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A function $f:X \to Y$ between Banach spaces is said to be compactly differentiable if there is a function $f'_x:X\rightarrow Y$ such that $$\lim_{t \to 0} \frac{f(x+th) - f(x) }{t} -f'_x(h) =0$$ where the limit holds uniformly in $h \in K \subset X$, where $K$ is a compact set.

We say $f$ has a Hadamard derivative if $$\lim_{n \to \infty} \frac{f(x+t_nh_n) - f(x)}{t_n} - f'_x(h)=0$$ holds where $h_n \to h$ and $t_n \to 0$ are any sequences.

There is a result (Prop. 3.3 of this ) that states Hadamard diff. implies compact diff. The proof starts as follows.

Let $f$ be Hadamard diff., then in order to show it's also compact diff. it is enough to show that for any compact set $S$ and sequences $h_n \in S$ and $t_n \to 0$, we have $$\lim_{n \to \infty} \frac{f(x+t_nh_n) - f(x)}{t_n} -f'_x(h_n) = 0.$$ The rest is omitted.

Question: how is this limit enough to show it? I don't see where the uniformity comes in at all. Could someone explain it please?

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  • $\begingroup$ The paper of Shapiro quoted in the OP is more general as he does not make any assumptions on the topology of the spaces $X$ and $Y$, other that they are liner topological spaces. $\endgroup$ – Oliver Diaz Dec 28 '20 at 18:16
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Suppose that $f$ were not compactly differentiable. Then there would be some compact $S$ such that $\frac{f(x+th) - f(x) }{t} -f'(x)(h)$ does not converge uniformly to $0$ on $S$. This means that for some $\epsilon>0$, for each $\delta>0$ there exist $t$ such that $|t|<\delta$ and $h\in S$ such that $\left\|\frac{f(x+th) - f(x) }{t} -f'(x)(h)\right\|>\epsilon$. Letting $\delta$ range over some sequence converging to $0$, we get a sequence $(t_n)$ converging to $0$ and a sequence $(h_n)$ in $S$ such that $\left\|\frac{f(x+t_nh_n) - f(x) }{t_n} -f'(x)(h_n)\right\|>\epsilon$ for each $n$. In particular, the sequence $\frac{f(x+t_nh_n) - f(x) }{t_n} -f'(x)(h_n)$ does not converge to $0$.

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  • $\begingroup$ One also needs to use the fact that $f'_x$ is sequentially continuous. $\endgroup$ – Oliver Diaz Dec 28 '20 at 21:00
  • $\begingroup$ How so? Is there an error in my proof? $\endgroup$ – Eric Wofsey Dec 28 '20 at 21:05
  • $\begingroup$ It's only to complete the proof you sketched. The negation of compact directional differentiability implies $\|t^{-1}_n(f(+t_nh_n)-f(h_n))-f'_x(h_n)\|$ along a subsequence $(t_n,h_n)\in (0,\infty)\times S$ with $t_n\rightarrow0$ as pointed out. Then one may assume $t_n\rightarrow0$ and $h_n\rightarrow h$ for some $h\in S$. The continuity of $f'_x$ then allows to change the $f'_x(h_n)$'s for $f'_x(h)$ for all $n$ large enough. I wrote a proof that is close to your but also closer to the stamens of the paper mentionsed by the OP. If you have the time to comment on it, I would appreciate it too. $\endgroup$ – Oliver Diaz Dec 28 '20 at 21:18
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    $\begingroup$ I was just answering the question in the OP, which was about this specific step of the proof. $\endgroup$ – Eric Wofsey Dec 28 '20 at 21:39
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This is an old problem but since the source of the OP states a more general result that the stated in the OP, I think it is worthwhile to present the complete result and a proof as presented by the source:


In what follows, $X$ and $Y$ are linear topological spaces (linear spaces equipped with a topology in which addition and scalar product are continuous operations, and singletons are closed), and $f$ is a function from $x+U$ to $Y$, where $U$ is some neighborhood of $0\in X$.

Theorem: Hadamard directional differentiability implies compact directional differentiability. Conversely, if $f$ admits a compactly directional derivative $f'_x$ at $x$, and if $f_x':X\rightarrow Y$ is sequentially continuous on $X$, then $f$ is directional differentiable in the sense of Hadamard.

