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I have an exercise regarding a proof involving the implicit function theorem at which I'm stuck.

We have an even $n\in\mathbb N,n\geq 2$ and a function defined by $$f:\mathbb R^n\to\mathbb R,\ (x_1,...,x_n)\mapsto\sum_{k=1}^nx_k-\prod_{k=1}^nx_k$$ and we're supposed to prove that the conditions for the implicit function theorem for the function $$F(x):=f(x)-f(p)=0,\ p\in\mathbb R^n$$ for at least one of the coordinates $x_1,...,x_n$ in a surrounding $U$ of $p$ hold if and only if $p\neq (1,...,1)$.


In my understanding I have to proof that iff $p\neq (1,...,1)$ there exists a $q\in\mathbb R^n$ and an $l\in\mathbb R$, such that (eventually after reorganizing the parameters) $$F(q,l)=0$$ and the differential of the function $y\mapsto F(q,y)$ is invertible at $y=l$. Is this correct so far?

If it is, I have no idea how the actual proof is supposed to work. I've taken a look at the first condition, which yielded $$l-p_n+\sum_{k=1}^{n-1}(q_k-p_k)-(l-p_n)\prod_{k=1}^{n-1}(q_k-p_k)=0$$ I've also looked at the differential: $$\begin{aligned}\frac{d}{dy}&\left(y-p_n+\sum_{k=1}^{n-1}(q_k-p_k)-(y-p_n)\prod_{k=1}^{n-1}(q_k-p_k)\right)\\ =&1-\prod_{k=1}^{n-1}(q_k-p_k)\end{aligned}$$ As it is a $1\times 1$ matrix, not being invertible means that it's non zero. So if $p=(1,...1)$, at least one of both conditions has to fail. For the second condition, this means that, $$(1)\qquad\prod_{k=1}^{n-1}(q_k-p_k)=1$$ One could hope that if $ p=(1,...,1) $, this is implied by the first condition, but the contrary is the case - it yields a contradiction in the first condition too.

Because (1) strongly implies that $ q=(1,...,1) $ (I'm not sure if it does mathematically though). Assuming that this is true, the first condition would transform: $$l-1+n-(l-1)=0\implies n=0$$ which is a contradiction.

So maybe the first conditions failure is implied by the success of the second condition, or the first condition is never true if $p=(1,...,1)$ - but why should this be the case? And why should (for the reverse direction) a $ p\neq (1,...,1) $ imply that the conditions are satisfieable?

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Sometimes it's better to write things out explicitly in low dimensions to see what's going on.

Let's start with $n=2$. Then $f(x,y) = x+y-xy$, and $Df(x,y) =\begin{bmatrix} 1-y & 1-x \end{bmatrix}$. The derivative vanishes only when $1-y=1-x=0$, i.e., precisely when $x=y=1$. Near any other point, the implicit function theorem guarantees you can solve for some variable in terms of the remaining variables.

What about $n=3$? We have $f(x,y,z) = x+y+z - xyz$. Now you can see that $DF(x,y,z)$ vanishes only when $1-xy=1-xz=1-yz=0$, which means that $x=y=z=1$.

In the general case, you have $f(x_1,\dots,x_n) = x_1+x_2+\dots+x_n - x_1x_2\cdots x_n$. Now the derivative will vanish only when $$1-\prod_{i\ne j} x_i = 0 \quad\text{for all } j=1,2,\dots,n.$$ Check that this happens if and only if all $x_i=1$.

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  • $\begingroup$ This makes sense, but I think I misunderstood the theorem itself then. Am I not supposed to look at the function $f(x)-f(p)$? And isn't the differential of the function without the parameter I'm solving for supposed to be invertible? And in your solution, when did $p$ become $x$? $\endgroup$
    – MetaColon
    Aug 9, 2019 at 18:08
  • $\begingroup$ If I understood correctly, the condition that $F(x)=0$ implies that $x=p$, which in the end results exactly in the case you described. Does this make sense? $\endgroup$
    – MetaColon
    Aug 9, 2019 at 18:29
  • $\begingroup$ The derivative of $f(x)-f(p)$ is just $Df(x)$. :) Only when $p=(1,\dots,1)$ does the hypothesis of the implicit function theorem fail. $\endgroup$ Aug 9, 2019 at 18:47
  • $\begingroup$ My problem with your arguing is in the fact, that you proof that $(x_1,...,x_n)=(1,...,1)$, instead of proofing that $p=(1,...,1)$. If I'm not mistaken, this is the same because $x$ and $y$ are forced to be the same, because of $f(x)-f(p)=0$ - is this correct? $\endgroup$
    – MetaColon
    Aug 9, 2019 at 19:23
  • $\begingroup$ If $p=(1,\dots,1)$, the hypotheses fail at $x=p$, yes. The question is asking specifically if the implicit function theorem will work in a neighborhood of $p$ when $p\ne (1,\dots,1)$. My work shows that the derivative fails to have maximum rank only at $(1,\dots,1)$, so the hypotheses of the implicit function theorem certainly hold at (and, indeed, near) $p$. $\endgroup$ Aug 9, 2019 at 19:41

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