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Consider a continuously differentiable function $f : \mathbb{R}^{n} \to \mathbb{R}^{n}$ such that $f(0) = 0$ and $\langle f(x), x \rangle \ge 0$ for all $x \in \mathbb{R}^{n}$. Does there exist a continuous function $A : \mathbb{R}^{n} \to \mathbb{R}^{n \times n}$ such that $A(x)x = f(x)$ and $\langle A(x) y, y \rangle \ge 0$ for all $x,y \in \mathbb{R}^{n}$?

It is always possible to write such a function $f$ in terms of a continuous matrix function $A$. In particular, \begin{equation} A(x) := \int_{0}^{1} \nabla f(sx) \, ds. \end{equation} But this matrix is not necessarily positive semidefinite. $A$ is unique up to an additive continuous matrix function $L$ satisfying $L(x)x = 0$ for all $x \in \mathbb{R}^{n}$ since $\big( A(x) + L(x) \big)x = A(x)x = f(x) $.

Edit: Let me elaborate further and also address diracdeltafunk's comment. On $\mathbb{R}$ the problem at hand is straightforward. The desired function $a$ exists if and only if $f(x)/x$ is continuous at $x = 0$ (which it is if $f$ is $C^{1}$, and the above integral formula will then give the unique answer).

In $\mathbb{R}^{2}$, take for instance $f(x) = (x_{2}^{2}, -x_{1}x_{2} )$. Then $\langle f(x) , x \rangle = 0 $ for all $x \in \mathbb{R}^{2}$. However, using the above integral formula, \begin{equation} A(x) = \begin{pmatrix} 0 & x_{2} \\ -\tfrac{1}{2} x_{2} & -\tfrac{1}{2} x_{1} \end{pmatrix} \end{equation} which is not P.S.D. But we can choose \begin{equation} A(x) = \begin{pmatrix} 0 & x_{2} \\ - x_{2} & 0 \end{pmatrix} \end{equation} which is P.S.D.

Consider a continuously differentiable function $f : \mathbb{R}^{n} \to \mathbb{R}^{n}$ such that $f(0) = 0$ and $\langle f(x), x \rangle \ge 0$ for all $x \in \mathbb{R}^{n}$. Does there exist a continuous function $A : \mathbb{R}^{n} \to \mathbb{R}^{n \times n}$ such that $A(x)x = f(x)$ and $\langle A(x) y, y \rangle \ge 0$ for all $x,y \in \mathbb{R}^{n}$?

It is always possible to write such a function $f$ in terms of a continuous matrix function $A$. In particular, \begin{equation} A(x) := \int_{0}^{1} \nabla f(sx) \, ds. \end{equation} But this matrix is not necessarily positive semidefinite. $A$ is unique up to an additive continuous matrix function $L$ satisfying $L(x)x = 0$ for all $x \in \mathbb{R}^{n}$ since $\big( A(x) + L(x) \big)x = A(x)x = f(x) $.

Edit: Let me elaborate further and also address diracdeltafunk's comment. In $\mathbb{R}$ the problem at hand is straightforward. The desired function $a$ exists if and only if $f(x)/x$ is continuous at $x = 0$ (which it is if $f$ is $C^{1}$, and the above integral formula will then give the unique answer).

In $\mathbb{R}^{2}$, take for instance $f(x) = (x_{2}^{2}, -x_{1}x_{2} )$. Then $\langle f(x) , x \rangle = 0 $ for all $x \in \mathbb{R}^{2}$. However, using the above integral formula, \begin{equation} A(x) = \begin{pmatrix} 0 & x_{2} \\ -\tfrac{1}{2} x_{2} & -\tfrac{1}{2} x_{1} \end{pmatrix} \end{equation} which is not P.S.D. But we could also choose \begin{equation} A(x) = \begin{pmatrix} 0 & x_{2} \\ - x_{2} & 0 \end{pmatrix} \end{equation} which is P.S.D.

Consider a continuously differentiable function $f : \mathbb{R}^{n} \to \mathbb{R}^{n}$ such that $f(0) = 0$ and $\langle f(x), x \rangle \ge 0$ for all $x \in \mathbb{R}^{n}$. Does there exist a continuous function $A : \mathbb{R}^{n} \to \mathbb{R}^{n \times n}$ such that $A(x)x = f(x)$ and $\langle A(x) y, y \rangle \ge 0$ for all $x,y \in \mathbb{R}^{n}$?

