f^1_x,f^1_y,f^2_x,f^2_y)$$ which are generically independent and so will leave us with one degree of freedom (ignoring questions of integrability). Now, let's analyze a specific example:

Take $$h(x,y) = (x,y)$$. Then we get the system $$f^1_x x + f^2_x y = 0, \\ f^1_y x + f^2_y y = 0, \\ f^2_x = f^1_y.$$ Plugging the third equation into the first two allows us to "decouple" the system and get two identical equations for $$f^1,f^2$$: $$f^1_x x + f^1_y y = 0, \\ f^2_x x + f^2_y y = 0.$$ Let's see if we can find a global solution. Geometrically, the first equation says that $$\nabla(f^1)$$ is perpendicular to $$(x,y)$$. Hence, on $$\mathbb{D}^2 \setminus \{ (0,0) \}$$, we must have that $$\nabla(f^1)(x,y) = a(x,y)(-y,x)$$ for some smooth uniquely determined function $$a$$. That is, $$\nabla(f^1)$$ is a multiple of $$\partial_{\theta}$$ (or, dually, $$df^1$$ is a multiple of the famous $$d\theta$$). However, not all possible multiples are legal -- the mixed second partial derivatives of $$f^1$$ should agree and we get an equation for $$a$$: $$f^1_{yx} = -a_y y - a = a_x x + a = f^1_{xy} \iff 2a = -(a_x \cdot x + a_y \cdot y).$$ This is a linear first order PDE for $$a$$ which can be solved explicitly using the method of characteristics. Fix $$(x_0,y_0) \in \partial{\mathbb{D}^2}$$ and set $$u(t) := a(e^{-t}(x_0,y_0))$$. Differentiating, we get $$u'(t) = a_x(e^{-t}(x_0,y_0))(-e^t x_0) + a_y(e^{-t}(x_0,y_0))(-e^t y_0) = 2a(e^{-t}(x_0,y_0)) = 2u(t)$$ which implies that $$u(t) = e^{2t} u(0) = e^{2t} a(x_0,y_0).$$ Hence, we see that $$a(x,y) = a \left( \frac{(x,y)}{\| (x,y)\|} \right) \frac{1}{\| (x,y) \|^2}, \\ (\nabla f^1)(x,y) = \frac{-(y,x)}{\| (x,y) \|^2} a \left( \frac{(x,y)}{\| (x,y) \|} \right).$$ On each ray through the origin, the length of $$(\nabla f^1)$$ decays like $$\frac{1}{r}$$ and so in order to have a limit at the origin, we must have $$a \equiv 0$$ and so $$f^1$$ must be constant (and similarly for $$f^2$$).

Note that over $$\mathbb{D}^2 \setminus \{ (0,0) \}$$ there is an infinite dimensional family of solutions to your equation. One non-constant solution is the "obvious" solution $$f = \frac{h}{\| h \|} = \left( \frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}} \right).$$

In general, if you follow the details of my analysis, you can show that any solution on (an open subset or the whole of) $$\mathbb{D}^2 \setminus \{ (0,0) \}$$ has the form $$f^1 = -\int \varphi(\theta) \sin \theta \, d\theta, \,\,\, f^2 = \int \varphi(\theta) \cos \theta \, d\theta.$$

If $$\varphi \equiv 1$$ then you get the "obvious" solution $$f^1 = \cos \theta = \frac{x}{\sqrt{x^2 + y^2}}, \,\,\, f^2 = \sin \theta = \frac{y}{\sqrt{x^2 + y^2}}$$

but you can take any other $$\varphi$$ and obtain infinitely many other solutions. If the resulting integrals are periodic, you get a solution on the whole of $$\mathbb{D}^2 \setminus \{ (0,0) \}$$ but none of the solutions will extend to the whole of $$\mathbb{D}^2$$.

• Thank you. I just wanted to mention that you can anaylze your example more simply: The condition on the gradient immediately implies that $f^1$ depends only on $\theta$ (since $0=df^1(\frac{\partial}{\partial r})=\langle \nabla f ^1, \frac{\partial}{\partial r} \rangle$), and any continuous function on the entire disk which depends only on $\theta$ must be constant (as the limits along different rays converging to the origin should all agree). – Asaf Shachar Aug 26 '19 at 16:11
• @AsafShachar: I agree, although you should formulate it carefully as $\theta$ is not a global function on the whole (or even the punctured) disk. – levap Aug 26 '19 at 23:39