# Don't follow proof that Hadamard differentiable implies compactly differentiable

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?

• 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. – Oliver Diaz Dec 28 '20 at 18:16

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 $0$.

• One also needs to use the fact that $f'_x$ is sequentially continuous. – Oliver Diaz Dec 28 '20 at 21:00
• How so? Is there an error in my proof? – Eric Wofsey Dec 28 '20 at 21:05
• 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. – Oliver Diaz Dec 28 '20 at 21:18
• I was just answering the question in the OP, which was about this specific step of the proof. – Eric Wofsey Dec 28 '20 at 21:39

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 $$0 \begin{align} \frac{f(x+th)-f(x)}{t}-f'_x(h)\in W,&\qquad (t,h)\in(0,\delta)\times H\tag{4}\label{four}\\ f'_x(h)-f'_x(h_n)\in W,&\qquad n\geq n_0\tag{5}\label{five}\\ t_n\in(0,\delta),&\qquad n\geq n_0\tag{6}\label{six} \end{align} Combining \eqref{four},\eqref{five}, and \eqref{six}, we obtain that for all $$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$$.

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 $$0 such that $$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.

• Could you please give a hint to prove your two remarks? – StopUsingFacebook Feb 23 at 9:33
• @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. – Oliver Diaz Feb 23 at 15:41
• @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 $t$. – Oliver Diaz Feb 23 at 15:46
• Thanks @Oliver Diaz – StopUsingFacebook Feb 24 at 9:25