Let $f=(f_1,f_2,\cdots,f_n)$ be a function from an open set $U$ in $\mathbb{R}^m$ to $\mathbb{R}^n$. Now $f$ is $C^1$ iff all partial derivatives $D_1,D_2,\cdots,D_m$ of all the components of $f$ are continuous. Can we say the similar where instead of taking partial derivatives of the components of $f$ we take directional derivatives $D_{v_1},D_{v_2},\cdots,D_{v_k}$ of components of $f$. What should be the value of $k$ and do we need any other conditions ?
Now my intuition is that for directional derivatives $k=m$ and $v_1,\cdots,v_m$ should be linearly independent. But i can not prove it. Can you help me
Best Answer
Let $\mathcal L(\mathbb R^m ,\mathbb R^n)$ denote the vector space of all linear maps $\mathbb R^n \to \mathbb R^n$. Representing $\phi \in \mathcal L(\mathbb R^m ,\mathbb R^n)$ by its associated $(n \times m)$-matrix $M(\phi)$ allows to identify $\mathcal L(\mathbb R^m ,\mathbb R^n)$ with $\mathbb R^{m\cdot n}$. This gives us a topology on $\mathcal L(\mathbb R^m ,\mathbb R^n)$. Using the identification $\mathcal L(\mathbb R^m ,\mathbb R^n) \cong \mathbb R^{m\cdot n}$, each map $F : X \to \mathcal L(\mathbb R^m ,\mathbb R^n)$ can be described via its component-functions $F_{ij} : X \to \mathbb R$. Clearly $F$ is continuous iff all $F_{ij}$ are continuous.
$f$ is defined to be differentiable at $x \in U$ with derivative $df(x) \in \mathcal L(\mathbb R^m ,\mathbb R^n)$ if $$\lim_{h \to 0}\frac{\lVert f(x+h) - f(x) - df(x)(h) \rVert}{\lvert h \rVert} = 0 .$$
The following facts are well-known:
If $f$ is differentiable at $x$, then all $D_jf_i(x)$ exist and $M(df(x))$ is the Jacobian matrix $Jf(x) = (D_jf_i(x))$.
If all $D_jf_i$ exist and are continuous on $U$, then $f$ is differentiable at all $x \in U$ and $df : U \to \mathcal L(\mathbb R^m ,\mathbb R^n)$ is continuous (which means that $f$ is $C^1$).
Now let $v_1,\ldots, v_m$ be linearly independent vectors of $\mathbb R^m$ (i.e. they form a basis of $\mathbb R^m$). We have $$D_{v_j}f_i(x) = \lim_{t \to 0} \frac{f_i(x+tv_j) - f_i(x)}{t} . $$ Thus the existence and continuity of all $D_{v_j}f_i$ is equivalent to the existence and continuity of all $$D_{v_j}f(x) = \lim_{t \to 0} \frac{f(x+tv_j) - f(x)}{t} .$$
There exists a unique isomorphism $L : \mathbb R^m \to \mathbb R^m$ such that $L(v_j) = e_j$, where the $e_j$ are the standard basic vectors of $\mathbb R^m$. It is well-known that $L$ is differentiable at all points $x$ with derivative $dL(x) = L$. Thus $L$ is $C^1$. Let $V = L(U)$; this is an open subset of $\mathbb R^m$. Consider the map $g = f \circ L^{-1} : V \to \mathbb R^n$. We have $$D_jg(y) = \lim_{t \to 0} \frac{g(y+te_j) - g(y)}{t} = \lim_{t \to 0} \frac{f(L^{-1}(y+te_j)) - f(L^{-1}(y))}{t} \\= \lim_{t \to 0} \frac{f(L^{-1}(y) +tL^{-1}(e_j))) - f(L^{-1}(y))}{t} = \lim_{t \to 0} \frac{f(L^{-1}(y) +tv_j)) - f(L^{-1}(y))}{t} =D_{v_j}f(L^{-1}(y))$$ which shows that all $D_jg$ exist and are continuous. This proves that $g$ is $C^1$. Now the chain rule shows that also $f = g \circ L$ is $C^1$.