# Does existence and continuity of partial derivatives imply differentiability in Normed Vector Spaces?

I'm wondering if the 'Normed Vector Space version' of the following theorem holds:

Theorem: let $$A\subseteq \mathbb{R}^n$$ be open and let $$f:A\to \mathbb{R}^m$$ have continuous partial derivatives $$\partial f_i/\partial x_j$$ on $$A$$. Then $$f$$ is differentiable on $$A$$.

In particular, the total derivative can be extended to NVS by considering the Fréchet Derivative. If we extend the definition of partial derivatives as follows:

Throughout the post let $$V$$ and $$W$$ be NVSs, with $$V'\subseteq V$$ open and $$f$$ a function $$V'\to W$$.

Definition: given $$a\in V'$$ and a vector $$v\in V$$, we define (assuming the limit exists) $$\frac{\partial f}{\partial v}(a) := \lim_{t\rightarrow 0}\frac{f(a+tv)-f(a)}{t},$$

and, in the case $$v=e_i$$ for a basis vector $$e_i$$,

$$\frac{\partial f}{\partial x_i}(a) := \frac{\partial f}{\partial e_i}(a).$$

then, is either of the following statements true?

Theorem (?): if $$f:V'\to W$$ has continuous partial derivatives $$\partial f/\partial v$$ (for any $$v\in V$$) on $$A$$, then $$f$$ is differentiable on $$A$$.

Theorem (?): let $$\dim V = n$$ and $$\dim W = m$$. If $$f:V'\to W$$ have continuous partial derivatives $$\partial f_i/\partial x_j$$ on $$A$$, then $$f$$ is differentiable on $$A$$.

Let $$V=c_0(\mathbb{N},\mathbb{R})$$ endowed with the maximum norm $$\|\cdot\|$$ and consider $$f=(f_n):V \to V$$ defined as $$f(x)=\left(\frac{1}{n}\sin(nx_n)\right) \quad (x=(x_n)\in V).$$ Fix any $$v=(v_n) \in V$$ and let $$x=(x_n) \in V$$. Note that for each $$n$$ $$\frac{d}{dt}\frac{1}{n}\sin(n(x_n+tv_n)) = \cos(n(x_n+tv_n))v_n,$$ hence by the mean value theorem $$\frac{1}{n}\sin(n(x_n+tv_n)) - \frac{1}{n}\sin(nx_n) = \cos(n(x_n+\tau_n(t)v_n))v_n t$$ for some $$\tau_n(t)$$ between $$0$$ and $$t$$, so $$|\tau_n(t)| \le |t|$$.
Let $$\varepsilon > 0$$. Choose $$n_0 \in \mathbb{N}$$ such that $$|v_n| \le \varepsilon/3$$ $$(n \ge n_0+1)$$. Now $$\|\frac{f(x+tv)-f(x)}{t} - (\cos(nx_n)v_n)\| = \| (\cos(n(x_n+\tau_n(t)v_n))v_n- \cos(nx_n)v_n)\|$$ $$\le \max_{n=1,\dots, n_0} | \cos(n(x_n+\tau_n(t)v_n))v_n- \cos(nx_n)v_n)| + 2\frac{\varepsilon}{3}$$ Since $$\tau_n(t) \to 0$$ $$(t \to 0)$$ for each $$n$$ there is some $$\delta >0$$ such that $$\max_{n=1,\dots, n_0} | \cos(n(x_n+\tau_n(t)v_n))v_n- \cos(nx_n)v_n)| \le \frac{\varepsilon}{3} \quad (0<|t|< \delta).$$ Summing up $$\|\frac{f(x+tv)-f(x)}{t} - (\cos(nx_n)v_n)\| \le \varepsilon \quad (0<|t|< \delta).$$ Thus $$\frac{\partial f}{\partial v}(x)= (\cos(nx_n)v_n).$$ For each fixed $$v$$ (w.l.o.g $$v\not=0$$) the function $$x \mapsto (\cos(nx_n)v_n)$$ is continuous on $$V$$: Let $$\varepsilon > 0$$ and again let $$n_0 \in \mathbb{N}$$ be such that $$|v_n| \le \varepsilon/3$$ $$(n \ge n_0+1)$$. For $$x,y \in V$$ we have $$\|(\cos(ny_n)v_n)-(\cos(nx_n)v_n)\| = \max_{n=1,\dots, n_0}|\cos(ny_n)v_n-\cos(nx_n)v_n| + 2\frac{\varepsilon}{3}$$ $$\le n_0 \|x-y\| \|v\| + 2\frac{\varepsilon}{3}.$$ If $$x$$ is fixed then for each $$y$$ with $$\|x-y\| \le \varepsilon/(3n_0 \|v\|)$$ we have $$\|(\cos(ny_n)v_n)-(\cos(nx_n)v_n)\| \le \varepsilon.$$ Finally we show that $$f$$ is not Frechet differentiable in $$0=(0)$$: Assume that $$f$$ is Frechet differentiable in $$0$$. Then $$f'(0)h=(\cos(n0)h_n)=(h_n)$$ $$(h \in V)$$, thus $$f'(0)=id_V$$. Now consider the sequence $$h^{(k)} := \frac{2\pi}{k}e_k \in V$$ (with $$e_k=(\delta_{nk})$$). Then $$f(h^{(k)})=0$$ for each $$k$$ and $$\|h^{(k)}\| \to 0$$ $$(k \to \infty)$$. Hence $$1=\frac{\|f(h^{(k)})-f(0)-h^{(k)}\|}{\|h^{(k)}\|} \to 0 \quad (k \to \infty),$$ a contradiction.