# Can we approximate harmonic functions with harmonic functions with non-vanishing differential?

Let $$\mathbb{D}^2$$ be the closed $$2$$-dimensional unit disk, and let $$g:\mathbb{D}^2 \to \mathbb{R}$$ be a non-constant harmonic function (in particular smooth up to the boundary).

Does there exist a sequence of smooth harmonic functions $$g_n$$ on $$\mathbb{D}^2$$, such that $$g_n \to g$$ in $$W^{1,2}$$ and $$dg_n \neq 0$$ everywhere on $$\text{int}(\mathbb{D}^2)$$?

Since we can add additive constants to the $$g_n$$, we can arrange $$\int_{\mathbb D^2} g_n=\int_{\mathbb D^2} g$$, so the $$W^{1,2}$$ convergence of the $$g_n$$ is essentially equivalent to $$dg_n \to dg$$ in $$L^2$$. (via Poincare inequality).

Thinking on $$dg$$ as a vector field, I think that we can always approximate it with a non-zero vector field in $$L^2$$. However, the only procedure I know for doing that does not produce approximating vector fields which are gradients of harmonic functions (or gradients of anything, really).

No, even if we only ask $$g_n\to g$$ in $$L^2$$ (or even just in the sense of distributions, but that gets more technical).

The important property is that $$L^2$$ convergence of harmonic functions on an open set implies uniform convergence of each derivative on each compact subset. This is a kind of topological version of Weyl's lemma: not only are harmonic functions smooth, but "convergent sequences of harmonic functions" are "convergent sequences of smooth functions" in a certain sense.

Set $$g(x,y)=x^2-y^2.$$

The mean value property of harmonic functions implies that $$\int f(z'-z)\psi(z)dx=f(z')\int \psi$$ whenever $$f$$ is harmonic on a ball $$B(z',r),$$ and $$\psi$$ is radially symmetric with support in the ball $$B(0,r).$$ Take a radially symmetric smooth function $$\psi$$ supported in the ball $$B(0,1/2)$$ and with $$\int \psi = 1.$$ For any $$z\in B(0,1/2)$$ we get

$$dg(z) = \int dg(z'-z)\psi(z) = -\int g(z'-z)d\psi(z)dz$$ and similarly for $$g_n.$$ So $$|(dg-dg_n)(z)|\leq |\int (g(z'-z)-g_n(z'-z))d\psi(z)dz|\leq \|g-g_n\|_2\|d\psi\|_2$$ which tends to zero uniformly in $$z$$ as $$n\to\infty.$$

$$dg$$ has winding number $$-1$$ around $$0$$ as $$z$$ goes anticlockwise around along the circle $$|z|=1/2.$$ So for large enough $$n,$$ by continuity $$dg_n$$ gets the same winding number and therefore must have a zero in the disc $$|z|\leq 1/2.$$

(I think a similar idea should answer https://mathoverflow.net/questions/330947/can-we-approximate-harmonic-maps-which-are-a-e-orientation-preserving-with-maps: take $$f(x,y)=(x^2-y^2,2xy)$$ and apply the above argument to $$f_1(x,y)=x^2-y^2.$$)

• Thank you! I just woke up and saw your answer. Actually, just before I went to sleep, I saw two theorems that say that uniformly bounded harmonic functions converge (up to a subsequence) uniformly to a harmonic function, and that uniformly convergent harmonic sequences converge in $C^{\infty}$. This are theorems $1.23$ and $2.6$ in the book referred here. I guessed that this means that now convergence is impossible , due to the known topological obstructions. This was late at night, so I did not have time to update the question... – Asaf Shachar Nov 21 '19 at 5:55
• I will take a thorough look at your answer when I will have the time, but from a brief glance, it looks perfect (and rather self-contained). Thank you again. I was hoping initially for a positive answer rather than negative, but I was naive. However, I still have (some) hope for my wider goal. I might post another question about that some time... – Asaf Shachar Nov 21 '19 at 5:57