Cauchy sequence under a uniform continuous function

Question

Let , $f:(1,4)\to \mathbb R$ be uniformly continuous and $\{a_n\}$ be a Cauchy sequence in $(1,2)$.

Consider: $x_n=a_n^2f(a_n^2)$ and $y_n=\frac{1}{1+a_n^2}f(a_n^2)$

Then which is correct ?

(A) Both $\{x_n\}$ and $\{y_n\}$ are Cauchy sequences in $\mathbb R$.

(B) $\{x_n\}$ is Cauchy sequence in $\mathbb R$ but $\{y_n\}$ is not Cauchy sequence in $\mathbb R$.

(C) $\{y_n\}$ is Cauchy sequence in $\mathbb R$ but $\{x_n\}$ is not Cauchy sequence in $\mathbb R$.

(D) Neither $\{x_n\}$ nor $\{y_n\}$ are Cauchy sequence in $\mathbb R$.

Since $\{a_n\}$ is Cauchy sequence so $\{a_n^2\}$ is also Cauchy sequence . As $f$ is uniformly continuous so, $f(a_n^2)$ is also Cauchy sequence.

So, both are convergent and so their product is also convergent and hence $\{x_n\}$ is Cauchy sequence.

But what about $\{y_n\}$ ?

Answer 1

Hint: for $\{y_n\}$,

$$|y_n-y_m|=\left| \dfrac{f(a_n^2)}{(1+a_n^2)}-\dfrac{f(a_m^2)}{(1+a_m^2)}\right| \le \left| \dfrac{f(a_n^2)-f(a_m^2)}{(1+a_n^2)}\right|+|f(a_m^2)|\left| \dfrac{1}{1+a_n^2}-\dfrac{1}{1+a_m^2}\right| \le \dfrac{1}{2}|f(a_n^2)-f(a_m^2)|+\dfrac{|f(a_m^2)||a_m^2-a_n^2|}{(1+a_n^2)(1+a_m^2)}\le \dfrac{1}{2}|f(a_n^2)-f(a_m^2)|+\dfrac{|f(a_m^2)|}{4}|a_m^2-a_n^2|$$

and $\{a_n^2\}, \{f(a_n^2)\}$ are Cauchy sequences.