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.

  • $\begingroup$ 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

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 $\vert t\vert<1/R$, $x\in D$, $$ \lim_{h\rightarrow 0} \frac1h\bigl(f(t,x+h)-f(t,x)\bigr) =\sum_{k\ge 0}f'_k(x) t^k. $$

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  • $\begingroup$ 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

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