It is always possible to write such a function $f$ in terms of a continuous matrix function $A$. In particular, \begin{equation} A(x) := \int_{0}^{1} \nabla f(sx) \, ds. \end{equation} But this matrix is not necessarily positive semidefinite. $A$ is unique up to an additive continuous matrix function $L$ satisfying $L(x)x = 0$ for all $x \in \mathbb{R}^{n}$ since $\big( A(x) + L(x) \big)x = A(x)x = f(x) $.

Consider a continuously differentiable function $f : \mathbb{R}^{n} \to \mathbb{R}^{n}$ such that $f(0) = 0$ and $\langle f(x), x \rangle \ge 0$ for all $x \in \mathbb{R}^{n}$. Does there exist a continuous function $A : \mathbb{R}^{n} \to \mathbb{R}^{n \times n}$ such that $A(x)x = f(x)$ and $\langle A(x) y, y \rangle \ge 0$ for all $x,y \in \mathbb{R}^{n}$?

It is always possible to write such a function $f$ in terms of a continuous matrix function $A$. In particular, \begin{equation} A(x) := \int_{0}^{1} \nabla f(sx) \, ds. \end{equation} But this matrix is not necessarily positive semidefinite. $A$ is unique up to an additive continuous matrix function $L$ satisfying $L(x)x = 0$ for all $x \in \mathbb{R}^{n}$ since $\big( A(x) + L(x) \big)x = A(x)x = f(x) $.

Edit: Let me elaborate further and also address diracdeltafunk's comment. On $\mathbb{R}$ the problem at hand is straightforward. The desired function $a$ exists if and only if $f(x)/x$ is continuous at $x = 0$ (which it is if $f$ is $C^{1}$, and the above integral formula will then give the unique answer).

In $\mathbb{R}^{2}$, take for instance $f(x) = (x_{2}^{2}, -x_{1}x_{2} )$. Then $\langle f(x) , x \rangle = 0 $ for all $x \in \mathbb{R}^{2}$. However, using the above integral formula, \begin{equation} A(x) = \begin{pmatrix} 0 & x_{2} \\ -\tfrac{1}{2} x_{2} & -\tfrac{1}{2} x_{1} \end{pmatrix} \end{equation} which is not P.S.D. But we can choose \begin{equation} A(x) = \begin{pmatrix} 0 & x_{2} \\ - x_{2} & 0 \end{pmatrix} \end{equation} which is P.S.D.

Consider a continuously differentiable function $f : \mathbb{R}^{n} \to \mathbb{R}^{n}$ such that $f(0) = 0$ and $\langle f(x), x \rangle \ge 0$ for all $x \in \mathbb{R}^{n}$. Does there exist a continuous function $A : \mathbb{R}^{n} \to \mathbb{R}^{n \times n}$ such that $A(x)x = f(x)$ and $\langle A(x) y, y \rangle \ge 0$ for all $x,y \in \mathbb{R}^{n}$?

It is always possible to write such a function $f$ in terms of a continuous matrix function $A$. In particular, \begin{equation} A(x) := \int_{0}^{1} \nabla f(sx) \, ds. \end{equation} But this matrix is not necessarily positive semidefinite. $A$ is unique up to an additive continuous matrix function $L$ satisfying $L(x)x = 0$ for all $x \in \mathbb{R}^{n}$ since $\big( A(x) + L(x) \big)x = A(x)x = f(x) $.

Edit: The claim as it stands is false, but I have only found a counter example constructed such that $\langle f(x), x \rangle = 0$ for all $x \in \mathbb{R}^{n}$. I am still looking for a counter example for which $\langle f(x), x \rangle > 0$ for all $x \ne 0$.

Consider a continuously differentiable function $f : \mathbb{R}^{n} \to \mathbb{R}^{n}$ such that $f(0) = 0$ and $\langle f(x), x \rangle \ge 0$ for all $x \in \mathbb{R}^{n}$. Does there exist a continuous function $A : \mathbb{R}^{n} \to \mathbb{R}^{n \times n}$ such that $A(x)x = f(x)$ and $\langle A(x) y, y \rangle \ge 0$ for all $x,y \in \mathbb{R}^{n}$?

It is always possible to write such a function $f$ in terms of a continuous matrix function $A$. In particular, \begin{equation} A(x) := \int_{0}^{1} \nabla f(sx) \, ds. \end{equation} But this matrix is not necessarily positive semidefinite. $A$ is unique up to an additive continuous matrix function $L$ satisfying $L(x)x = 0$ for all $x \in \mathbb{R}^{n}$ since $\big( A(x) + L(x) \big)x = A(x)x = f(x) $.