Let $g_n(x)= g(x+\frac{1}{n})$. Prove that $(g_n)$ converges uniformly to $g$.

Question

Let $g: \mathbb{R}\longrightarrow \mathbb{R}$ be a uniformly continuous function and, for every $n\in\mathbb{N}$, let $g_n(x)= g(x+\frac{1}{n})$. Prove that $(g_n)$ converges uniformly to $g$.

My try:

Because $g$ is uniformly continous then we know that $\forall \epsilon>0 \quad\exists \delta>0$ tal que $\forall x,y\in \mathbb{R}, |x-y|<\delta$, then $|g(x)-g(y)|<\epsilon$. Now let's take $n\in\mathbb{N}$ and $x+\frac{1}{n}, x \in \mathbb{R}$. Then $|x+\frac{1}{n}-x|<\delta$, then $|g(x+\frac{1}{n})-g(x)|<\epsilon$. Then $|g_n(x)-g(x)|<\epsilon$, $\forall x\in\mathbb{R}$.

I'm not sure if my poof is correct. Any suggestions would be great!

Answer 1

Since $g$ is uniformly continuous, given $\varepsilon>0$ there exists $\delta>0$ such that

$$\forall x, y\in \mathbb R; |x-y|<\delta\Rightarrow |g(x)-g(y)|<\varepsilon.$$ By the Archimedian property of $\mathbb R$ we can find $n_0\in\mathbb N$ such that $n_0\cdot \delta>1$. Then,

$$n\geq n_0 \Rightarrow \left|x-\frac{1}{n}-x\right|=\frac{1}{n}\leq \frac{1}{n_0}<\delta.$$

Hence,

$$n\geq n_0\Rightarrow |g_n(x)-g(x)|<\varepsilon$$

for all $x\in \mathbb R$. Since this holds for every $x\in\mathbb R$, it follow that

$$n\geq n_0 \Rightarrow \sup_{x\in\mathbb R} |g_n(x)-g(x)|\leq \varepsilon.$$

This means, $(g_n)_{n\in\mathbb N}$ is uniformly continuous.