The following theorem is in Drabek, Milota's Nonlinear Analysis. Like Drabek and Milota, I won't assume a priori Gateaux differentials are continuous nor linear.

Theorem. Let $X,Y$ be normed spaces, and $f: X \rightarrow Y$ a map (perhaps not continuous). Fix $a,b \in X$. Suppose that the Gateaux differential $df(a+t(b-a); b-a)$ exists for all $t \in [0,1]$. Then, we have the estimate $$ ||f(b)-f(a)||_Y \leq \sup_{t \in [0,1]} ||df(a+t(b-a) ; b-a) ||_Y.$$

The proof uses the Hahn-Banach Theorem (or a corollary thereof) by taking a linear functional $\varphi \in X^*$ such that $||\varphi||_{X^*}=1$ and $\varphi(f(b)-f(a)) = ||f(b)-f(a)||$.

I would like to know if the use of the Hahn-Banach Theorem - or more generally, the Axiom of Choice - is needed to prove the theorem (that is, does there exist a proof that does not make use of choice principles)? I would also be interested in weaker choice principles, such as dependent or countable choice.

I have also tried to modify the argument I saw in Cartan's Differential Calculus in the proof of the Mean Value Theorem for differentiable maps $[a,b] \rightarrow Y$, $Y$ a normed space (the argument requires no choice principles); see Theorem 3.1.1, page 37-39 in Cartan. Somewhat in analogy to Cartan's argument, I have tried to define

$$E= \{t \in [0,1] | \text{ for every neighbourhood } U \text{ of } 0, \text{ there is } p \in U \text{ such that } |p| ||f(b)-f(a)||_Y > ||f(a+t(b-a) + p(b-a)) - f(a+t(b-a)) ||_Y \}.$$

Then perhaps one could show $E$ is open in $\mathbb R$ and one notes that $\inf E \notin E$. I am not sure exactly how to proceed to obtain a contradiction (as in Cartan's proof), however. My ideas for this attempt aren't well-formed, but I thought it would be useful to mention Cartan's proof.

Perhaps this stackexchange question is also useful: Does one need the Hahn-Banach theorem to prove the mean value inequality for maps into a normed space?, though I'm interested in the version of the Mean Value theorem involving Gateaux derivatives as stated above.

Thanks in advance.

Edit: for reference, I'll include Cartan's Theorem here and a sketch of the proof. (The theorem is stated for Banach $Y$, but completeness is in fact unnecessary). It is the proof of Theorem 3.1.1 I have tried to modify to my particular case, but it seems that I am unsure how to proceed.

Theorem (Cartan, 3.1.1). Let $Y$ be normed, let $f:[c,d] \rightarrow Y, g:[c,d] \rightarrow \mathbb R$ be maps. Suppose that the right derivatives $f'_{+}, g'_{+}$ exists at all points in $(c,d)$, and assume that $||f'_{+}(x)|| \leq g'_{+}(x)$ for all $x \in (c,d)$. Then, $$||f(d)-f(c)|| \leq g(d)-g(c).$$

Sketch: Following Cartan, one fixes $\varepsilon>0$ and puts $$U = \{x \in [c,d] | ||f(x)-f(c)|| > g(x)-g(c)+ \varepsilon(x-c) \}.$$ Aiming for a contradiction, suppose $U$ is not empty. Then, $U$ is open and let us define $\alpha= \inf U$; then, $\alpha \notin U$, and in fact, $\alpha>c$. By the existence of the right derivatives, there is a $\delta>0$ such that for any $\alpha<t<\alpha+\delta$, one has $$ - \frac{\varepsilon}{2} + \left | \left | \frac{f(t)-f(\alpha)}{t-\alpha} \right | \right | \leq || f'_{+}(\alpha) || \leq g'_{+}(\alpha) < \frac{g(t)-g(\alpha)}{t-\alpha} + \frac{\varepsilon}{2}.$$ It follows that $||f(t)-f(\alpha)|| < g(t)-g(\alpha) + \varepsilon(t-\alpha)$ for every $\alpha<t<\alpha+\delta$.

