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The answer is yes. We may assume wlog that $\alpha$ is increasing. Take any $x_0$ and define $x_{n+1}=f(x_n)$. Then $|x_{n+1}-x_n|\leq k^n|x_1-x_0|$ where $k=\alpha(|x_1-x_0|)<1.$ Indeed, we have $$|x_{n+1}-x_n|\leq\alpha(|x_{n}-x_{n-1}|)|x_{n}-x_{n-1}|.$$ In particular, $|x_{n+1}-x_n|\leq|x_n-x_{n-1}|$, from which follows by induction that $|x_{n+1}-x_n|\leq|x_1-x_0|.$ So we can continue the previous inequality: $$|x_{n+1}-x_n|\leq k|x_{n}-x_{n-1}|,$$ and again by induction $$|x_{n+1}-x_n|\leq k^n|x_1-x_0|,$$ Therefore the series $$\sum|x_{n+1}-x_n|$$ is majorized by a geometric progression, and it follows that $x_n$ is a Cauchy sequence.

(I use notation $|x-y|$ for the distance, even if $x-y$ is not defined.)

The answer is yes, if $\alpha$ is increasing. Take any $x_0$ and define $x_{n+1}=f(x_n)$. Then $|x_{n+1}-x_n|\leq k^n|x_1-x_0|$ where $k=\alpha(|x_1-x_0|)<1.$ Indeed, we have $$|x_{n+1}-x_n|\leq\alpha(|x_{n}-x_{n-1}|)|x_{n}-x_{n-1}|.$$ In particular, $|x_{n+1}-x_n|\leq|x_n-x_{n-1}|$, from which follows by induction that $|x_{n+1}-x_n|\leq|x_1-x_0|.$ So we can continue the previous inequality: $$|x_{n+1}-x_n|\leq k|x_{n}-x_{n-1}|,$$ and again by induction $$|x_{n+1}-x_n|\leq k^n|x_1-x_0|,$$ Therefore the series $$\sum|x_{n+1}-x_n|$$ is majorized by a geometric progression, and it follows that $x_n$ is a Cauchy sequence.

(I use notation $|x-y|$ for the distance, even if $x-y$ is not defined.)

The answer is yes. Take any $x_0$ and define $x_{n+1}=f(x_n)$. Then $|x_{n+1}-x_n|\leq k^n|x_1-x_0|$ where $k=\alpha(|x_1-x_0|)<1.$ Indeed, we have $$|x_{n+1}-x_n|\leq\alpha(|x_{n}-x_{n-1}|)|x_{n}-x_{n-1}|.$$ In particular, $|x_{n+1}-x_n|\leq|x_n-x_{n-1}|$, from which follows by induction that $|x_{n+1}-x_n|\leq|x_1-x_0|.$ So we can continue the previous inequality: $$|x_{n+1}-x_n|\leq k|x_{n}-x_{n-1}|,$$ and again by induction $$|x_{n+1}-x_n|\leq k^n|x_1-x_0|,$$ Therefore the series $$\sum|x_{n+1}-x_n|$$ is majorized by a geometric progression, and it follows that $x_n$ is a Cauchy sequence.

(I use notation $|x-y|$ for the distance, even if $x-y$ is not defined.)

The answer is yes. We may assume wlog that $\alpha$ is increasing. Take any $x_0$ and define $x_{n+1}=f(x_n)$. Then $|x_{n+1}-x_n|\leq k^n|x_1-x_0|$ where $k=\alpha(|x_1-x_0|)<1.$ Indeed, we have $$|x_{n+1}-x_n|\leq\alpha(|x_{n}-x_{n-1}|)|x_{n}-x_{n-1}|.$$ In particular, $|x_{n+1}-x_n|\leq|x_n-x_{n-1}|$, from which follows by induction that $|x_{n+1}-x_n|\leq|x_1-x_0|.$ So we can continue the previous inequality: $$|x_{n+1}-x_n|\leq k|x_{n}-x_{n-1}|,$$ and again by induction $$|x_{n+1}-x_n|\leq k^n|x_1-x_0|,$$ Therefore the series $$\sum|x_{n+1}-x_n|$$ is majorized by a geometric progression, and it follows that $x_n$ is a Cauchy sequence.

(I use notation $|x-y|$ for the distance, even if $x-y$ is not defined.)

The answer is yes. Take any $x_0$ and define $x_{n+1}=f(x_n)$. Then $|x_{n+1}-x_n|\leq k^n|x_1-x_0|$ where $k=\alpha(|x_1-x_0|)<1.$ Therefore the series $$\sum|x_{n+1}-x_n|$$ is majorized by a geometric progression, and it follows that $x_n$ is a Cauchy sequence.

(I use notation $|x-y|$ for the distance, even if $x-y$ is not defined.)

The answer is yes. Take any $x_0$ and define $x_{n+1}=f(x_n)$. Then $|x_{n+1}-x_n|\leq k^n|x_1-x_0|$ where $k=\alpha(|x_1-x_0|)<1.$ Indeed, we have $$|x_{n+1}-x_n|\leq\alpha(|x_{n}-x_{n-1}|)|x_{n}-x_{n-1}|.$$ In particular, $|x_{n+1}-x_n|\leq|x_n-x_{n-1}|$, from which follows by induction that $|x_{n+1}-x_n|\leq|x_1-x_0|.$ So we can continue the previous inequality: $$|x_{n+1}-x_n|\leq k|x_{n}-x_{n-1}|,$$ and again by induction $$|x_{n+1}-x_n|\leq k^n|x_1-x_0|,$$ Therefore the series $$\sum|x_{n+1}-x_n|$$ is majorized by a geometric progression, and it follows that $x_n$ is a Cauchy sequence.

(I use notation $|x-y|$ for the distance, even if $x-y$ is not defined.)