A norm $\|\cdot\|$ on $\mathbb R^N$ such that $\|x\|^2$ is $C^2$ near $x=0$ is induced by an inner product. Let $\|\cdot\|$ be a norm on $\mathbb R^N$ and define $f: \mathbb R^N\to\mathbb R$ by $f(x)=\|x\|^2$. Suppose that $f$ is $C^2$ near $x=0$. Prove that there is an inner product $(\cdot,\cdot)$ on $\mathbb R^N$ such that $\|x\|^2=(x,x)$ for all $x\in \mathbb R^N$.
Here is my attempt. I guess that the inner product is given by a positive-definite matrix. Therefore, I define $a_{ij}=\frac12\frac{\partial^2f}{\partial x_i\partial x_j}(0)$ for all $1\leq i,j\leq N$ and let $A=(a_{ij})$. Since $f$ has a minimum at $x=0$ and is $C^2$ near $x=0$, I deduced that the matrix $A$ is positive-definite. It suffices to show that $f(x)=x^TAx$ for all $x\in \mathbb R^N$; if so, then $(x,y):= x^TAy$ gives the desired inner product.
But I don't know how to prove $f(x)=x^TAx$.
Any hints are welcome！
 A: A trick that I have gradually learned is to fully use that $f\in C^2(\mathbb{R}^n)$ with Taylor's Theorem with remainder. In particular, that
$$f(x) = f(0) + \sum_{i=1}^n \frac{\partial f(0)}{\partial x_i}(x_i-0)  +\sum_{i,j=1}^n(x_i-0)(x_j-0)\int_0^1(1-t)\frac{\partial^2 f}{\partial x_i\partial x_j}(tx)dt$$
Now since $||\cdot||$ is a norm, then $f(0) = ||0||^2 = 0$. We also have that if $e_i$ are the standard basis, then
$$
    \frac{\partial f(0)}{\partial x_i} = \lim_{h\to 0}\frac{||he_i||^2 - ||0||^2}{h} = \lim_{h\to 0} \frac{h^2}{h} = \lim_{h\to 0} h = 0
$$
Hence we can now write that
$$
   f(x) = \sum_{i,j=1}^n x_i x_j\int_0^1(1-t)\frac{\partial^2 f}{\partial x_i\partial x_j}(tx)dt
$$
If we set $A(x)$ to be the matrix with components $a_{ij}(x) = \int_0^1(1-t)\frac{\partial^2 f}{\partial x_i\partial x_j}(tx)dt$. Then it follows that we can write $f(x) = x^T A(x)x$ where $A(x)$ is a symmetric matrix for all $x$. Since $||\cdot||$ is a norm, then for all $\alpha\in\mathbb{R}$ we will have that $f(\alpha x) = ||\alpha x||^2 = \alpha^2||x|| = \alpha^2 f(x)$. Thus
$$
 \alpha^2 f(x) = f(\alpha x) = (\alpha x)^T A(\alpha x)(\alpha x) = \alpha^2(x^T A(\alpha x)x)
$$
If we take $\alpha_n = \frac{1}{2^n}$, then the expression above can be simplified to yield $f(x) = x^T A(\frac{x}{2^n})x$ for all $n\in \mathbb{N}$. Since $f\in C^2$, then $x\to A(x)$ will be a continuous operation and so if we take the limit as $n\to +\infty$ we get that $f(x) = x^T A(0)x$. Well $a_{ij}(0) = \int_0^1(1-t)\frac{\partial^2 f}{\partial x_i\partial x_j}(0)dt = \int_0^1 (1-t)dt\frac{\partial^2 f}{\partial x_i\partial x_j}(0) = \frac{1}{2}\frac{\partial^2 f}{\partial x_i\partial x_j}(0)$ which is what you initially had. To see that $A(0)$ is symmetric positive definite, just use that $0\geq ||x||^2 = x^TA(0)x$ since $||\cdot||$ is assumed to be a norm.
A: There is no need to use Taylor with special remainders. By the condition, we have $$f(x)= f(0) + Jx + \frac{1}{2}x^t H x + o(|x|^2)$$ where $J, H$ are the Jacobian and Hessian of $f(x)$ around $0$ separately, and $|x|$ stands for the usual Euclidean norm of $x$.
Fix $x$, for small $h$, we have from the above $$f(hx) = f(0) + (Jx) h + (\frac{1}{2}x^t H x) h^2 + o(h^2)$$
Meanwhile $$f(hx) = h^2 f(x) = 0 + 0 + f(x)h^2 + o(h^2)$$
Since the 2nd order Taylor approximation of $f(hx)$ around $0$ must be unique, we must have $f(x) = \frac{1}{2}x^tHx$.
This holds for any $x$, hence $f(x)$ is induced by the inner product $(x, y):=x^t (\frac{1}{2}H)y$.
