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I am seeing the generalization of the MVT to $2$ variable functions (Rudin PMA $9.40$).

excerpt of PMA Rudin theorem 9.40, generalization of Mean Value Theorem 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.

niobium
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1 Answers1

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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

Let $f$ be defined on $[a,b]$. If $f$ is differentiable at a point $x \in [a,b]$, then $f$ is continuous at $x$.

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 $[a,a+h]$.

Walid
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  • More generally, can we say that if a partial derivative of $f$ with respect to its first variable $\frac{\partial f}{\partial x}$ exists at a point, then the function $f$ is continuous at this point with respect to its first variable ? – niobium Jul 23 '23 at 18:59
  • @niobium In general this is not true. Have a look at older posts showing counterexamples. See here and here – Walid Jul 23 '23 at 21:22