Is $f$ $p$-times differentiable at $x^0 $？

Question

A weird exercise that puzzles me a lot.

Let $f$ be a real-valued funcation defined on an open subset $O$ of $\mathbb{R}^n$, all of whose partial derivatives up to order $p$ are defined in $O$. $\partial_{x_{j_{1}}}\cdots\partial_{x_{j_{p}}}f：O\mapsto \mathbb{R}$ be continuous at $x^0\in O$, for arbitrary collections(with repetition allowed) of $p$ indices $1\leq j_1,\cdots,j_p\leq n$. Show $f$ is $p$-$th$ differentiable at $x^0\in O.$

In the case $p=1$, $f$ is (first-)differentiable at $x^0\in O$.From the definition of higher differentials,we say that $f$ is $p$-times differentiable at $x^0\in O$ on the premise of $f^{\left(p-1\right)}$ is defined on some neighbourhood of $x^0$ contained in $O$. Only just by the above condition , maybe $f^{\left(p-1\right)}(p\geq 2)$ is not defined on any neighbourhood of $x^0$ contained in $O$, let alone $f$ is $p$-$th$ ($p\geq 2$) differentiable at $x^0\in O$ . Are there any counterexamples here？

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Definition

Let $O\subseteq \mathbb R^n$ be an open set,and let $f：O\mapsto \mathbb{R}$ be a mapping.Whenever $f：O\mapsto \mathbb{R}$ is (first-)differentiable in $O$,we can consider the first derivative of $f$, $i.e.$the mapping $f^{′}：O\mapsto \mathcal{L}(\mathbb{R}^n,\mathbb{R}),$which to $x\in O$ associates the differential $f^{′}(x)$ of $f$ at $x$. Suppose $f^{′}(x)$ is differentiable at $x^0$;that is ,suppose there is a linear map $\Lambda \in \mathcal{L}(\mathbb{R}^n,\mathcal{L}(\mathbb{R}^n,\mathbb{R}))$ such that $f^{′}(x)-f^{′}(x^0)=\Lambda(x-x^0)+o(x-x^0)$ as $ x\rightarrow x^0$,then we say that $f$ is twice-differentiable at $x^0$,denoted $\Lambda$ as $f^{"}(x^0)$. And so on,the definition of $f$ is $p$-$th$ differentiable at $x^0$,denoted as $f^{(p)}(x^0)\in \mathcal{L}(\mathbb{R}^n,\mathcal{L}(\mathbb{R}^n,\cdots,\mathcal{L}(\mathbb{R}^n,\mathbb{R}),\cdots))$ where there are $p$ iterated $\mathcal{L}^{′}s$, likewise.

Answer 1

Let us do it a bit more general and consider a function $f : O \to \mathbb R^m$ all of whose partial derivatives $\partial_{j_{1} \ldots j_r} f_i = \partial_{x_{j_{1}}}\cdots\partial_{x_{j_{r}}}f_i$ up to order $p$ are defined in $O$. Here $i = 1,\ldots, m$, $r = 1,\ldots,p$ and $j_k = 1,\ldots, n$. We assume moreover that they are continuous on $O$ for $r < p$ and continuous at $x_0$ for $r = p$. Note that the continuity of $\partial_{j_{1} \ldots j_r} f_i$ for $r < p$ does not follow from the fact that $\partial_{j_{1} \ldots j_r} f_i$ is partially differentiable. See A function whose partial derivatives exist at a point but is not continuous.

Then $f' : O \to \mathcal L(\mathbb R^n,\mathbb R^m)$. The range is a normed linear space, and we can of course define derivatives for maps $g : O \to L$ into any normed linear space $L$. We obtain $g' : O \to \mathcal L(\mathbb R^n,L)$. This provides the higher derivatives of $f$ as explained in your question. In fact, let $\mathcal L_0 = \mathbb R^m$ and $\mathcal L_{r+1} = \mathcal L (\mathbb R^n, \mathcal L_r)$. Then $f^{(r)} : O \to \mathcal L_r$.

However, to apply the known machinery of multivariable calculus it is better to identify $\mathcal L(\mathbb R^n,\mathbb R^m)$ with the "standard" Euclidean space of dimension $n\cdot m$. This is $\mathbb R^{n\cdot m}$ with points $x = (x_k)$ whose coordinates $x_k \in \mathbb R$ are indexed by $k=1,\ldots , n\cdot m$. We shall use an equivalent description in the form $\mathbb R^{[m,n]}$ with points $x = (x_{ij})$ whose coordinates $x_{ij} \in \mathbb R$ are indexed by pairs $(i,j)$ with $i = 1,\dots,m$ and $j = 1,\ldots, n$.

Since all partial derivatives $f'_{ij} = \partial_j f_i = \dfrac{\partial f_i}{\partial x_j}$ exist and are continuous on $O$, we know that $f$ is differentiable on $O$ and its derivative $f' : O \to \mathbb R^{[m,n]} = \mathcal L(\mathbb R^n,\mathbb R^m) = \mathcal L_1$ has coordinate functions $f'_{ij}$. The latter simply reflects the fact that the linear map $f'(x) \in \mathcal L_1$ is represented by the Jacobian matrix $\left( f'_{ij}(x) \right)$.

We can now iterate this procedure. Define $\mathbb R^{[m,n;r]}$ as the set of all points $x = (x_{ij_1\ldots j_r})$ whose coordinates $x_{ij_1\ldots j_r} \in \mathbb R$ are indexed by $(r+1)$-tuples $(i,j_1,\ldots, j_r)$ with $i = 1,\dots,m$ and $j_k = 1,\ldots, n$. Then $$\mathbb R^{[m,n;r]} = \mathcal L (\mathbb R^n, R^{[m,n;r-1]} = \mathcal L (\mathbb R^n, \mathcal L_{r-1}) = \mathcal L_r .$$ We see that $f' : O \to \mathbb R^{[m,n]}$ is differentiable on $O$ because all its coordinate functions are continuously partially differentiable. Its derivative $f'' : O \to \mathbb R^{[m,n;2]} = \mathcal L_2$ has coordiante functions $f''_{ij_1j_2} = \partial_{j_1j_2} f_i$. Proceeding inductively we see that $f$ is $(p-1)$-times differentiable on $O$ and $p$-times differentiable at $x_0$. In fact we get for $r < p$ $$f^{(r)} : O \to R^{[m,n;r]}, f^{(r)}_{{ij_1\ldots j_r}} = \partial_{j_{1} \ldots j_r} f_i$$ and for $r = p$ $$f^{(p)}_{ij_{1} \ldots j_p}(x_0) = \partial_{j_{1} \ldots j_p} f_i(x_0). $$