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Let a function $f:X \times \mathbb{R} \rightarrow \mathbb{R}$ continuous, with $X \subset \mathbb{R}$ compact, and supose that $\partial_2 f(x,t)$ exists for all $x \in X$ and is continuous. (here $\partial_2$ is the derivative with respect to the second coordinate)

I would like to know if $$\displaystyle \lim_{(x,t) \to (x_0,0)} \frac{f(x,t)-f(x,0)}{t} = \partial_2 f(x_0,0)$$

I think it is not true, but i can't find a counterexample. Observe that if we take the limits separately (first $x\to x_0$ and then $t \to 0$, or in the reverse order), these limits are both equal to $\partial_2 f(x_0,0)$. It seems to me like the classical examples of the functions whose limits in each coordinate exists separately but the total limit does not exist.

Can someone help me with this?

Thanks

EDITED (generalizing the question)

Supose the folowing situation

$f:X \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ continuous, with $X$ a compact metric space. Denote by $f_x$ the function $f_x(v) = f(x,v)$, and supose that for each $x \in X$, $f_x: \mathbb{R}^m \rightarrow \mathbb{R}^m$ is differentiable and $$ (x,v) \mapsto (f_x)'(v) \in Gl(\mathbb{R}^m)$$ is continuous.

Is it true that $$\lim_{(x,v)\to (x_0,0)} \frac{f(x,v)-f(x,0)}{\|v\|} = (f_{x_0})'(0)\cdot v ?$$

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    $\begingroup$ Hey, sorry if this question is too elementary. It came from a question about a differential structure for the space of the sections of a vector bundle $\endgroup$
    – Ferraiol
    Commented Feb 11, 2011 at 18:59
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    $\begingroup$ This looks too elementary to me, but maybe I'm missing something. By the mean value theorem, the quotient on the left is $\partial_2f(x,u)$ for some $u$ between 0 and $t$. Since you're assuming continuity of this partial derivative, it will approach $\partial_2f(x_0,0)$. (I'm assuming that your continuity assumption for the partial derivative means continuity with respect to both variables together, not some sort of separate continuity.) $\endgroup$
    – Andreas Blass
    Commented Feb 11, 2011 at 19:26
  • $\begingroup$ What do you mean by "$\partial_2 f(x,t)$ exists for all $x\in X$ and is continuous"? Note that $f$ and $\partial_2 f$ are functions of two variables. $\endgroup$
    – Christian Blatter
    Commented Feb 11, 2011 at 19:31
  • $\begingroup$ Thank you, Andreas. This solve the problem. Sorry to post this elementary question. Thank you too for the answer, Willie. but in you example the quotient $\frac{f(x,t)-f(x,0)}{t}$ is always zero, so this is not a counterexample. $\endgroup$
    – Ferraiol
    Commented Feb 11, 2011 at 19:38
  • $\begingroup$ Christian, I mean that the partial derivative $\partial_2 f(x,t)$ exist and the function $(x,t) \mapsto \partial_2 f(x,t)$ is continuous. $\endgroup$
    – Ferraiol
    Commented Feb 11, 2011 at 19:41

1 Answer 1

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For $t\ne 0$ one has $${f(x,t)-f(x,0) \over t}- \partial_2 f(0,0)= \int_0^1 (\partial_2 f(x,\tau \thinspace \thinspace t) - \partial_2 f(0,0))\thinspace d\tau ,$$ and here the right side is $<\epsilon$ when $(x,t)$ is in a suitable neighbourhood of $(0,0)$.

For an $f:X\times {\bf R}^n\to {\bf R}^m$ it is enough to consider the $i$-th coordinate function $f_i:X\times {\bf R}^n\to {\bf R}$, again denoted by $f$, and for the latter consider the auxiliary function $\phi(t):=f(x,t \thinspace v)$ on the interval $[0,1]$. One gets $$f(x,v)-f(x,0) =\int_0^1 \phi'(t)\thinspace dt = \int_0^1 \nabla f_2(x,t\thinspace v)\cdot v \thinspace dt,$$ whence $$f(x,v)-f(x,0) = \nabla f_2(x_0,0)\cdot v + o(\|v\|) \qquad ((x,v)\to(x_0,0)).$$

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