Let $X$, $Y$ be real Banach spaces. Consider a map

$F\colon X\rightarrow C^1(Y,\mathbb{R})$

such that for every $y\in Y$ the map $F(\cdot,y)\colon X\rightarrow \mathbb{R}$ is smooth.

Claim: Then $F\colon X\rightarrow C^1(Y,\mathbb{R})$ is smooth.

Is the claim true or false?

My attempt was to set $((DF)_{x_0}x)(y):=(DF(\cdot,y))_{x_0}x$. However, I think this need not to be well defined, since $(DF(\cdot,y))_{x_0}x$ may not be $C^1$ in $y$.


Your function $F$ can be identified with a function $F : X \times Y \to \mathbb{R}$. Your assumptions read:

  • $F(x, \cdot) \in C^1(Y,\mathbb{R})$ for all $x \in X$
  • $F(\cdot, y) \in C^1(X,\mathbb{R})$ for all $y \in Y$

Hence, you know that your function has continuous partial derivatives (w.r.t. $x$ and $y$). If you additionally assume that $F$ is continuous, then you can show that $F \in C^1(X \times Y, \mathbb{R})$, see, e.g., Section 3.7 in "Differential Calculus" by Cartan (the original is in French, and there exist further translations).

You, however, asked whether $F \in C^1(X, C^1(Y,\mathbb{R}))$ and this amounts to second (mixed) derivatives. A counterexample should be given by $$F(x)(y) = (x^2 + y^2)^{3/4}.$$


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