I am seeing the generalization of the MVT to $2$ variable functions (Rudin PMA $9.40$).
The proof uses the one-dimensional MVT twice, and I was checking if the hypothesis of the $1$-dimension MVT were gathered. These are, in one dimension, that the function must be continuous on an interval $[a,b]$ and differentiable on the open interval $(a,b)$. Applied to a $2$ variable function ($f(x,y)$) I think this translate as (considering for example the first variable):
- $f$ must be continuous with respect to the first variable on $[a,b]$ (that means: if we fix the second variable, $f$ considered as a function of its first variable only must be continuous).
- $D_1 f$ must exist on an interval $(a,b)$.
Back to our theorem $9.40$, we can see from the hypothesis of the theorem, that $D_1 f$ exists on the interval of interest for the MVT ($[a, a+h]$) since the rectangle is a closed set of $E$.
Hence my question: "if $D_1 f$ exists on $[a, a+h]$, does it implies that $f$ is continuous with respect to the first variable on $[a, a+h]$" ?
Note I am only focusing on the first application of the MVT. The second one (which involves $D_{21} f$) is similar.
Thanks in advance.
We see that $u(t) = f(t, b+k) - f(t,b)$ is a function in one variable. The existence of $D_1 f$ on $Q$ implies the existence of the derivatives of $f(t, b+k)$ and $f(t, b)$ on $[a,a+h]$; therefore, $u(t)$ is a differentiable function on $[a,a+h]$.
Theorem 5.2 in Rudin states that
In our case, $u(t)$ is a real function in one variable that is differentiable on $[a,a+h]$, therefore it must be continuous on $[a,a+h]$. This means we can apply the mean value theorem to $u(t)$. So yes, if we fix the second variable such that it is in $[b,b+k]$ then $f$ would be continuous in its first variable in $[a,a+h]$.