For a Banach space $E$, is it sufficient to check differentiability of $F: U \to E$ on $\phi \circ F$ for all linear functionals $\phi$? Let $E$ be a complex Banach space and and $U \subset \mathbb{C}$ an open set containing $0$. Consider a function $F: U \to E$. If for every linear functional $\phi \in E^*$, the function $\phi \circ F: U \to \mathbb{C}$ is (complex) differentiable at $0$, does this imply that $F$ is differentiable at $0$?
related questions/generalizations: What about the case of real Banach spaces and real differentiability?
Also, what happens if $U$ is also taken to be a subset of a complex Banach space and we consider Frechet derivatives?
My ideas so far: We are assuming that for every $\phi \in E^*$ there exists a $g(\phi) \in E$ s.t.
$$
\phi(F(h)) = \phi(F(0)) + h \cdot g(\phi) + o(h)
$$
It's easy to see that $\phi \to g(\phi)$ is linear so $g \in E^{**}$. If we assume reflexivity of $E$ then we can say $g \in E$and obtain that $\phi(F(h) - h\, g) = o(h)$ for any $\phi \in E^*$.
However, the $o(h)$ might not be uniform in $\phi$, so I don't know how to continue from here.
 A: Negative Result
You can't continue because it's false. Assuming complex differentiability at a single point is not a strong enough condition, because what you're assuming is that a certain collection of elements in $E$ (namely the difference quotient $\frac{f(z)-f(0)}{z}$) converges weakly. However, at this stage you should already be suspicious because weak-convergence alone doesn't guarantee norm-convergence.
As an explicit counter example, consider the complex Hilbert space $E=L^2([0,1];\Bbb{C})$ (the inner product is $\langle \phi,\psi\rangle:=\int_0^1\phi(x)\overline{\psi}(x)\,dx$), where we use the Lebesgue measure. Consider the mapping $f:\Bbb{C}\to E$ defined as
\begin{align}
f(z)&:=
\begin{cases}
x\mapsto z\sin\left(\frac{x}{|z|}\right) & \text{if $z\neq 0$}\\
x\mapsto 0 & \text{if $z=0$}
\end{cases}
\end{align}
Since $E$ is a Hilbert space, its dual is isomorphic to itself. $f$ is weakly complex-differentiable at the origin with derivative $0$, because for each $z\neq 0$, and $g\in E$, we have
\begin{align}
\left\langle \frac{f(z)-f(0)}{z}-0_E\,\,\,,\,\,\,g\right\rangle&=\int_0^1\sin\left(\frac{x}{|z|}\right)\overline{g(x)}\,dx,
\end{align}
and by the Riemann-Lebesgue lemma, this approaches $0$ as $z\to 0$. If $f$ was differentiable at the origin, then it's derivative must be $0$, but this is not the case because for each $n\in\Bbb{N}$, we have
\begin{align}
\left\|\frac{f\left(\frac{1}{2\pi n}\right)-f(0)}{\frac{1}{2\pi n}}-0\right\|_E^2&=\int_0^1\sin^2(2\pi nx)\,dx = \frac{1}{2}
\end{align}
so it clearly doesn't converge to $0$ when $n\to \infty$. Thus, $f$ is not complex-differentiable at the origin.
Since the claim fails for a complex Hilbert space, it also fails for complex Banach spaces, and hence also for real Banach spaces. If you change the domain to a more general Banach space, it will thus also fail.

Positive Results

*

*Suppose $E,F$ are Banach spaces (doesn't matter real or complex), $U\subset E$ open and $f:U\to F$ a mapping which is weakly Frechet differentiable at a point $a\in U$ (i.e for every $\phi\in F^*$, $\phi\circ f$ is Frechet-differentiable at $a$). If $F$ is finite-dimensional then $f$ is Frechet-differentiable at $a$. (this is a trivial result which you probably already know)


*Suppose $F$ is a complex Banach space, $U\subset \Bbb{C}$ a non-empty open set and $f:U\to F$ a mapping which is weakly holomorphic on $U$ (i.e for each $\phi\in F^*$, $\phi\circ f:U\to\Bbb{C}$ is holomorphic, meaning complex differentiable at every point of $U$). Then, $f$ is holomorphic on $U$. This follows from Cauchy's theorem extended to Banach-space-valued functions; see this answer of mine for more details (and the link there also proves the equivalence of holomorphy and analyticity for Banach-valued mappings).


*Building on the previous result, we have the following. Let $E,F$ be complex Banach spaces, $U\subset E$ open, and $f:U\to F$ a continuous mapping such that for each $a\in U$ and $v\in E$, the mapping $z\mapsto f(a+zv)$ is holomorphic as a mapping of an open subset of $\Bbb{C}$ into $F$. Then, $f:U\to F$ is infinitely often continuously Frechet differentiable on $U$.
Actually one can strengthen this to say $f$ is analytic on $U$, but then one would have to develop the notion of polynomials (i.e essentially sums of multilinear maps), hence series of polynomials when the domain is an arbitrary Banach space; this is a little annoying to do, but for reference, take a look at Mujica's book on complex analysis. Also, I'm pretty sure one can drop the bolded continuity hypothesis in (3); this is a (difficult) theorem due to Hartogs I believe, and I think it's also proven in Mujica's book. Anyway, assuming continuity of $f$ from the beginning isn't that strong of an assumption, and it makes life simple :)
Anyway, the way I have stated (3) follows from (2) rather simply. By induction, it suffices to show that $f:U\subset E\to F$ is once Frechet-differentiable (i.e holomorphic) on $U$. Now, fix a direction $v\in E$ and let $a\in U$. Then, the directional derivative of $f$ at $a$ in the direction of $v$ exists and
\begin{align}
(D_vf)(a)&:=\frac{d}{dz}\bigg|_{z=0}f(a+zv)=\frac{1}{2\pi i}\int_{|\zeta|=r}\frac{f(a+\zeta v)}{\zeta^2}\,d\zeta,
\end{align}
where I used one of the standard consequences of Cauchy's integral formula (again, see the link for the extension to the vector-valued case). Now, on the RHS, continuity of $f$ implies that after integrating we still have a continuous function of $a$. This means for each $v\in E$, the directional derivative $D_vf$ is a continuous function on $U$, and thus $f$ is continuously Frechet differentiable on $U$ (this latter theorem is a standard result which can be proven using the mean-value inequality in Banach spaces; see Henri Cartan's differential calculus, or Loomis and Sternberg's advanced Calculus etc. This is the natural generalization of the familiar fact that if a function $g:U'\subset \Bbb{R}^n\to \Bbb{R}^m$ is such that all partial derivatives exist and are continuous on the open set $U'$, then $g$ is continuously Frechet-differentiable on $U'$).
