0
$\begingroup$

I'm reading this lecture note about differentiability of convex function.


Let $X$ be a normed space, $A \subset X$ an open set, $f: A \rightarrow \mathbb{R}$ a function, and $a \in A$ a point. For a "direction" $v \in X$ (not necessarily of norm one), we shall consider the right directional derivative $f_{+}^{\prime}(a, v)$, the left directional derivative $f_{-}^{\prime}(a, v)$, and the (bilateral) directional derivative $f^{\prime}(a, v)$, which are defined by: $$ \begin{aligned} f_{+}^{\prime}(a, v) &=\lim _{t \rightarrow 0+} \frac{f(a+t v)-f(a)}{t} \\ f_{-}^{\prime}(a, v) &=\lim _{t \rightarrow 0-} \frac{f(a+t v)-f(a)}{t} \\ f^{\prime}(a, v) &=\lim _{t \rightarrow 0} \frac{f(a+t v)-f(a)}{t} \end{aligned} $$

We shall say that $f$ is:

  • Gâteaux differentiable at $a$ if there exists $x^{*} \in X^{*}$ such that $f^{\prime}(a, v)=x^{*}(v)$ for each $v \in X$ (that is, $f^{\prime}(a, \cdot)$ is everywhere defined, real-valued, linear and continuous);
  • Fréchet differentiable at $a$ if there exists $x^{*} \in X^{*}$ such that $$ \lim _{\|h\| \rightarrow 0} \frac{f(a+h)-f(a)-x^{*}(h)}{\|h\|}=0 . $$ The functional $x^{*}$ is called the Gâteaux/Fréchet differential (or derivative) of $f$ at $a$, and it is denoted by $f^{\prime}(a)$.

Observation 0.3. The following assertions are equivalent: (i) $f$ is Fréchet differetiable at $a$; (ii) there exists $x^{*} \in X^{*}$ such that $$ f(a+h)=f(a)+x^{*}(h)+o(\|h\|) \quad \text { as } h \rightarrow 0 ; $$ (iii) $f$ is Gâteaux differentiable at $a$ and the limit $$ \lim _{t \rightarrow 0} \frac{f(a+t v)-f(a)}{t}=f^{\prime}(a)(v) $$ is uniform for $\|v\|=1$.

Could you explain what it means by "the limit ... uniform for $\|v\|=1$"?

$\endgroup$
2
  • 1
    $\begingroup$ I think it means that for every $\varepsilon > 0$, there is some $\delta > 0$ small enough such that, for every $v$ satisfying $\|v\|=1$ one has $\big|\frac{f(a+tv)-f(a)}{t} - f'(a,v) \big| \le\varepsilon$ whenever $0<|t|\le \delta$. $\endgroup$
    – Célestin
    May 30, 2022 at 15:57
  • $\begingroup$ Thank you so much @Célestin. You're right. I have found related information here. $\endgroup$
    – Akira
    May 31, 2022 at 7:28

1 Answer 1

0
$\begingroup$

As @Célestin pointed out in a comment and as in here, the limit being uniform means

$\forall \varepsilon > 0, \exists \delta>0, \forall t \text{ s.t. } |t| < \delta, \forall v \text{ s.t. } \|v\|=1$, we have $$ \left | \frac{f(a+t v)-f(a)}{t} - f^{\prime}(a)(v) \right | < \varepsilon. $$

I also present the proof of the equivalence below.


  • (i) $\implies$ (ii)

Assume the Fréchet differential of $f$ at $a$ is $x^* \in X^*$. It is well-known that $f'(a) = x^*$. Assume the contrary that there is $\varepsilon>0$ such that for each $n \in \mathbb N$, there is $(t_n, v_n)$ such that $|t_n| < 1/n, \|v_n\|=1$, and $$ \left | \frac{f(a+t_n v_n)-f(a)}{t_n} - f^{\prime}(a)(v_n) \right | \ge \varepsilon. $$

We have $t_nv_n \to 0$, so by definition of Fréchet derivative, we get $$ \lim _{n} \frac{f(a+t_n v_n)-f(a)-x^{*}(t_nv_n)}{\|t_nv_n\|} = \lim_n \frac{f(a+t_nv_n)-f(a)- x^*(t_nv_n)}{t_n} \frac{t_n}{|t_n|} =0 . $$

Then $$ \lim_n \left [\frac{f(a+t_nv_n)-f(a)}{t_n} - x^*(v_n) \right ]=0. $$

This is a contradiction.

  • (ii) $\implies$ (i)

Assume that the Gâteaux differential of $f$ at $a$ is $f'(a) \in X^*$. Let $(h_n) \subset X$ such that $h_n \to 0$. We want to prove $$ \lim_n \frac{f(a+h_n)-f(a)-f'(a)(h_n)}{\|h_n\|} = 0 . $$

Let $v_n := h_n / \|h_n\|$ and $t_n :=\|h_n\|$. Then $t_n \to 0^+$ and $\|h_n\|=1$. The problem reduces to prove $$ \lim_n \left [\frac{f(a+t_n v_n)-f(a)}{t_n} - f'(a)(v_n) \right ] = 0 . $$

On the other hand, then the limit $$ \lim_{t \to 0} \frac{f(a+t v_n)-f(a)}{t}=f^{\prime}(a)(v_n) $$ is uniform for all $n$, i.e., $\forall \varepsilon > 0, \exists \delta>0, \forall t_m \text{ s.t. } |t_m| < \delta, \forall v_n$, we have $$ \left | \frac{f(a+t_m v_n)-f(a)}{t_m} - f^{\prime}(a)(v_n) \right | < \varepsilon. $$

This in turn implies $\forall t_m \text{ s.t. } |t_m| < \delta$, we have $$ \left | \frac{f(a+t_m v_m)-f(a)}{t_m} - f^{\prime}(a)(v_m) \right | < \varepsilon. $$

This completes the proof.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .