Let $\phi : [a,b] \to \mathbb{R}^m$ be continues and differentiable on $( a,b)$ and let $\lVert x\rVert = \sqrt{\langle x,x\rangle}$ the euclidean norm in $\mathbb{R}^m$, show that $\lVert\phi(b)-\phi(a)\rVert \leq \sup_{t \in (a,b)} \lVert\phi'(t)\rVert (b-a)$ ?
I did used Lagrange mean value theorem on $g(t) = \langle\phi(t),\phi(b)-\phi(a)\rangle$ but ended up with two sup on the interval $(a,b)$
Any hints or proof or ideas are appreciated
You have just to apply the Mean Value Theorem to your function $$ g(t) := \langle \phi(t), \phi(b) - \phi(a)\rangle, \qquad t \in [a,b]. $$ Namely, $$ g(b) - g(a) = \|\phi(b) - \phi(a)\|^2, $$ whereas, for $t\in (a,b)$, by the Cauchy-Schwarz inequality we get $$ g'(t) = \langle \phi'(t), \phi(b) - \phi(a)\rangle \leq \|\phi'(t)\|\, \|\phi(b)-\phi(a)\| \leq \sup_{s\in (a,b)}\|\phi'(s)\|\, \|\phi(b) - \phi(a)\|. $$ On the other hand, by the MVT, there exists $\xi \in (a,b)$ such that $$ g(b) - g(a) = g'(\xi) (b-a) $$ so that $$ \|\phi(b) - \phi(a)\|^2 = g'(\xi) (b-a) \leq \sup\|\phi'(t)\|\, \|\phi(b) - \phi(a)\|\, (b-a). $$