Differentiability on normed spaces My Advanced Calculus homework have the following problem: define differentiability on normed spaces $E,F$. Show that if $U \subset E$ is open and $f: U \to F$ is differentiable at $x \in U$ then continuity of $f$ at $x$ is equivalent to continuity of the derivative (linear transformation) $f'(x): E \to F$.
Now, the definition of differentiability on normed spaces should be the same as the one on euclidian spaces: $f: U \subset E \to F$ is differentiable at $x \in U$ if there exists a linear transformation $T: E \to F$ such that:
\begin{eqnarray}
f(x+h) = f(x) + T(h) + r(h) \tag{1}\label{1}
\end{eqnarray}
where:
$$\lim_{h\to 0}\frac{r(h)}{||h||} = 0,$$ provided $x+h \in U$.
Now, write $\rho(h) = r(h)/||h||$ for every $h \neq 0$ and set $\rho(0) := 0$. Then, (\ref{1}) becomes:
\begin{eqnarray}
f(x+h) = f(x) + f'(x)(h) + \rho(h)||h|| \tag{2}\label{2}
\end{eqnarray}
where $\lim_{h\to 0}\rho(h)=0$ (I wrote $f'(x)$ instead of $T$). Thus, $\lim_{h\to 0}\rho(h)||h|| = 0$ and we have:
\begin{eqnarray}
f(x+h)-f(x) = f'(x)(h) \tag{3}\label{3}
\end{eqnarray}
Now, if we write $x+h = y$, (\ref{3}) becomes
\begin{eqnarray}
f(y) - f(x) = f'(x)(y)-f'(x)(x) \Rightarrow \lim_{y\to x}f(y) - f(x) = 0 \iff \lim_{y\to x}f'(x)(y)-f'(x)(x) = 0 \tag{4}\label{4}
\end{eqnarray}
Question: Is (\ref{4}) sufficient to prove the second statement of the problem, i.e. that continuity of $f$ at $x$ is equivalent to continuity of the derivative $f'(x): E \to F$?
 A: It is sufficient. Linear operators $T: E \rightarrow F$ are continuous everywhere iff they are continuous in one point $x \in E$: Let $T$ be continuous in $x \in E$. Now let $z \in E$ be arbitrary and $z_n \rightarrow z$ for some sequence $(z_n)_{n \in \mathbb{N}} \subseteq Z$. Then $z_n+x-z \rightarrow x$. So:
$$
\lvert T(z_n) - T(z) \rvert = \lvert T(z_n +x-z) -T(x) \rvert \rightarrow 0
$$
We used linearity and then continuity in $x$.
Your other arguments are correct, too.
A: $(3)$ is an equation containing the variables $x$ and $h$. If it were true for all $h$, then subtraction of $(3)$ from the correct equation $(2)$ would give $r(h)= \rho(h)/\lVert h \rVert = 0$ for all $h$. This is true if and only if $f(x + h) = f(x) + T(h)$ which is a very special case ($f$ is an affine map). In other words, $(3)$ and $(4)$ are in general false. This does not matter, you do not need $(4)$. Since $r(h) \to 0$ as $h \to 0$, we see that $\lim_{h \to 0} f(x+h) = f(x)$ if and only if $\lim_{h \to 0} T(h) = 0$. But $\lim_{h \to 0} f(x+h) = f(x)$  means that $f$ is continuous at $x$ and $\lim_{h \to 0} T(h) = 0$ means that $T$ is continuous at $0$ which is equivalent to the continuity of $T$ (since $T$ is linear).
