How can I linearize for the continuous function $f$ the functional equation $f^2(x)+1=2f(x+1)$
I' ve been studying recently the book for functional equations by Christopher G. Small and I' ve encountered the term "Linearization" of functional equations. I cannot figure out how to apply this method in this equation but if I had the equation $f^2(x)=2f(x+1)$ instead I know I could apply logarithms in both sides (thought I don't know if $f(x)>0$) such that $2\ln f(x)=\ln f(x+1)+ \ln2$.
Is it possible to do something like this in the first functional equation?
I am not sure that linearization is applicable to this functional equation, but I will depict another way to solve this. First, lets consider trivial solutions. $$f^2(x)+1=2f(x+1)$$ $$f(x)=c$$ $$c^2+1=2c$$ $$c^2-2c+1=0$$ $$(c-1)^2=0$$ $$c=1$$ $$f(x)=1$$ I assume that the function is meant to be continuous and defined everywhere. Lets find some restrictions we can put on $f(x)$. The domain must be all real numbers. $$f(x+1)=\frac{f^2(x)}{2}+\frac{1}{2}$$ $$f(x-1)= \pm \sqrt{2f(x)-1}$$ Both of these operations should be valid if we want $f(x)$ to have a domain of all real numbers. There is no problem scaling a value of $f(x)$ up, but there may be a problem scaling down due to the domain of square roots. $$2f(x)-1 \geq 0$$ $$2f(x) \geq 1$$ $$f(x) \geq \frac{1}{2}$$ This means that we only need to consider the positive branch of the square root. $$f(x-1)= \sqrt{2f(x)-1}$$ We will further bound $f(x)$ by analyzing a sequence constructed from the equation. $$a_{n+1}= \sqrt{2a_n-1}$$ We must ensure that $a_n$ never goes below $\frac{1}{2}$. We will first establish that $a_n$ is a monotonically decreasing sequence. $$a_n>a_{n+1}$$ $$a_n>\sqrt{2a_n-1}$$ $$a_n^2>2a_n-1$$ $$a_n^2-2a_n+1>0$$ $$(a_n-1)^2>0$$ This is true for all $a_n$ except at $a_n=1$, which is the only solution to $a_n=a_{n+1}$. Since $a_n$ is monotonically decreasing, all $a_n<1$ can be ruled out as they will eventually go below $\frac{1}{2}$. We have tightened the bound on $f(x)$ to: $$f(x) \geq 1$$ For all $a_n>1$, we must prove that $a_n$ always stays above $1$. $$\sqrt{2a_n-1}>1$$ $$2a_n-1>1$$ $$2a_n>2$$ $$a_n>1$$ Since $a_n=1$ is the only solution to $a_n=a_{n+1}$, $a_n$ is monotonically decreasing, and $a_n$ stays above $1$ for all $a_n>1$, $a_n$ converges to $1$ for all $a_0>1$. Convergence of this sequence to $1$ and the bound on $f(x)$ means that: $$\lim_{x \to -\infty} f(x)=1$$ $$\text{From the main identity:}$$ $$f^2(x)+1=2f(x+1)$$ It is clear that the entirety of $f(x)$ can be defined if $f(x)$ is known over a domain interval of length $1\text{.}$ To find the other values $f(x)$ can be shifted using the identity. Let this interval be $x \in [0, 1]\text{.}$ Let $S(x)$ be a continuous function defined on this interval. $S(x)$ can be used contrive other values of $f(x)\text{.}$ For $f(x)$ to be continuous at $x=1$, S(x) must satisfy: $$S^2(0)+1=2S(1)$$ Generally, any continuous seed function on the $[0, 1]$ interval will allow for computation of a continuous function $f(x)$ given that the seed function satisfies: $$S(x) \geq 1$$ $$S^2(0)+1=2S(1)$$ Conclusions that can be drawn about the resulting continuous function $f(x)$ include: $$f^2(x)+1=2f(x+1)$$ $$f(x) \geq 1$$ $$\lim_{x \to -\infty} f(x)=1$$ Additionally, there are a few things that can be said about $f(x)$ as $x$ tends to $\infty\text{.}$
If $S(x)=1$, then $\lim_{x \to \infty}f(x)=1$
If $S(x)>1$, then $\lim_{x \to \infty}f(x)=\infty$
If $S(x)=1$ for some $x$ in the interval but not all $x$, then: $$\liminf_{x \to \infty}f(x)=1 \quad\quad \limsup_{x \to \infty}f(x)=\infty$$ This occurs because the sequence $b_n$ is monotonically increasing for all $b_n>1$ where: $$b_{n+1}=\frac{b_n^2}{2}+\frac{1}{2}$$ $$b_{n+1}>b_n$$ $$\frac{b_n^2}{2}+\frac{1}{2}>b_n$$ $$b_n^2+1>2b_n$$ $$b_n^2-2b_n+1>0$$ $$(b_n-1)^2>0$$ Note that using the seed function method to generate $f(x)$ can produce all possible $f(x)$ that are continuous everywhere. The next step would be to determine if any solutions are differentiable everywhere. I will come back to edit this answer if I find any.