Let $X$ be a Banach space. $F$ is Gateux-differentiable at all points $x\in X$. For any pair of points $x,x+t\in X$ there exists $\beta\in(0,1)$ such that $$ F(x+t) - F(x) = DF(x+\beta t)t $$
As a demonstration I define a function $$ \tau : [0,1] \longrightarrow \mathbb{R}, \quad \alpha \mapsto \tau(\alpha) = F(x+\alpha t) $$ I want to try $\tau$ which is continuous in $[0,1]$. I tried using the definition but I had problems. Could you suggest some other way?
Sea $X$ un espacio de Banach. $F$ es Gateux-diferenciable en todo punto $x \in X$. Para cualquier par de puntos $x, x+t \in X$ existe $\beta \in (0,1)$ tal que $$ F(x+t) - F(x) = DF(x+\beta t)t $$
Para la demostración definí una función $$ \tau : [0,1] \longrightarrow \mathbb{R}, \quad \alpha \mapsto \tau(\alpha) = F(x+\alpha t) $$ Quiero probar $\tau$ que es continua en $[0,1]$. intente por definición pero tuve problemas, ¿podrían sugerirme algún otro camino?.
$\tau$ is actually not only continuous but even differentiable. Just write down the definition of the derivative of $\tau$ at $\alpha$ and of the Gateaux derivative of $F$ at $x+\alpha t$ and compare.