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Suppose $f:\mathbb R^n\to \mathbb R^n$ is differentiable and that $Df$ is positive-definite at every point. As homework, I need to prove $f$ is injective.

I thought about proving by contradiction. If $f(a) = f(b) ,a\neq b$, consider the straight line $\gamma:a\to b$ in $\mathbb R^n$, the composition $f\circ \gamma:\mathbb R\to \mathbb R$ (we take the codomain as $\mathbb R$ only) allows using the mean value theorem to find some $c\in (a,b)$ for which $(f\circ \gamma)^\prime (c)=0$. Then, by the chain rule $(f\circ \gamma)^\prime(c)=Df|_{\gamma c} \gamma^\prime(c)=0$. At this point I'm stuck. I don't see how to get to an inner product to contradict positive definiteness.

Positive-definiteness implies invertibility whence $\gamma^\prime(c)=0$, which can't happen for a straight line, so it looks like the weaker hypothesis that $Df$ is everywhere invertible implies $f$ is injective. But that doesn't sound right...

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Hint: You are almost there. You just have to take the scalar product with $(b-a)$, i.e. look at $\langle f(b)-f(a),b-a \rangle$, apply your argument and show that this quantity is strictly positive whenever $b\neq a$.

More details: Let $\gamma(t)=a+t(b-a)$, $t\in [0,1]$ be our path from $a$ to $b$ and consider the real-valued function: $$F(t)=\langle f(\gamma(t))-f(a), b-a \rangle$$$\langle f(b)-f(a),b-a \rangle$ Suppose for a contradiction that $f(b)=f(a)$, whence implying that $F(1)=F(0)=0$. Then by the MVT there is $0<c<1$ so that: $$ F'(c) = \langle Df(\gamma(c)).(b-a), b-a \rangle = 0 $$ But $Df(c)$ is supposed positive definite at every point so we conclude that $b-a=0$ and that the function $f$ is injective. The proof shows in fact that for distinct $a$ and $b$ we have $F(1)>0$, i.e. that $$\langle f(b)-f(a),b-a \rangle > 0$$

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  • $\begingroup$ Thank you. What about the last paragraph I wrote? What's the error there? $\endgroup$
    – Calc
    Feb 1 '17 at 15:02
  • $\begingroup$ The derivative of $\gamma$ does not vanish (which is precisely why the quantity I mention must be strictly positive). In fact, $\gamma(t) = a + t (b-a)$, $0\leq t\leq 1$ describes the path from $a$ to $b$ and $\gamma'(t) = b-a$ (which is non-zero). It is btw just an ordinary derivative, not a gradient. $\endgroup$
    – H. H. Rugh
    Feb 1 '17 at 15:13
  • $\begingroup$ Ah, I missed the fact differentiation commutes with inner product. Thanks! $\endgroup$
    – Calc
    Feb 1 '17 at 16:20

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