Let \(f:\mathbb R^2\setminus\{(0,0)\}\to\mathbb R\) be \(C^1\). Suppose there is \(M>0\) such that \[ \left|\frac{\partial f}{\partial x}\right|\le M, \qquad \left|\frac{\partial f}{\partial y}\right|\le M \] throughout the punctured plane. Prove that \[ \lim_{(x,y)\to(0,0)}f(x,y) \] exists.
Solution
Set \[ C\coloneqq\sqrt2\,M. \]
1. At every point of the punctured plane,\[ \norm{\nabla f}\leq C. \]
The hypotheses give \[ \abs{f_x}\leq M, \qquad \abs{f_y}\leq M. \] Therefore \[ \norm{\nabla f} = \sqrt{f_x^2+f_y^2} \leq \sqrt{M^2+M^2} = \sqrt2\,M =C. \]
2. If \(p,q\in\RR^2\setminus\{0\}\) satisfy\[ \norm{p}<r, \qquad \norm{q}<r, \]then there is a piecewise \(C^1\) path in the punctured disk\[ \{z:0<\norm{z}<r\} \]joining \(p\) to \(q\) whose length is less than\[ (2+\pi)r. \]
Put \[ R\coloneqq\max\{\norm{p},\norm{q}\}<r. \] Move radially from \(p\) to the point on the same ray having norm \(R\), then along the shorter circular arc of radius \(R\) to the ray through \(q\), and finally radially to \(q\).
The two radial pieces have total length at most \(2R\), and the shorter arc has length at most \(\pi R\). Thus the total length is at most \[ (2+\pi)R < (2+\pi)r. \] Every point on the path has positive norm at most \(R\), so the path avoids the origin.
3. For any \(p,q\) as in step 2,\[ \abs{f(p)-f(q)} < C(2+\pi)r. \]
Let \(\gamma\) be the path from step 2, parametrized piecewise by arc length. Along each smooth piece, \[ \frac{d}{dt}f(\gamma(t)) = \inner{\nabla f(\gamma(t))}{\gamma'(t)}. \] Hence, by step 1, \[ \abs{\frac{d}{dt}f(\gamma(t))} \leq C\norm{\gamma'(t)}. \] Integrating along all pieces gives \[ \abs{f(p)-f(q)} \leq C\,\operatorname{length}(\gamma) < C(2+\pi)r. \]
4. The sequence\[ a_n\coloneqq f(1/n,0) \]is Cauchy and therefore converges to some \(L\in\RR\).
Let \(\varepsilon>0\). Choose \(N\) so large that \[ \frac{C(2+\pi)}{N}<\varepsilon. \] If \(m,n>N\), then both points \[ (1/m,0), \qquad (1/n,0) \] have norm less than \(1/N\). Step 3, with \(r=1/N\), gives \[ \abs{a_m-a_n}<\varepsilon. \] Thus \((a_n)\) is Cauchy. Completeness of \(\RR\) gives a limit \(L\).
5. One has\[ \boxed{ \lim_{p\to0}f(p)=L }. \]
Let \(\varepsilon>0\). Choose \(r>0\) such that \[ C(2+\pi)r<\frac{\varepsilon}{2}. \] By step 4, choose \(n\) large enough that \[ \frac1n<r \] and \[ \abs{f(1/n,0)-L}<\frac{\varepsilon}{2}. \] If \[ 0<\norm{p}<r, \] then step 3, applied to \(p\) and \((1/n,0)\), gives \[ \abs{f(p)-f(1/n,0)} < \frac{\varepsilon}{2}. \] Therefore \[ \begin{aligned} \abs{f(p)-L} &\leq \abs{f(p)-f(1/n,0)} + \abs{f(1/n,0)-L}\\ &< \varepsilon. \end{aligned} \] This is precisely the definition of the limit at the puncture.
6. Q.E.D.
Step 5 proves that the required finite limit exists.