Example 4 in Section 9 Chapter 2 of Analysis on manifolds, J. R. Munkres

Question

The fourth example listed in Section 9 of Chapter 2 of Analysis on manifolds (1991, I don't if there are more editions) is about a particular case of the implicit function theorem. The aim of this example is to show that the function you find is unique in a neighborhood but may not be if you consider a larger one. Then, the author defines $f(x,y)=y^2-x^4$ and study the thesis of the theorem around the point $(1,2)$. Here it comes the first problem, because $f(1,2)\neq 0$, so I guess the correct $f$ is

$$f(x,y)=y^2-4x^4.$$

Now, Munkres solves for $y$:

$$y=g(x)=2x^2$$.

The function $g$ is not computed explicitly actually, but certainly $f(x,g(x))=0$ and $g(1)=2$.

Now he draws the following picture:

From this picture I conclude that, when you consider the origin, you are including other possible solutions, namely, the above $g$ and also

$$ z(x)=\begin{cases} 2x^2, & x\geq 0 \\\ -2x^2, & x<0 . \end{cases} $$

However, the function $f$ is $C^\infty$ so, in particular it is $C^3$. However, the above function is not, and the theorem ensures you that it will be. Hence the $z$ is not the the one that the theorem provides. And even if it is not $z$ the function Munkres is thinking about, I cannot imagine how you can paste $x^2$ smoothly to get different ones.

Then, my question is: Is there some detail I am misunderstanding or is it actually a mistake?

Thanks

Answer 1

It seems like you are actually asking if there is a mistake in mathematics, rather than a mistake in Munkres. Indeed, forget about what Munkres is saying, look at the graph of $f(x,y) = y^2-4x^4$ and note that for $g(x) = 2x^2$ for $x \ge 0$ and $g(x) = -2x^2$ for $x < 0$, it holds that $f(x,g(x)) = 0$ and $g(1) = 2$, but $g$ is not $C^\infty$, supposedly contradicting the implicit function theorem.

It turns out that there is no mistake in mathematics and that there is some detail you are misunderstanding.

The implicit function theorem says that there exists a neighborhood of $1$ for which there is a unique continuous function $h$ satisfying $h(1) = 2$ and $f(x,h(x)) = 0$, and it turns out that this unique function $h$ is $C^\infty$. [The implicit function theorem does not directly say there is a unique $C^\infty$ function $h$, though it of course implies there is.] However, there is no comment about the uniqueness or smoothness of a $g$ satisfying $g(1) = 2$ and $f(x,g(x)) = 0$ in other neighborhoods! And this is what Munkres is demonstrating. The lack of uniqueness comes from the fact that also $h(x) \equiv 2x^2$ (for $x \ge 0$ and $x < 0$) also satisfies $h(1) = 2$ and $f(x,h(x)) = 0$, and the lack of smoothness comes from the fact (that you noted) that $g$ is not even $C^3$.

Answer 2

Yes, there is some misunderstanding:

First: Implicit Function Theorem state that one variable can be expressed as a unique function over other variables. In this case $y=g(x)$, so there is no $f$ or $z$ can explain by this theorem! Just a relation between variables.

Second: The example and its figure try to show that the uniqueness of this relation is not guaranteed if you large your neighbourhood. In this case theorem fails and no valid relation can be expressed any more. In fact when $x$ reaches $0$, by $y=g(x)=2x^2$ we get $y=0$ too. But in this case $\partial f / \partial y$ also become $0$, that result in theorem criteria fails, so after that any relation no more remain valid. The figure is $yx$-axis and shows both two function that claims express $y$ over $x$. Don't confuse it by graph of $f$.