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$f:\mathbb{R}\to \mathbb{R}, ~ f'(x)<0 ~\forall x\in \mathbb{R}$. I have to prove that $f(x)=x$ has unique solution.

$K(x)=f(x)-x$ is strictly decreasing $\implies K$ is 1-1 so $f(x)=x$ has at most one solution.

But how do I prove that it has solution?

I thought that $Df=R$ and $g(R)=R$ so they have. But is there any algebra solution?

Thanks!!!

erfink
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Tom Avram
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  • Let's try to prove it doesn't need a solution. As $x$ goes right, $f(x)$ will have to either increase or be less than $x$, and since it has to decrease going right, it will have to be less than $x$. As it goes right. As $x$ goes left, $f(x)$ either has to decrease or be greater than $x$, and since it has to increase going left, it has to be greater than $x$. We can see $f(x)$ has to be less than $x$ as it goes to the right and greater than $x$ as it goes left, and you can show that through the same philosophy as the intermediate value theorem you need at least one unique solution to $f(x)=x$. – Jacob Claassen May 07 '17 at 20:30

1 Answers1

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$f(x)\,$ must be continuous, given that it is differentiable. If $f(x) = x$ had no solutions then, by continuity, either $\,f(x) \gt x\,$ or $\,f(x) \lt x\,$ for all $\forall x \in \mathbb{R}$. Suppose that $f(x) \gt x\,$, and let $\,y_0=f(0)\,$, but then $\lim_{x \to \infty} f(x) \ge \lim_{x \to \infty} x = + \infty\,$ implies that there exists some $x_0$ such that $f(x) \gt y_0 = f(0)$ for all $\forall x \gt x_0$. The latter, however, contradicts the decreasing monotonicity of $f(x)$. The case $f(x) \lt x$ can be disproved with a similar argument for $x \to -\infty\,$. Therefore, at least one solution to $f(x)=x$ must exist and, by injectivity, there can only be one such solution.

dxiv
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