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The following is somewhat unclear to me. Let $X$, $Y$ be locally convex vector spaces, let $f: X \supseteq U \longrightarrow Y$ be a (nonlinear) continuous map. Then one can say that $f$ is $C^1$ if

a) the difference quotient $$ Df(x)\xi := \lim_{\epsilon \longrightarrow 0} \frac{f(x + \epsilon \xi) - f(x)}{\epsilon}$$ exists for all $x \in U$, $\xi \in X$ and if $$Df: U \times X \longrightarrow Y$$ is continuous; or

b) $f$ maps $C^1$ curves in $X$ to $C^1$ curves in $Y$.

Question: Are the notions a), b) equivalent?

As far is understood, they are not in general, right? But for which spaces are they equivalent?

Addition: If $X$ and $Y$ are not Banach spaces, it seems too strong a notion to require the map $Df$ to be continuous from $X$ to $L(X, Y)$. Why is this so?

Kofi
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  • Regarding the addition, why does it seem too strong a notion to require the continuity of $Df \colon U \to L(X,,Y)$? That's what I'd require for a $C^1$ map. Note that that is a weaker requirement than the continuity of $\varphi \colon U\times X \to Y;; \varphi(x,,\xi) = Df\lvert_x(\xi)$ because for non-normable spaces the pairing $L(X,,Y)\times X \to Y$ is (in general at least) not continuous. – Daniel Fischer Jul 16 '13 at 21:18
  • Let $X$ be the Banach space of continuous periodic functions on some interval. Then the Translation operator $T_sf(x) = f(x+s)$, is continuous $\mathbb{R}\times X \rightarrow X$, $(s,f) \mapsto T_sf$, but not $\mathbb{R} \rightarrow L(X)$, $s \mapsto T_s$. – Kofi Jul 17 '13 at 08:35
  • Sure, weaker notions of continuity are reasonable too, but $C^1$ means "differentiable function with continuous derivative" for me. – Daniel Fischer Jul 17 '13 at 08:44

1 Answers1

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b) does not imply a) even when $X=Y=\mathbb R^2$. It's easier to describe the counterexample in complex notation: $$f(r\exp(i\theta)) = r\exp(i\theta+i\sin 2\theta),\quad r,\theta\in\mathbb R \tag1$$ It is important to notice that (1) applies both when $r<0$ and when $r>0$, giving consistent result because $\sin 2\theta$ is $\pi$-periodic.

The map (1) is not differentiable at the origin, because it is degree 1 homogeneous but not linear.

On the other hand, it maps $C^1$ curves to $C^1$ curves. Indeed, let $\Gamma:(-1,1)\to \mathbb R^2\approx \mathbb C$ be a $C^1$-curve. For definiteness assume $\Gamma(0)=0$. Let $\rho e^{i\phi}=\Gamma'(0)$; here $\rho>0$. Write $\Gamma(t)=t \rho e^{i\phi}+o(t)$ and plug this into $f$. Since $f$ is Lipschitz, we gave $$f\circ \Gamma(t)=f(t \rho e^{i\phi})+o(t) = t\rho e^{i\phi+i\sin 2\phi}+o(t) \tag2 $$ Hence, $f\circ \Gamma$ has a tangent at $t=0$. Its tangent at the points where $\Gamma\ne 0$ has argument $\arg \Gamma'(t)+2\sin \arg \Gamma(t)$ which is consistent with (2): that is, the tangent direction is continuous along the curve.

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