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Let $f:\mathbb{R}\times[0,1)\to \mathbb{R}$ be a continuous function $f(x,y) = (x^2+y^2)(ye^{|x|}-1)$. Prove that, for each $x\in \mathbb{R}$, there is a unique $y=\eta(x)\in[0,1)$ such that $f(x,\eta(x)) = 0$ but the function $\eta:\mathbb{R}\to[0,1)$ is not continuous

First of all, I tried to suppose $2$ differente $y$ and $y_1$ functions and did:

$$(x^2+y^2)(ye^{|x|}-1) = (x^2+y_1^2)(y_1e^{|x|}-1)$$

but I cannot end with $y = y_1$.

Also, I'm studying the implicit function theorem, which kinda says:

If a function $f$ is of class $C^k$ and there exists $(x_0,y_0)$ such that $f(x_0,y_0) = 0$, and $\frac{\partial f}{\partial y}(x_0,y_0) \neq 0$ then there will exists a ball $B(x_0,\delta)$ and $J = [y_0-\epsilon,y_0+\epsilon]$ and a unique $\eta(x)$ where $f(x,\eta(x))=0$ for all $x\in B(x_0,\delta)$

We can take $(x_0,y_0) = (0,0)$, so $f(0,0) =0$ and $\frac{\partial f}{\partial y} = 2y\cdot y\cdot e^{|x|} + y^2e^{|x|}-2y = 3y^2e^{|x|}-2y = y(3ye^{|x|}-2)$ which is $0$ at $(0,0)$, so I don't know if I can apply the theorem and say that there exists such function $\eta(x)$. Also, why $\eta(x)$ is not continuous?

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Lets look at your problem first. In order to make sure your function is zero at some point, one of the factors has to be $0$. So in this case: $$ f(x)=(x^2+\eta(x)^2)(\eta(x)e^{|x|}-1) $$ If you want your first factor to vanish, you need both $x^2$ and $\eta(x)^2$ to be zero. So your function has to statisfy $\eta(0)=0$.
For the other one, you get that $$ \eta(x)e^{|x|}=1 $$ Or in a more simple way:
$$ \eta(x)=e^{-|x|} $$ Combining the pieces, you get the function $\eta(x)$ is discontinuous. If you look at the function, it is differentiable for any open set $U \subset \mathbb{R} \setminus0$. If you try to apply the implicit function theorem there, you will get a non-vanishing derivative in the $y$ component and that gives you the local(!) existence of a smooth(!) parametrisation $y=\eta(x)$. In your case, however, you got a vanishing derivative whenever $y=0$

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  • $\begingroup$ why is this function unique? $\endgroup$ – Paprika May 31 '17 at 1:11
  • $\begingroup$ Take a look how the function was constructed and remember multiplicative inverses are unique. $\endgroup$ – F. Conrad May 31 '17 at 1:22

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