Explicit exemple of a derivative which changes when we change the norm I know that if $E$ and $F$ are finite-dimensional Banach spaces and $f:E\to F$ is differentiable, then $\mathrm{d}f$ does not depend on the choice of norms in $E$ and $F$ (since all the norms are equivalent). However, that does not seem to be the case when $E$ or $F$ are infinite-dimensional.
I would like to find an explicit exemple of infinite-dimensional Banach spaces $E,F$ and a function $f:E\to F$ such that $\mathrm{d}f$ is different when we choose one norm for $E,F$ when compared to another norm in $E,F$. It would be nice too to see an exemple where the $f$ is differentiable for one norm but not differentiable for another.
 A: First, note that if you are only interested in what the differential $D f$ is
(and not in whether $f$ is differentiable to begin with),
then the norm on $E$ is immaterial.
Indeed, you have
$$
  D f (x_0) v
  = \lim_{t \to 0} \frac{f(x_0 + t v) - f(x_0)}{t} \, ,
$$
which is a limit in $F$, completely independent of the topology on $E$.
Next, let us see that if $\|\cdot\|$ and $|\cdot|$ are two norms on $F$
such that $F$ is complete for any of the two norms
and such that the two norms are not equivalent, then there is a function
$g : (-1,1) \to F$ such that $g$ is differentiable at $0$ for both choices of the norm,
but with different differentials.
For brevity, let us write $F_1$ for the vector space $F$ equipped with the norm $\|\cdot\|$,
and $F_2$ for the vector space $F$ equipped with the norm $|\cdot|$.
The following lemma will be crucial:
Lemma. There is a sequence $(x_n)_n$ in $F$ and $x,y \in F$ with $x \neq y$
such that $\|x_n - x\| \to 0$ and $|x_n - y| \to 0$.
Proof. Assume towards a contradiction that the claim fails.
Consider the linear operator
$$
  T : F_1 \to F_2, x \mapsto x \, .
$$
I claim that $T$ has closed graph.
Indeed, let $(x_n, y_n)_{n \in \mathbb{N}} \subset \mathrm{graph}(T) \subset F_1 \times F_2$
be a sequence in the graph of $T$, with $(x_n, y_n) \to (x,y) \in F_1 \times F_2$.
Since $(x_n, y_n) \in \mathrm{graph}(T)$, we have $x_n = y_n$ for all $n \in \mathbb{N}$,
and thus $\|x_n - x\| \to 0$ and $|x_n - y| = |y_n - y| \to 0$.
Since we assume that the claim is false, we must have $x = y$,
and thus $(x,y) \in \mathrm{graph}(T)$, as desired.
By the closed graph theorem, $T$ is thus a bounded linear operator.
Clearly, $T$ is bijective, so that $T^{-1}$ is also a bounded operator,
by the bounded inverse theorem.
Since $T$ and $T^{-1}$ are bounded, the norms $\|\cdot\|$ and $|\cdot|$ are equivalent,
contradicting our assumption that they are not.$\square$
Given a sequence $(x_n)_{n \in \mathbb{N}}$ and $x,y$ as provided by the lemma,
we can now construct the map $g$.
First, define
$$
  f_1 : (-1,1) \to F_1,
        t \mapsto \begin{cases}
                    x_{n+1} + \frac{|t| - (n+1)^{-1}}{n^{-1} - (n+1)^{-1}} \cdot (x_n - x_{n+1}),
                    & \text{if } |t| \in [(n+1)^{-1}, n^{-1}) \text{ for some } n \in \mathbb{N}, \\
                    x, & \text{if } t = 0 \, .
                  \end{cases}
$$
It is not hard to check that $f_1$ is continuous (but in fact, we will not even need this).
Furthermore, for $|t| \in [(n+1)^{-1}, n^{-1})$,
we have $\gamma := \frac{|t| - (n+1)^{-1}}{n^{-1} - (n+1)^{-1}} \in [0,1]$ and thus
$$
  \|f_1 (t) - f_1(0)\|
  = \| \gamma \cdot (x_n - x) + (1-\gamma) \cdot (x_{n+1} - x) \|
  \leq \max \{ \|x_n - x\| , \|x - x_{n+1}\| \} \, ,
$$
which goes to zero as $t \to 0$, since then $n \to \infty$.
Thus, $g_1$ is continuous at $0$.
In exactly the same way, one sees that
$$
  f_2 : (-1,1) \to F_2 ,
        t \mapsto \begin{cases}
                    x_{n+1} + \frac{|t| - (n+1)^{-1}}{n^{-1} - (n+1)^{-1}} \cdot (x_n - x_{n+1}),
                    & \text{if } |t| \in [(n+1)^{-1}, n^{-1}) \text{ for some } n \in \mathbb{N}, \\
                    y, & \text{if } t = 0
                  \end{cases}
$$
is continuous at $0$.
Thus, if we define $g_1 : (-1,1) \to F_1, t \mapsto t \cdot f_1 (t)$, then $g_1$ is differentiable
at $t = 0$, with $D g_1 (0) = x$, since
$$
  \lim_{t \to 0} \frac{g_1(t) - g_1 (0)}{t}
  = \lim_{t \to 0} \frac{t \cdot f_1(t)}{t}
  = \lim_{t \to 0} f_1 (t)
  \to f_1 (0)
  = x \, ,
$$
with convergence of the limit in $F_1$.
In precisely the same way, we see that $g_2 : (-1,1) \to F_2, t \mapsto t \cdot f_2 (t)$ is
differentiable at $0$ with $D g_2 (0) = y \neq x$.
Finally, note that $g_1 (t) = g_2 (t)$ for all $t \in (-1,1)$;
for $t \neq 0$ this is clear and for $t = 0$, we have $g_1 (0) = 0 = g_2 (0)$.$\square$
Remark 1 Above, I committed the usual abuse of notation of identifying
an element $x \in F$ of a vector space with the linear map $\mathbb{R} \to F, t \mapsto t \cdot x$.
Remark 2 By slightly modifying the construction (in fact, things get easier),
one can show that there is a function $g : (-1,1) \to F$
such that as a map into $F_1$, $g$ is differentiable at $0$,
but as a map into $F_2$, it is not.
