Vtyjkyuk

Suppose $f:mathbb R^nto mathbb R^n$ is a $mathcal C^1$ map such that $langle df_x(v),vrangle>0$ for all $xin mathbb R^n$ and $vin R^nsetminus 0$. Prove that $f$ is injective. Hint: For $xne 0, $ consider $g:mathbb Rto mathbb R^n$ given by $g(t)=f(tx)$. Find $g'(t)$ and show that $f(x)ne f(0)$.

Let $x=(x_1,dots,x_n)$. To compute $g'(t)=dg_t$, note that $g$ is the composition of $h:mathbb Rto mathbb R^n$ given by $tmapsto tx$ and $f$. By the chain rule, $dg_t=df_txcdot dh_t$ as matrices. Now $dh_t=[x_1,dots,x_n]^t,df_tx=[D_if_j]$, where $f=(f_1,dots,f_n)$, so $$dg_t=[D_1f_1(tx)x_1+dots+D_1f_n(tx)x_n,dots, D_nf_1(tx)x_1+dots +D_nf_n(tx)x_n]^t.$$

Injectivity means that $g(0)=f(x)ne 0$ if $xne 0$. What does it have to do with injectivity? How to use the given inequality?

So $g=f circ h$. $h$ is linear. Hence its derivative at each point is itself. Which means that for each $t$, $h^prime(t)$ is the real map $h^prime(t): umapsto ux$.

Consequently using the chain rule, we have:

$$g^prime(t) =f^prime(tx).x$$

Now consider $G(t)= langle g(t), x rangle$. $G$ derivative is
$$G^prime(t) = langle g^prime(t),xrangle = langle f^prime(tx).x,xrangle $$

which is strictly positive for $x neq 0$ according to the hypothesis $langle f^prime(y).v,vrangle >0$

Then
$$langle f(x)-f(0),xrangle = G(1)-G(0)=int _0^1 G^prime(t) dt>0$$

allowing to conclude that $f(x)neq f(0)$. The injectivity of $f$ can be deduced by changing $g(tx)$ into $g(t(y-x)+x)$.