This limit converges to the partial derivative? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T11:00:01Z http://mathoverflow.net/feeds/question/55152 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55152/this-limit-converges-to-the-partial-derivative This limit converges to the partial derivative? Ferraiol 2011-02-11T18:51:04Z 2011-02-13T11:08:38Z <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#55198 Answer by Christian Blatter for This limit converges to the partial derivative? Christian Blatter 2011-02-12T12:37:36Z 2011-02-13T11:08:38Z <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 $&lt;\epsilon$ 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>