Show that a function $g$ is uniformly continuous

Question

Let $g:\mathbb{R}\to\mathbb{R}$ be defined by $g(x)=\lim_{y\to x}f(y)$. Show that $g$ is uniformly continuous and that for all $y\in\mathbb{Q}$, $g(y)=f(y)$.

Answer 1

This statement is clearly false. Let $f$ be a continuous function which is not uniformly continuous on $\mathbb{R}$ (for example $e^{x}$). Then $g$ is continuous and equal to $f$ at every point. So $g$ cannot be uniformly continuous as well.

Answer 2

Given that $g:\mathbb{R}\to\mathbb{R}$ is defined by $g(x)=\displaystyle\lim_{y\to x}f(y)$, suppose $g$ is not uniformly continuous on $\mathbb{R}$. Then there exists two sequences $(x_n)$ and $(u_n)$ in $\mathbb{R}$ such that $\lim(x_n-u_n)=0$ and there exists $\varepsilon_0>0$ such that $|f(x_n)-f(u_n)|\geq\varepsilon_0$ for all $n\in\mathbb{N}$. Since $g(x)=\displaystyle\lim_{y\to x}f(y)$, then for any two sequences $(y_n)$ and $(z_n)$ in $\mathbb{Q}\subseteq\mathbb{R}$, $(y_n)$ converges to $x$ and $(z_n)$ converges to $x$, so that $\lim(y_n-z_n)=0$. Now, since $(y_n)$ and $(z_n)$ are convergent, then according to the Cauchy Convergence Criteria, $(y_n)$ and $(z_n)$ are Cauchy sequences. Thus, according to Part (1) and the fact that $f$ is a uniformly continuous function, $(f(y_n))$ and $(f(z_n))$ are Cauchy sequences in $\mathbb{R}$, and thus convergent according to the Cauchy Convergence Criteria. The sequence $(f(y_n))-(f(z_n))$ is convergent, and since $(f(y_n))$ converges to $g(x)$ and $(f(z_n))$ converges to $g(x)$, then $(f(y_n))-(f(z_n))$ converges to 0. Thus, for $\varepsilon_0>0$, $K(\varepsilon_0)\in\mathbb{N}$ is such that for all natural numbers $n\geq K(\varepsilon_0)$, $|f(y_n)-f(z_n)|<\varepsilon_0$. However, since $g$ is not uniformly continuous on $\mathbb{R}$, then $|f(y_n)-f(z_n)|\geq\varepsilon_0$ for all $n\geq K(\varepsilon_0)$. Thus a contradiction has been reached. Therefore, $g$ is uniformly continuous on $\mathbb{R}$.

Part (1) refers to the statement: If $f:\mathbb{Q}\to\mathbb{R}$ is a uniformly continuous function and $(x_n)$ is a Cauchy sequence in $\mathbb{Q}$, then $f(x_n)$ is a Cauchy sequence in $\mathbb{R}$.

Please let me know if I went wrong anywhere. Thanks.