If $f$ is continuous, then $(f_n)_n$ converges pointwise to $f$

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function. Consider for every $n \in \mathbb{N}_0$ the function $$ f_n : \mathbb{R} \rightarrow \mathbb{R}: x \mapsto f(x + 1/n). $$ I need to determine if the following is false or true:

if $f$ is continuous, then the sequence $(f_n)_n$ converges pointwise to $f$ on $\mathbb{R}$.

I think it is true. To prove this, we need to show that $$ \forall x \in \mathbb{R}, \forall \epsilon > 0, \exists n_0 \in \mathbb{N}, \forall n \in \mathbb{N}: n \geq n_0 \Rightarrow | f_n(x) - f(x) | < \epsilon. $$ I was thinking we need to use the triangle inequality to get $$ | f_n(x) - f(x) | $$ smaller then epsilon, by using the continuity of $f$. I'm not sure how to do this though. Help would be appreciated.

Answer 1

For episolon-delta:

We supppose $f$ is continuous. Let $x_0 \in \mathbb{R}$ and $\epsilon \gt 0$. As $f$ is continuous, there is a $\delta \gt 0$, so that for all $x \in \mathbb{R}$ with $\lvert x-x_0\rvert \lt \delta$ we have $\lvert f(x) - f(x_0)\rvert \lt \epsilon$.
Now we find an $n_0 \in \mathbb{N}: \frac{1}{n_0} \lt \delta$. Let $n \geq n_0$. Then $$\lvert(x_0 + \frac{1}{n}) - x_0\rvert = \lvert \frac{1}{n}\rvert = \frac{1}{n} \leq \frac{1}{n_0} \lt \delta \implies \lvert f(x_0 + \frac{1}{n}) - f(x_0)\rvert = \lvert f_n(x_0) - f(x_0)\rvert \lt \epsilon$$

Answer 2

Let $x\in\mathbb R$, and $\varepsilon >0$. Since $f$ is continuous, there exists a $\delta>0$ s.t. $$\forall t\in (x-\delta,x+\delta)\quad |f(t)-f(x)|\leq \varepsilon. $$

As the sequence $(x+1/n)_n$ converges to $x$ as $n\to+\infty$, there exists $N\in\mathbb N$ s.t. $$\forall n\in\mathbb N\quad n\geq N\Rightarrow x+1/n \in (x-\delta,x+\delta)$$ thus $|f_n(x)-f(x)|=|f(x+1/n)-f(x)|\leq \varepsilon\quad \forall n\geq N. $