Since $\alpha \notin U$, we also have $||f(\alpha)-f(c)|| \leq g(\alpha)-g(c)-\varepsilon(\alpha-c)$. If $t \in (\alpha, \alpha+\delta)$, then \begin{align*} ||f(t)-f(c)|| &\leq ||f(t) - f(\alpha)|| + ||f(\alpha)-f(c) || \\ & \leq (g(t)-g(\alpha) + \varepsilon(t-\alpha) ) + (g(\alpha)-g(c)+ \varepsilon(\alpha-c) \\ &= g(t)-g(c)+\varepsilon(t-c). \end{align*}

So, $(\alpha, \alpha+\delta) \cap U = \emptyset$ and supposedly $\alpha = \inf U$, which is impossible. So in fact $U=\emptyset$, and thus $b \in U^c$, from which it follows $||f(b)-f(a)|| \leq g(b)-g(a)+\varepsilon(b-a)$ and $\varepsilon$ was arbitrary. $\blacksquare$

Remark: the Theorem remains true if right differentiability is assumed for all but countably many points in $(c,d)$.

Corollary (Cartan, 3.3.1). Let $X,Y$ be normed, let $f:U \rightarrow Y $ be a map (where $U \subseteq X$ is open and not empty) and fix $x,y \in U$. Suppose the line segment between $x$ and $y$ is contained in $U$, and suppose that $f$ is Frechet differentiable on $U$. Then, $$||f(x)-f(y)||_Y \leq \sup ||f'(\zeta)||_Y ||x-y||_X,$$ where the sup is taken over the line segment between $x$ and $y$.

(The corollary follows from applying Theorem 3.1.1 to $g:= f \circ \psi$, where $\psi:[0,1] \rightarrow X, t \mapsto tx+(1-t)y$).

  • $\begingroup$ Do I get it correctly that the only difference between the Theorem in your question and Cartan's Theorem 3.1.1 is that you do not have the continuity of $f$? $\endgroup$
    – gerw
    Jan 16, 2019 at 7:45
  • $\begingroup$ What is precisely your definition of Gateaux derivative? Does it involve a one-sided or a two-sided limit? $\endgroup$
    – gerw
    Jan 16, 2019 at 7:51
  • $\begingroup$ Let $f:X \rightarrow Y$ be a map of real or complex normed spaces, and let $x_0, h \in X$. Then, if the limit $\lim_{t \to 0} \frac{f(x_0+th)-f(x_0)}{t}$ exists, I will either say the Gateaux differential exists or that the directional derivative of $f$ at $x_0$ in direction $h$ exists (I use both terms synonymously), and I denote this limit by $df(x_0 ; h)$. The limit here is a two sided limit if $X$ is a real vector space. I will state it explicitly if I assume that the map $h \mapsto df(x_0; h)$ is continuous or linear, but I don't take this as part of the definition. $\endgroup$ Jan 16, 2019 at 23:44
  • $\begingroup$ (Comment continued) Cartan's Theorem is a result for differentiable maps on the compact interval $[c,d] \subseteq \mathbb R$ that map into a normed space. Cartan proves as a corollary of 3.1.1 the mean value theorem for Frechet derivatives (Thm. 3.3.1 I believe). Yes, the theorem in my question is only concerned with the line segment with endpoints $a,b \in X$, and yes I do not assume continuity of $f$. Though I am unsure how or if it is possible to give a choiceless proof. I'm not sure if I answered your question, so don't hesitate to let me know if you need more clarification. $\endgroup$ Jan 16, 2019 at 23:50

1 Answer 1


Let us define the function $\varphi : \mathbb{R} \to Y$ via $$ \varphi(t) := f( a + t \, (b-a)).$$

Since the Gateaux differential $df ( a + t \, (b-a); b-a)$ exists for all $t \in [0,1]$, we have that $\varphi$ is differentiable at $t \in [0,1]$ with $$ \varphi'(t) = df( a + t \, (b-a); b-a).$$ This follows directly from the definition. From this, we infer that $\varphi$ is continuous on $[0,1]$. Now, you can apply Theorem 3.1.1 from Cartan.


This site is temporarily in read-only mode and not accepting new answers.

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