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Let $U\subset \mathbb R^n$ be an open subset, $f:U\to \mathbb R^n$ a differentiable function and $\varphi:\mathbb R^n\to \mathbb R$ of class $C^1$. Suppose $\varphi(f(x))=0$ for every $x\in U$.

I'm trying to prove:

If $a\in U$ and $\nabla \varphi(f(a))\neq 0$, then $\det f'(a)=0$.

Using the chain rule we know that $\varphi(f(x))'=\varphi'(f(x))\cdot f'(x)\equiv 0$ for every $x\in U$ because $\varphi\circ f$ is constant. I also know that since $\varphi\in C^1$, $\varphi'$ is continuous, thus

$$\varphi'(f(a)+v)-\varphi'(f(a))\to 0$$ as $v\to 0$.

My problem is I don't know how to use these facts to prove this question.

2 Answers 2

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The thing you've written as $$ \varphi'(f(x)) $$ (or better, $\varphi'(f(a))$ ) can also be written $$ \nabla \phi (f(a)). $$ So you have $$ \nabla\varphi(f(a))\cdot f'(a) = 0. $$ And you are given that $\nabla\varphi(f(a)) \ne 0$. But when you multiply that on the right by $f'(a)$, you get zero. So the transformation described by "multiply on the right by $f'(a)$" must be singular, hence the determinant of that matrix must be zero.

General hint: when you get stuck on a problem like this, try to build a counterexample. That's what I did. I said "What if, for an open disk in the plane, for instance, $\varphi$ is identically zero? Then the claim is false for $f(x, y) = x$." And then I looked and saw that there was a condition that $\nabla \varphi(f(a)) \ne 0$, and that led me to the answer. As an alternative: look through the hypotheses and see whether there's one you haven't used yet. In your case, that missing one was that the gradient of $\varphi$ at $a$ is nonzero.

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    Thank you for your hint, mainly the example stuff. The problem is time, I spend a lot of time building such examples. – 2017-01-08
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    I didn't understand why the singularity of the transformation described by "multiply on the right by $f'(a)$" implies the determinant of $f'(a)$ must be zero. Thank you again – 2017-01-08
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    If a linear transformation sends a nonzero vector to zero, then its kernel is nontrivial (by definition). That means that $0$ is an eigenvalue (any nonzero vector in the kernel is an eigenvector for $0$), and since the determinant is the product of the eigenvalues, this means that determinant must be zero. – 2017-01-08
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    Let me myself clear, I didn't understand what this transformation "multiphy on the right by $f'(a)$" looks like and what this transformation has to do with the determinant of $f'(a)$. For me your argument would have been true if we had found a non-zero vector $v$ such that $f'(a)\cdot v=0$, in this case, I would agree with you, the $\det f'(a)$ must be zero. Thank you again! – 2017-01-08
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    $\nabla \varphi (f(a))$ is an $1 \times n$ vector. Consider the vector space $V$ of all such $1 \times n$ vectors ($V$ is isomorphic to $\Bbb R^n$, of course!). Let $A = f'(a)$. Define $T : V \to \Bbb R^n: v \mapsto vA$. Then $T$ is a linear transformation on an $n$-dimensional space, and $\nabla \varphi (f(a)) \ne 0$ is in its kernel. Hence the matrix for the xform must (by the logic of the previous comment) have determinant zero. – 2017-01-08
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    I know that you're about to ask "Where did you use $C^1$?" and the answer is "It appears that I did not need to use that." – 2017-01-08
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    I got it! Then $\det T=\det f'(a)=0$, since the matrix of $T$ is equal to the matrix of $f'(a)$ in the canonical bases. – 2017-01-08
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Suppose $\det (df(a)) \ne 0.$ Then $df(a)$ is nonsingular, hence surjective. Thus $df(a)[v] = \nabla \varphi (f(a))$ for some $v\in \mathbb R^n.$ Thus

$$\tag 1\langle \nabla \varphi (f(a)), df(a)[v]\rangle = \langle \nabla \varphi (f(a)), \nabla \varphi (f(a))\rangle = | \nabla \varphi (f(a))|^2 \ne 0.$$

But the left of $(1)$ is precisely $d(\varphi \circ f)(a)[v],$ and we know $d(\varphi \circ f)(a)$ is the zero linear transformation, contradiction.

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    Thank you for your answer. i have a question, I would like to know where you are using $\varphi\in C^1$. – 2017-01-08