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Let $X$ and $Y$ be Banach spaces.

Definition: A function $f:X\rightarrow Y$ is called Hadamard differentiable at $x\in X$ tangentially to $U\subseteq X$ iff $x\in U$ and there exists a continuous linear function $f'_x:U\rightarrow Y$ such that for all $u\in U$ it holds that $$\left\Vert\frac{f(x + \delta_nu_n)-f(x)}{\delta_n} - f'_x(u)\right\Vert_Y\rightarrow 0$$ as $n\rightarrow\infty$ for every sequence $(u_n)_{n\in\mathbb N}$ in $X$ converging to $u$ and every sequence $(\delta_n)_{n\in\mathbb N}$ in $\mathbb R$ converging to $0$.

Problem: In my case, $X = \textit{BV}([a,b])\times\textit{BV}([a,b])$, where $\textit{BV}([a,b])$ is the set of all real-valued functions on $[a,b]$ that are of bounded variation, $Y = \mathbb R$, and $$f(F,G) = \int_a^b F\,\mathrm dG.$$ Here the integral is the Lebesgue-Stieltjes integral; note that since real-valued functions on a compact set that are of bounded variation are also bounded and real-valued functions are measurable, $f$ is well-defined. Is $f$ Hadamard differentiable (maybe tangentially to some subset $U$) for each $(F,G)\in X$?

I know that if $X = \big(\textit{BV}([a,b])\cap\textit{D}([a,b])\big)\times\big(\textit{BV}([a,b])\cap\textit{D}([a,b])\big)$, where $D([a,b])$ is the space of real-valued cadlag functions on $[a,b]$, then $f$ is Hadamard differentiable with derivative $$f_{(F,G)}'(A,B) = B(b)F(b) - B(a)F(a) - \int_a^b B_{-}\,\mathrm dF + \int_a^b A\,\mathrm dG,$$ where $B_{-}$ is the left-continuous version of the right-continuous function $B$. <Edit: In an earlier version of this post, there was an error. I wrote $B(F(a))$ instead of $B(a)F(a)$ (and similar for $b$). The error is also in the comments. I hope this error didn't cause any confusion.>

However, since I am operating on the larger space $\textit{BV}([a,b])\times\textit{BV}([a,b])$, I suppose that this result no longer holds. Therefore my question: is there any hope that $f$ is still Hadamard differentiable on this larger space? Unfortunately, I have no idea how to tackle this problem. So any help is appreciated.

The functions I want apply $f$ to are the of the form $\frac{L + R}2$, where $L$ and $R$ are left- and right-continuous functions with the same points discontinuities (i.e., if $L$ is discontinuous at $t\in[a,b]$, then so is $R$). So I have some form of continuity and an at most countable set of discontinuities. Maybe that's enough to have Hadamard differentiability tangentially to the set of these functions?

Alex Ravsky
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  • I'm not too familiar with the properties of these spaces, but the intuitive answer using linearity of integrand and the differential term in the L-S integral seems like $f'_{(F,G)}(A, B)$ should be $\int_a^b F dB + \int_a^b A dG$. But this doesn't seem to match the result you mentioned in the smaller space. Is there something about this guess that fails (e.g. continuity)? – jet457 Oct 12 '23 at 05:26
  • if there is continuity, then indeed $\int_a^b F,\mathrm dB = B(F(b)) - B(F(a)) - \int_a^b B,\mathrm dF$ (integration by parts).

    PS: this is not related to my question, but proving that the derivative is given by the above expression was very tedious without any intution (imo). Since you are talking about intution here, what's the intuition for this result?

    – Syd Amerikaner Oct 12 '23 at 14:39
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    Aah I see, so IBP is the connection. My intuition is coming from if you take $A,B$ in your space and $\delta>0$, then you should have $\int_a^b (F+\delta A) d (G + \delta B) = \int_a^b F d G + \delta \int_a^b A d G + \delta \int_a^b Fd B + \delta^2 \int_a^b Ad B$. Then taking the difference with $f(F,G)$, the "linear" part remaining is $\delta \left[ \int_a^b A d G + \int_a^b Fd B \right]$. – jet457 Oct 12 '23 at 20:26

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