Proof: The direct part (which is the one the OP is concerned) is based on the fact (see the comment section) that if $f$ has Hadamard derivative $f'_x$, the $f'_x:X\rightarrow Y$ is sequentially continuous. Assume $f$ is not compactly differentiable at $x$. This means that there exists a compact set $K\subset X$ and an open neighborhood $V$ of $0\in Y$ such that for any $n\in\mathbb{N}$, there is $(t_n,h_n)\in (0,\tfrac1n)\times K$ such that $$ \begin{align} \frac{f(x +t_nh_n)-f(x)}{t_n}-f'_x(h_n)\notin V\tag{0}\label{zero} \end{align} $$ Let $W$ be a symmetric neighborhood of $0\in Y$ such that $W+W\subset V$. By the compactness of $K$, we may assume that $h_n\xrightarrow{n\rightarrow\infty} h$ for some $h\in K$ (the fact that $X$ is $T_1$ separable linear space -hence Hausdorff- implies that every sequence in a compact subset of $X$ admits a convergent subsequence). As $f$ is Hadamard differentiable, we have that there is $n_0\in\mathbb{N}$ such that $$ \begin{align} \frac{f(x +t_nh_n)-f(x)}{t_n}-f'_x(h)&\in W \quad n\geq n_0\tag{1}\label{one}\\ f'_x(h_n) -f'_n(h)&\in W\quad n\geq n_0\tag{2}\label{two} \end{align} $$ It follows from \eqref{one} and \eqref{two} that \begin{align} \frac{f(x +t_nh_n)-f(x)}{t_n}-f'_x(h_n)&=\Big(\frac{f(x +t_nh_n)-f(x)}{t_n}-f'_x(h)\Big) +\\ &\quad\quad \Big(f'_x(h) -f'_n(h_n)\Big)\in W+W\subset V \end{align} for all $n\geq n_0$ in contradiction to \eqref{zero}.

Conversely, assume that $f$ admits a compactly directional derivative $f'_x:X\rightarrow Y$ which is sequentially continuous. Let $(t_n,h_n)\in (0,\infty)\times X$ a sequence such that $t_n\xrightarrow{n\rightarrow\infty}0$ and $h_n\xrightarrow{n\rightarrow\infty}h$. The set $H=\{h_n,h:n\in\mathbb{N}\}$ is a compact subset of $X$. For any open neighborhood $V$ of $0\in Y$ choose a symmetric neighborhood $W$ of $0\in Y$ such that $W+W\subset V$. By assumption, there are $\delta>0$ and $n_0\in\mathbb{N}$ such that for all $0<t<\delta$ $$ \begin{align} \frac{f(x+th)-f(x)}{t}-f'_x(h)\in W,&\qquad (t,h)\in(0,\delta)\times H\tag{4}\label{four}\\ f'_x(h)-f'_x(h_n)\in W,&\qquad n\geq n_0\tag{5}\label{five}\\ t_n\in(0,\delta),&\qquad n\geq n_0\tag{6}\label{six} \end{align} $$ Combining \eqref{four},\eqref{five}, and \eqref{six}, we obtain that for all $n\geq n_0$ $$\begin{align} \frac{f(x+t_nh_n)-f(x)}{t_n}-f'_x(h)&=\Big(\frac{f(x+t_nh_n)-f(x)}{t_n}-f'_x(h_n)\Big)+\\ &\quad\quad \Big(f'_x(h_n)-f'_x(h)\Big)\in W+W\subset V \end{align} $$ This shows that $f$ has Hadamard directional derivative $f'_x$.


Comments:

Definition 0: $f$ is directionally differentiable at $x$ in the sense of Gâteaux if there is a function $f'_x:X\rightarrow Y$ such that for any $h\in X$ \begin{align} \lim_{t\rightarrow0+}\frac{f(x+th)-f(x)}{t}-f'_x(h)=0 \end{align}

Definition 1: $f$ is directionally differentiable at $x$ in the sense of Hadamard, if there is a function $f'_x:X\rightarrow Y$ such that for any $h\in X$ and any $h_t\xrightarrow{t\rightarrow0+}h$
\begin{align} \lim_{t\rightarrow0+}\frac{f(x+th_t)-f(x)}{t}-f'_x(h)=0 \end{align} (Notice that $x+th_t\in x+U$ for all sufficiency small $t$.)

