Exercise on the implicit function theorem.

Question

Prove that the equation $x^2+y^2+\sin y=0$ defines a unique function $y=f(x)$ in a neighbourhood of $(0,0)$. Prove also that in $x=0$ there's a maxima for $f$.

I tried to do this exercise in two different ways and I'm asking you to tell me if I did something wrong, if the procedure is correct and if I could have done something better.

To prove that the equation defines a unique function $y=f(x)$ I want to use the implicit function theorem with the function $F(x,y)=x^2+y^2+\sin y$.

We have that $F(0,0)=0$ and $F$ is clearly $C^{\infty}$ so we don't have regularity problems. It's easy to verify that $F_y(x,y)=2y+\cos y$ and so $F_y(0,0)=1 \neq 0$.

For the implicit function theorem we have that there exists a unique function $f \colon \mathbb{R} \to \mathbb{R}$ at least defined in a neighbourhood of $0$ such that $y=f(x)$. We also remind that as $F$ is $C^{\infty}$ even $f$ is $C^{\infty}$.

Now we have the two different solutions for the second question, proving that in $x=0$ there is a maxima.

1. We know from the implicit function theorem that $$f'(x)=-\frac{F_x(x,f(x))}{F_y(x,f(x))}$$ and so $$f'(x)=0$$ which tells us that $x=0$ is a stationary point for $f$. We can now use the chain rule to say that $$f''(x)=-\frac{[F_{xx}(x,f(x))+F_{xy}(x,f(x))f'(x)]F_y(x,f(x))-F_x(x,f(x))[F_{yx}(x,f(x))+F_{yy}(x,f(x))}{F_y(x,f(x))^2}$$ from which $f''(0)<0$ and so $f$ has a maxima in $0$.

2. We know that $f$ is $C^{\infty}$ so we have that $$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+o(x^2)$$ and for what we observed in the first part we have $f(0)=f'(0)=0$. We know also that $F(x,f(x))=0$ and so, by plugging in the last equation the expression we found for $f(x)$ we have that $$x^2+\frac{f''(0)}{2}x^2+o(x^2)=0 \implies f''(0)=-2$$ and so the thesis.

Is it alright? Suggestions?

Answer 1

Your reasoning looks correct to me, and the first idea seems to work fine (I didn't check the details of your computation of the second derivative, though).

I am a little subconscious about the second idea, since little o only makes sense "in the limit" as $x\rightarrow0$. I guess you can make it work, but you would have to be careful to make it rigorous.

In the same spirit, however, I can suggest the following: you know that $f(0)=0$ and that
$$ F(x,f(x))=x^2+(f(x))^2+\sin (f(x))=0 $$ holds in a neighborhood of the origin, correct? In particular, close to $0$ $$ \sin (f(x))=-(x^2+(f(x))^2)<0 $$ so $f(x)$ has to be negative there (otherwise you would reach a contradiction). Therefore, $0$ is a local maximum. Does that make sense?