# $f$ 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 $\|v\|=1$

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 $$\|v\|=1$$"?

• 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 $0<|t|\le \delta$. May 30, 2022 at 15:57
• Thank you so much @Célestin. You're right. I have found related information here. May 31, 2022 at 7:28

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

• (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 $$\left | \frac{f(a+t_m v_m)-f(a)}{t_m} - f^{\prime}(a)(v_m) \right | < \varepsilon.$$