If directional derivatives are bounded, then function is continuous 
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 $\frac{\partial f}{\partial x}$?
 A: 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 $C\|A^{-1}\|_1$.
