# How can I show that f is twice differentiable if $f$ is continuously partially differentiable twice

Let $$f$$:$$\mathbb{R}^n \rightarrow \mathbb{R}^m$$ be a map .How can I show that $$f$$ is twice differentiable if $$f$$ is continuously partially differentiable twice?

So I started with : first derivative : $$Df:\mathbb{R}^n \rightarrow L(\mathbb{R}^n,\mathbb{R}^m)$$

and the second one $$D^{2}f:\mathbb{R}^n\rightarrow L(\mathbb{R}^n,L(\mathbb{R}^n,\mathbb{R}^m))$$

In the end I get that the first direction goes when $$D^{2}f(x)(v,w)$$ exist and continuous , but i am stuck on how to prove it .

First we have the following proposition, which is basically just an identification.

Proposition. Let $$f: \mathbb{R}^n \rightarrow \mathbb{R}^m$$ be differentiable. Then $$f$$ is twice differentiable if and only if each partial derivative $$D_if_j$$ ($$1\leq i\leq n$$, $$1\leq j\leq m$$) is differentiable.

Proof. Fix $$x\in\mathbb{R}^n$$. Suppose $$Df$$ is differentiable at $$x$$, and let $$A=D^{2}f(x)$$. Then by definition $$A\in L(\mathbb{R}^n,L(\mathbb{R}^n,\mathbb{R}^m))$$ and

$$\frac{\|f'(x+h)-f'(x)-Ah \|_{\text{op}}}{\|h\|}\to 0 \quad \text{as}\quad h\to 0 \quad \quad (1)$$

The norm in the numerator is the operator norm on $$L(\mathbb{R}^n,\mathbb{R}^m)$$, while the norm in the denominator is the euclidean norm on $$\mathbb{R}^n$$.

Now, since the vector space $$L(\mathbb{R}^n,\mathbb{R}^m)$$ is finite dimensional, all norms on this space are equivalent, and therefore $$(1)$$ holds with any such norm in the numerator. We will construct a norm on $$L(\mathbb{R}^n,\mathbb{R}^m)$$ as follows:

Let $$\alpha,\beta,\gamma$$ denote the standard bases for $$\mathbb R^n,\mathbb R^m,\mathbb R^{mn}$$ respectively. For $$T\in L(\mathbb{R}^n,\mathbb{R}^m)$$, write $$[T]_\alpha^\beta$$ for the $$m\times n$$ matrix representation of $$T$$ with respect to $$\alpha$$ and $$\beta$$. Define $$\varphi: L(\mathbb{R}^n,\mathbb{R}^m)\to\mathbb{R}^{mn}$$ by $$\varphi(T)=\text{vec } [T]_\alpha^\beta$$ where $$\text{vec}$$ denotes the vec operator, which stacks the columns of a matrix on top of one another. Note that $$\varphi$$ is the composition of two invertible linear maps, and hence is itself an invertible linear map. This allows us to define a norm on $$L(\mathbb{R}^n,\mathbb{R}^m)$$ via the rule

$$T\mapsto \|\varphi(T)\|$$

Therefore $$(1)$$ is equivalent to

$$\frac{\|(\varphi\circ f')(x+h)-(\varphi \circ f')(x)-(\varphi \circ A)h \|}{\|h\|}\to 0 \quad \text{as}\quad h\to 0 \quad \quad (2)$$

Since $$\varphi \circ A:\mathbb{R}^n \to \mathbb{R}^{mn}$$ is linear, we have

$$(\varphi \circ A)h=[(\varphi \circ A)h]_\gamma=[\varphi \circ A]_\alpha^\gamma [h]_\alpha=Bh$$

where $$B=[\varphi \circ A]_\alpha^\gamma$$ is $$mn\times n$$. Moreover, for any $$h\in\mathbb{R}^n$$, we have $$(\varphi\circ f')(h)=\text{vec } J(h)$$, where $$J(h)$$ denotes the Jacobian matrix of $$f'(h)$$. Therefore $$(2)$$ is equivalent to

$$\frac{\|\text{vec } J(x+h)-\text{vec } J(x)-Bh \|}{\|h\|}\to 0 \quad \text{as}\quad h\to 0 \quad \quad (3)$$

From $$(3)$$ it is clear that each row entry of $$\text{vec } J(x)$$ is a differentiable function, the derivative being the corresponding row of $$B$$.

Conversely suppose each partial derivative $$D_if_j$$ is differentiable. Then there exists an $$mn\times n$$ matrix $$B$$ such that $$(3)$$ holds, namely the matrix $$B$$ whose rows are the $$1\times n$$ vectors $$D D_if_j$$. Define $$A:\mathbb{R}^n\to L(\mathbb{R}^n,\mathbb{R}^m)$$ by the rule

$$Ah=(\varphi^{-1}\circ B) (h)$$

Then $$A\in L(\mathbb{R}^n,L(\mathbb{R}^n,\mathbb{R}^m))$$. Define a norm on $$\mathbb{R}^{mn}$$ by the rule

$$y\mapsto \|\varphi^{-1}(y)\|_{\text{op}}$$

Since all norms on $$\mathbb{R}^{mn}$$ are equivalent, it follows that $$(3)$$ is equivalent to

$$\frac{\|f'(x+h)-f'(x)-Ah \|_{\text{op}}}{\|h\|}\to 0 \quad \text{as}\quad h\to 0 \quad \quad (3)$$

and hence $$f'$$ is differentiable at $$x$$ with $$D^2f(x)=A$$.

Corollary. Let $$f:\mathbb{R}^n \rightarrow \mathbb{R}^m$$ be twice continuously differentiable, i.e. all second partial derivatives $$D_iD_{j}f_{k}$$ exists and are continuous. Then $$f$$ is twice differentiable.

Proof. Fix $$i,j$$. By assumption the partial derivatives of $$D_if_j$$ exists and are continuous, which implies that $$D_if_j$$ is differentiable. As $$i,j$$ are arbitrary, the previous proposition implies that $$f$$ is twice differentiable.