0
$\begingroup$

The theorem says that:

A function $f: \mathbb{R}^2 \to \mathbb{R}$ is differentiable at $(x_0, y_0)$ if its partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are continuous at $(x_0, y_0)$.

How do I prove this? I know that for a function to be differentiable, the condition is:

$$ \displaystyle \lim_{\lVert h,k \rVert \to 0} \dfrac{f(x_0+h, y_0+k) - f(x_0, y_0) - h\,\frac{\partial f}{\partial x}\rvert_{(x_0, y_0)} - k\,\frac{\partial f}{\partial y}\rvert_{(x_0, y_0)}}{\lVert h,k \rVert} = \lim_{\lVert h,k \rVert \to 0} \mathcal O(h,k) = 0 $$

The problem is that the definition of differentiability is using the values of partial derivatives at the point itself, and not the function. I am not understanding how I can "link" that statement to the definition of continuity of partial derivatives, which are functions themselves. I am not even getting where I could start!

$\endgroup$
  • $\begingroup$ Tip: This obviously is (to me at least) a standard result. Have you tried looking in a book or googling? $\endgroup$ – Git Gud Sep 10 '17 at 10:31
1
$\begingroup$

The way to link the function to the partial derivatives is to use the Mean Value Theorem.


Let us write $$ \begin{align*} f(x_0 + h, y_0 + k) - f(x_0,y_0) &= f(x_0 + h,y_0) - f(x_0,y_0)\\ &\qquad \quad {}+ f(x_0 + h, y_0 + k) - f(x_0 + h,y_0). \end{align*} $$ Now, we can apply the Mean Value Theorem to the two pairs of terms: $$ \begin{align*} f(x_0 + h, y_0) - f(x_0,y_0) &= h \cdot \frac{\partial f}{\partial x}(b_1,y_0)\\ f(x_0 + h, y_0 + k) - f(x_0 + h,y_0) &= k \cdot \frac{\partial f}{\partial y} (x_0+h,b_2) \end{align*} $$ for some $b_1 \in (x_0,x_0 + h)$ and $b_2 \in (y_0, y_0 + k)$. Therefore, $$ \begin{align*} & \frac{\left| f(x_0+h,y_0+k)-f(x_0,y_0)-h\frac{\partial f}{\partial x}(x_0,y_0)-k\frac{\partial f}{\partial y}(x_0,y_0) \right|}{\| (h,k)\|}\\ ={} & \frac{\left| h\left( \frac{\partial f}{\partial x}(b_1,y_0)-\frac{\partial f}{\partial x}(x_0,y_0) \right) - k \left( \frac{\partial f}{\partial y}(x_0+h,b_2) - \frac{\partial f}{\partial y}(x_0,y_0) \right) \right|}{\| (h,k)\|}\\ \leq{} & \frac{| h |}{\| (h,k) \|} \cdot \left| \frac{\partial f}{\partial x}(b_1,y_0)-\frac{\partial f}{\partial x}(x_0,y_0) \right| + \frac{|k |}{\| (h,k) \|} \cdot \left| \frac{\partial f}{\partial y}(x_0+h,b_2) - \frac{\partial f}{\partial y}(x_0,y_0) \right| \\ \leq{} & \left| \frac{\partial f}{\partial x}(b_1,y_0)-\frac{\partial f}{\partial x}(x_0,y_0) \right| + \left| \frac{\partial f}{\partial y}(x_0+h,b_2) - \frac{\partial f}{\partial y}(x_0,y_0) \right| \end{align*} $$ since $| h | / \| (h,k) \|$ and $| k | / \| (h,k) \|$ are less than or equal to $1$ for all $(h,k) \neq (0,0)$.

Now, taking $\lim_{\|(h,k)\| \to 0}$, we get $0$, because $(b_1,y_0) \to (x_0,y_0)$ and $(x_0+h,b_2) \to (x_0,y_0)$ as $(h,k) \to (0,0)$, and the partial derivatives are continuous at $(x_0,y_0)$. This is where we use the continuity of the partial derivatives at $(x_0,y_0)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.