4
$\begingroup$

Given a real analytic family of Lipschitz continuous functions $f_t:\overline{U}\rightarrow\mathbb{R}^n$, $t\in\mathbb{R}$, with $U\subset \mathbb{R}^n$ some open and bounded domain. For each $t_0\in \mathbb{R}$ there exists $\epsilon >0$ and Lipschitz functions $f^k:\overline{U}\rightarrow\mathbb{R}^n$ such that for all $t\in (t_0-\epsilon, t_0+\epsilon)$, $x\in \overline{U}:$

$$f_t(x)=\sum_{k=0}^{\infty} f^k(x)(t-t_0)^k.$$

By Rademacher theorem, we can find a subset $D\subset U$ of full measure, such that the derivatives $D_x f^k$ exist for all $x\in D$ and $k=0,1,2,\ldots$

I am wondering, if the derivative $D_x f_t$ does necessarily exist for all $x\in D$ and $t\in (t_0-\epsilon, t_0+\epsilon).$ Is that conclusion valid?

Any help is very much appreciated.

$\endgroup$
  • $\begingroup$ You should add $t_0$ as a lower index of $f^k$, given that $f^k$ depends on it. $\endgroup$ – Alex M. Jul 9 '18 at 18:06
  • $\begingroup$ This is OK because the OP only cares about a neighborhood of $t_0$. $\endgroup$ – Fan Zheng Jul 9 '18 at 18:13
  • $\begingroup$ Sorry, I misunderstood your question and my argument is not correct. I will think about the problem. $\endgroup$ – Piotr Hajlasz Jul 9 '18 at 20:17
1
$\begingroup$

Let me assume that $ f(t,x)=\sum_{k\ge 0} f_k(x) t^k, \quad \vert x\vert \le 1, \quad \vert t\vert < 1, $ with $f_k$ Lipschitz-continuous with an $L^\infty$ norm on $\vert x\vert \le 1$ bounded above by 1 and $\Vert f'_k\Vert_{L^\infty}\le C_0 R^k$. Then each $f_k$ is a.e. differentiable, i.e. $\forall k, \exists D_k, \vert D_k^c\vert=0$, $f_k$ is differentiable on $D_k$ (the complement is taken in the unit ball in $x$). Defining $ D=\cap_{k\ge 0} D_k, $ gives that $\vert D^c\vert=0$ and all $f_k$ are differentiable on $D$. We have in the distribution sense on the open cube $\{\vert x\vert < 1, \vert t\vert <R^{-1}\}$ $$ \frac{\partial f}{\partial x} (t,x)=\sum_{k\ge 0}f'_k(x) t^k, $$ and for $h\not=0$, \begin{multline} \frac1h\bigl(f(t,x+h)-f(t,x)\bigr)=\sum_{k\ge 0}\frac1h\bigl(f_k(x+h)-f_k(x)\bigr) t^k \\=\sum_{k\ge 0}\frac1h\int_{x}^{x+h} \bigl(f'_k(y)-f'_k(x)\bigr) dy t^k +\sum_{k\ge 0} f'_k(x)t^k. \end{multline} Thanks to the Lebesgue Differentiation Theorem, each $\frac1h\int_{x}^{x+h} \bigl(f'_k(y)-f'_k(x)\bigr) dy$ goes to zero with $h$ for $x\in D$. We have also (for $h>0$, $\vert t\vert<1/R$) $$ \sum_{k\ge N}\frac1{h}\int_{x}^{x+h} \bigl\vert f'_k(y)-f'_k(x)\bigr\vert dy \vert t\vert^k\le \sum_{k\ge N} 2\Vert f'_k\Vert_{L^\infty}\vert t\vert^k\le 2C_0(R\vert t\vert)^N\frac{1}{1-R\vert t\vert}. $$ We get that for all $N\ge 0$, $$ \limsup_{h\rightarrow 0}\left\vert \frac1h\bigl(f(t,x+h)-f(t,x)\bigr) -\sum_{k\ge 0}f'_k(x) t^k\right\vert \le \frac{ 2C_0(R\vert t\vert)^N}{1-R\vert t\vert}, $$ which implies for $\vert t\vert<1/R$, $x\in D$, $$ \lim_{h\rightarrow 0} \frac1h\bigl(f(t,x+h)-f(t,x)\bigr) =\sum_{k\ge 0}f'_k(x) t^k. $$

$\endgroup$
  • $\begingroup$ Thank you for your support. I do not quite understand your answer though. How can I prove the following statement: $\forall t\in (t_0-\epsilon, t_0+\epsilon)$ the function $f_t:\overline{U}\rightarrow \mathbb{R}^n$ is differentiable $\forall x\in D$ (I mean differentiable w.r.t. to $x$)? For example, if $t=t_0$ then $f_{t_0}=f^0$ and thus $f_{t_0}$ is differentiable on $D$. What if $t\neq t_0$? $\endgroup$ – Oliver Watt Jul 10 '18 at 3:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.