Proving F is Not Injective in R^n

[Lolyta]

Hello. I have the next problem:
Let $f: U\subset R^n\rightarrow R$ a class $C^1$ function in an open subset $U$ of $R^n$. Proff that f can't be injective.There are some idications: suppose that the vector ($\nabla f$)(p) is not zero (if it's zero the function is not injective WHY?) in $p\in U$ and that we can assume that $\frac{\partial f}{\partial x_i}$(p) $\neq$ 0; consider so $F(x_1,x_2,...,x_n) = (f(x),x_2,...x_n)$ with $x=(x_1,x_2,..,x_n)$ and get the result applying the inverse function theorem.

Thank you so much for any help!

 

[jvicens]

Hello there,

Thank you for bringing this problem to our attention. I can provide some clarification and a proof for why a function cannot be injective if the gradient at a point is zero.

Firstly, let's define what it means for a function to be injective. A function is said to be injective if every element in the domain has a unique image in the range. In other words, no two elements in the domain can have the same image in the range.

Now, let's consider the given function $f: U\subset R^n\rightarrow R$, which is a class $C^1$ function in an open subset $U$ of $R^n$. Since $f$ is a $C^1$ function, it is continuously differentiable in $U$. This means that the gradient of $f$ exists at every point in $U$.

Now, suppose that at a point $p\in U$, the gradient of $f$, denoted by ($\nabla f$)(p), is equal to zero. This means that all the partial derivatives of $f$ at the point $p$ are equal to zero. In other words, the function has a flat tangent plane at the point $p$.

Now, let's assume that the function $f$ is injective. This means that for every point $p\in U$, there exists a unique image in the range. However, since the gradient at $p$ is equal to zero, this contradicts the definition of injectivity. This is because if the function has a flat tangent plane at $p$, then any point on this tangent plane will have the same image as $p$. Therefore, $p$ cannot have a unique image in the range, and hence the function cannot be injective.

To prove this formally, we can follow the given indications. Let's consider the function $F(x_1,x_2,...,x_n) = (f(x),x_2,...x_n)$, where $x=(x_1,x_2,..,x_n)$. This function takes $n$ variables and maps them to $n+1$ variables. Now, by the inverse function theorem, if the Jacobian of $F$ is non-singular at a point, then $F$ is invertible at that point. However, since the gradient of $f$ at $p$ is