I am reading a book on differential calculus in normed linear space and something escapes me about how to define differentiability.
I consider $f : U\to E$ where U is an open subset of $V$ and $V$ and $E$ are two Banach spaces.
We say that $f$ is differentiable on $U$ if there exists $\Lambda\in L(V,E)$ such that $f(x+h) - f(x) - \Lambda(h) = o(h)$ . We want to show that this implies continuity
Consider $\epsilon>0$, take $0<\delta<1$ such that if $\lVert h\rVert_{V}\leq\delta$ then $x+h\in U$ and $\lVert f(x+h) - f(x) - \Lambda(h)\rVert_{E}\leq \lVert h\rVert_{V}\quad (1) $
This requirement seems clear for me. In fact it relies on the notion of tangent in Banach space and it is explained in the text as follows : $f$ and $g$ are said to be tangent to one another at $x_0$ if $\forall\epsilon>0, \exists r>0 : \lVert h\rVert < r \implies \lVert f(x_0+h) - g(x_0+h)\rVert\leq\epsilon\lVert h\rVert$
And in the definition of the differentiability introduce in the text this "tangent property" is satisfied so this inequality makes sense.
Now, here is the point I don't understand after $(1)$ : Consequently, $\lVert f(x+h) - f(x)\rVert_{E}\leq (\lVert\Lambda\rVert_{E} +1)\lVert h\rVert_{V}$
And I don't see where does it come from, someone has an idea ? Thank you a lot !
Using $\lVert f(x+h) - f(x) - \Lambda(h)\rVert_{E}\leq \lVert h\rVert_{V}\quad (1) $ and triangle inequality we get $\lVert f(x+h) - f(x) \leq \lVert h\rVert_{V} +\lVert\Lambda(h)\rVert_{E}$ and $\|\Lambda (h)\|_E\leq \|\Lambda \|_E \|h\|_V$.