This limit converges to the partial derivative? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-22T11:00:01Zhttp://mathoverflow.net/feeds/question/55152http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/55152/this-limit-converges-to-the-partial-derivativeThis limit converges to the partial derivative?Ferraiol2011-02-11T18:51:04Z2011-02-13T11:08:38Z
<p>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)</p>
<p>I would like to know if $$\displaystyle \lim_{(x,t) \to (x_0,0)} \frac{f(x,t)-f(x,0)}{t} = \partial_2 f(x_0,0)$$</p>
<p>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.</p>
<p>Can someone help me with this?</p>
<p>Thanks</p>
<p>EDITED (generalizing the question)</p>
<p>Supose the folowing situation</p>
<p>$f:X \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ 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.</p>
<p>Is it true that $$\lim_{(x,v)\to (x_0,0)} \frac{f(x,v)-f(x,0)}{\|v\|} = (f_{x_0})'(0)\cdot v ?$$</p>
http://mathoverflow.net/questions/55152/this-limit-converges-to-the-partial-derivative/55198#55198Answer by Christian Blatter for This limit converges to the partial derivative?Christian Blatter2011-02-12T12:37:36Z2011-02-13T11:08:38Z<p>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 $(0,0)$.</p>
<p>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
$$f(x,v)-f(x,0) = \nabla f_2(x_0,0)\cdot v + o(\|v\|) \qquad ((x,v)\to(x_0,0)).$$ </p>