Remark 1a: It is a simple exercise to show that this definition is equivalent to saying that for any sequence $(t_n,h_n)\in (0,\infty)\times X$ such that $(t_n,h_n)\xrightarrow{n\rightarrow\infty}(0,h)$, \begin{align} \lim_{n\rightarrow\infty}\frac{f(x+t_nh_n)-f(x)}{t_n}-f'_x(h)=0 \end{align}

Remark 1b: When $X$ is a normed space, it is an easy exercise to show that differentiability at $x$ in the sense of Hadamard is equivalent to \begin{align} \lim_{(t,k)\rightarrow(0+,h)}\frac{f(x+tk)-f(x)}{t}-f'_x(h)=0 \end{align} for all $h\in X$.

Remark 1c: Clearly differentiability in the sense of Hadamard implies differentiability in the sense of Gateaux. Furthermore, the function $f'_x$ is positive homogeneous, that is $f'_x(\alpha h)=\alpha f'_x(h)$ for all $\alpha\geq0$ and $h\in X$. In most applications, it is also required in the definitions above that $f'_x$ be a linear bounded operator from $X$ and $Y$.

Definition 2: $f$ is compactly differentiable at $x$ if there is a function $f'+x:X\rightarrow Y$ such that for any compact $K\subset X$ \begin{align} \lim_{t\rightarrow0+}\frac{f(x+th)-f(x)}{t}-f'_x(h)=0 \end{align} uniformly on $K$. (Notice that since $K$ is compact, it is bounded and so, $x+tK\subset x +U$ for all $t>0$ small enough.)

The main result used on the proof above is the following

Lemma: If $f$ has directional derivative $f'_x$ at $x$ in the sense of Hadamard, then $f'_x:X\rightarrow Y$ is sequentially continuous.

Proof: Let $\{h_n:n\in\mathbb{N}\}$ be a sequence in $X$ that converges to some $h\in X$. For any open neohborhood $V$ of $0\in Y$, let $W$ be a symmetric open neighborhood of $0\in Y$ such that $W+W\subset V$. Since Differentiability in the sense of Hadamard implies differentiability in the sense of Gâteaux differentiability, we have that for any $n$ there is $0<t_n$ such that $t_n\xrightarrow{n\rightarrow\infty}0$ and \begin{align} \frac{f(x+t_nh_n)-f(x)}{t_n}-f'_x(h_n)\in W\tag{7}\label{seven} \end{align} Differentiability in the sense of Hadamard on its own implies that there is $n_0\in\mathbb{N}$ such that \begin{align} \frac{f(x+t_nh_n)-f(x)}{t_n}-f'_x(h)\in W,\qquad n\geq n_0,\tag{8}\label{eight} \end{align} Combining \eqref{seven} and \eqref{eight} we obtain for all $n\geq n_0$ \begin{align} f'_x(h_n)-f'_x(h)&=\Big(\frac{f(x+t_nh_n)-f(x)}{t_n}-f'_x(h)\Big) -\\ &\qquad\quad\Big(\frac{f(x+t_nh_n)-f(x)}{t_n}-f'_x(h_n)\Big)\in W+W\subset V \end{align} This proves that $f'_x$ sequentially differentiable.

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  • $\begingroup$ Could you please give a hint to prove your two remarks? $\endgroup$ – StopUsingFacebook Feb 23 at 9:33
  • $\begingroup$ @StopUsingFacebook: That definition 1 implies the statement of remark 1a is obvious: given a sequence $(t_n,\boldsymbol{h}_n)\xrightarrow{n\rightarrow\infty}(0+,\boldsymbol{0})$ define $\boldsymbol{h}_t=\boldsymbol{h}_n$ for $t_n\leq t<t_{n-1}$. The converse can be obtained by contradiction: suppose there is $(t,\mathbf{h}_t)\xrightarrow{t\rightarrow0+}(0,\mathbf{h})$ for which statement in definition 1 fails. That allows one to construct a sequence of the type of remark 1a. $\endgroup$ – Oliver Diaz Feb 23 at 15:41
  • $\begingroup$ @StopUsingFacebook: That the assertion in Remark 1b implies definition 1 is obvious. The converse can be by contradiction and using the equivalence In remark 1a. Th statement of remark 1c is obvious: take $\mathbf{h}_t=\mathbf{h}$ for all $t$. $\endgroup$ – Oliver Diaz Feb 23 at 15:46
  • $\begingroup$ Thanks @Oliver Diaz $\endgroup$ – StopUsingFacebook Feb 24 at 9:25

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