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Suppose we have $f:S\to \mathbb R$ where $S$ is the open unit circle: $S=\lbrace (x,y)\in \mathbb R ^{2} \mid x^2 +y^2 <1\rbrace $, and suppose $\vec v, \vec w$ are two independent unit vectors.

Suppose $\exists C>0,\forall \vec z\in S,\Bigl|\frac {\partial f}{\partial v}(z)\Bigr|<C,\Bigl|\frac {\partial f}{\partial w}(z)\Bigr|<C$.

Show that $f$ is continuous in $S$.

Now, I know that if the partial derivatives exist and bounded, then $f$ is continous, but I'm having trouble expressing the partial derivatives in terms of the given directional derivatives. Of course $e_x=(1,0)$ can be expressed as a linear combination of $v,w$, but how can I formally apply it to $\frac{\partial f}{\partial x}$?

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  • $\begingroup$ Between every two points in $S$ there is a two segment path that joins them and the segments are parallel to the axes. Use this to estimate the difference in $f$. $\endgroup$ – copper.hat Jul 15 '18 at 18:32
  • $\begingroup$ @copper.hat I'm not sure how to construct this two segment path, I could only come up with a one-segment path $\phi(t)=t\cdot(\alpha \vec w +\beta \vec v)+(1-t)\cdot (\lambda \vec w +\gamma \vec v) $ for $t\in [0,1]$, where $z_1 ,z_2 \in S$ are represented by these two linear combinations. $\endgroup$ – gbi1977 Jul 15 '18 at 20:13
  • $\begingroup$ Draw a picture. Take two points in the disc. Join them with a Manhattan like path. $\endgroup$ – copper.hat Jul 15 '18 at 20:15
  • $\begingroup$ @copper.hat oh okay I think I see what you mean. Then the idea is that each point on this L path can be represented as a linear comb. of $v,w$. So along the path, the directional derivative is of course bounded, depending on the $v$ or $w$ component, but since it applies to any point on a line pararllel to the axes - it applies to the partial derivatives themselves? $\endgroup$ – gbi1977 Jul 15 '18 at 20:34
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Here is a more complete proof. However, I think the basic idea gets lost in the details.

Let $A= \begin{bmatrix}v & w \end{bmatrix}$ and define $\phi(z) = f(A z)$ on the open convex set $\Sigma=A^{-1} S$. Note that ${\partial \phi(z) \over \partial z_1} = {\partial f(Az) \over \partial u}$ and ${\partial \phi(z) \over \partial z_2} = {\partial f(Az) \over \partial w}$.

The key element of the proof is that for any $z_1,z_2 \in \Sigma$, we can choose (because $\Sigma$ is open & convex) a path $y_1=z_1 \to y_2\to \cdots \to y_{n-1} \to y_n = z_2$ such that each segment of the path is parallel to either axis and the direction of each segment is the same (with respect to the corresponding axis).

In particular, if $V_j(y) = [y]_j$ we have $V_j(z_2-z_1) = \sum_k V_j(y_k-y_{k-1})$, for a given $j$, all of the $V_j(y_k-y_{k-1})$ have the same sign and at most one of $V_1(y_k-y_{k-1}) $ or $V_2(y_k-y_{k-1})$ are non zero. Hence $|V_j(z_2-z_1)| = \sum_k |V_j(y_k-y_{k-1})|$.

Note that on each segment, we can apply the mean value theorem to get $|\phi(y_k)-\phi(y_{k-1})| \le C (|V_1(y_k-y_{k-1}) | + |V_2(y_k-y_{k-1})| )$ and so \begin{eqnarray} |\phi(z_2)-\phi(z_1)| &\le& \sum_k |\phi(y_k)-\phi(y_{k-1})| \\ &\le & C \sum_k(|V_1(y_k-y_{k-1}) | + |V_2(y_k-y_{k-1})| ) \\ &=& C(|V_1(z_2-z_1)| + |V_2(z_2-z_1)|) \\ &=& C \|z_2-z_1\|_1 \end{eqnarray}

Finally, we return to '$f$' space:

$|f(z_2)-f(z_1)| = |\phi(A^{-1} z_2) - \phi(A^{-1} z_1)| \le C \|A^{-1}(z_2-z_1)\|_1 \le C\|A^{-1}\|_1 \|z_2-z_1\|_1$

Hence $f$ is Lipschitz continuous with rank at most $C\|A^{-1}\|_1$.

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    $\begingroup$ Doesn’t this argument assume that $\vec v=\langle1,0\rangle$ and $\vec w=\langle0,1\rangle$ (or vice versa)? $\endgroup$ – Steve Kass Jul 15 '18 at 21:42
  • $\begingroup$ @SteveKass: It does, but exactly the same approach works except for the $\sqrt 2$ part and the $C need to be adjusted. $\endgroup$ – copper.hat Jul 16 '18 at 0:06
  • $\begingroup$ Yes, it just seemed to me that the details of how to make that approach work were at the heart of the OP’s question. (Or a little more bluntly, if I’d assigned this as a class exercise, I would expect those details to be worked out in a submitted answer.) $\endgroup$ – Steve Kass Jul 16 '18 at 0:27
  • $\begingroup$ @SteveKass: I guess my take was that the OP just needed to see the idea and could take it from there. I will sketch out the details. $\endgroup$ – copper.hat Jul 16 '18 at 0:50
  • $\begingroup$ @SteveKass: I elaborated the detail, but I think it has become a technical mess. $\endgroup$ – copper.hat Jul 16 '18 at 2:28

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