Let $g(x,y)=(e^x+1) y^2+2(e^{x^2}-e^{2x-1})y+(e^{-x^2}-1)$, I wish to prove that there exists a constant $\overline{y}>0$, such that for any fixed $y\in [0,\overline{y}]$,$g(x,y)=0$ has a solution.
Any help will be very appreciate. I have spent some hard time staring at it without coming up with a solution. Hints or solutions will both be great! Thank you in advance!
EDIT: By drawing picture with wolframalpha I have tested and calculated we can take $\overline{y}=0.01$. But the method is so ugly. Could someone please point out some approach that is not by trial?
EDIT': Someone point it out or confirm it if $0.01$ is correct, please?
EDIT'':
My attempt is:
First note that when $y=0$, we can pick $x(y)=0$ as follows from the definition of $g(x,y)$, so it suffices to find $\overline{y}>0$ such that for any fixed $y\in (0,\overline{y}]$, the equation $g(x,y)=0$ admits a solution $x(y)$.
Claim: We can take $y=0.01$.
Proof: Now fix any $y\in (0,0.01]$, so $y>0$.
$g(0,y)=(1+1)y^2+2(1-\frac{1}{e})y>0$.
Consider $g(-0.5,y)$, as $e^x+1>0,e^{x^2}-e^{2x-1}\ge 0$, we have once $x$ is fixed, $g$ is monotone increasing respect to $y$, so as $y\le 0.01$, $g(-0.5,y)\le g(-0.5,0.01)<-0.198<0$.
Obviously $g$ is continuous, so by intermediate value theorem exists $x\in (0,-0.5)$ such that$g(x,0.01)=0$.
Again, I find the solution above ugly. An elegant proof would be very appreciate. And could someone please tell me if the above proof is correct?
I try using the Implicit function theorem and the Intermediate value theorem.
We have $g(0,0) = 0$ and $\partial_y g(0,0) = 2(1-e^{-1}) \neq 0$. By the Implicit function theorem we get the existence of two open neighbourhoods $(-r,r),(-s,s)$, $r,s > 0$, and of a unique continuously differentiable function $f\colon (-r,r) \to (-s,s)$ such that $g(x,f(x))=0$ and $f(0)=0$. Now let us fix $$\overline{y}:= \max_{[-r/2,r/2]} f.$$ Each number $y$ such that $0 \leq y \leq \overline{y}$ sits in $\mathrm{Im} \,f_{|[-r/2,r/2]} \cap [0,s)$ as well, because $f$ is a continuous path connecting $0$ and $\overline{y}$, and then the claim follows from the Intermediate value theorem. In particular, since $y \in \mathrm{Im} \,f_{|[-r/2,r/2]}$ there exists $x \in [-r/2,r/2]$ such that $f(x)=y$. Thus for any $y \in [0,\overline{y}]$ there exists $x \in [-r/2,r/2]$ such that $g(x,y)=g(x,f(x))=0$.