Show that if the function $f: R \to R$ is uniformly continuous, then the sequence $f_{n}(x) = f(x + \frac{1}{n})$ converges uniformly.
I know that f(x + 1/n) approaches f(x) as $n \to \infty$. I feel like I could say $Sup |f_n(x) - f(x)| = Sup|x - x| \ as \ n \to \infty$ = 0. Therefore, the function is uniformly convergent.
My instructor told me that using this version of the definition would not be the best idea though. He said using the $\forall \epsilon > 0 \exists N ...$ would be easier. Does anyone agree/disagree and see a way of doing this?
Thank you!
I don't think the proof you wrote really makes sense. $Sup |x-x|$ should just be 0 but this doesn't seem to be what you mean.
Let $\varepsilon > 0$. By definition of uniform continuity, there is some $\delta > 0$ such that for all $x, y$ with $|x-y|<\delta$, $|f(x)-f(y)|<\varepsilon$. Take now some $M \in \mathbb N$ such that $1/N < \delta$. This exists by the Archimedean property. Take now any $n \geq N$. We have that $|f(x)-f_n(x)| = |f(x) - f(x+1/n)|$ for all $x$. As $|x+1/n - x| = 1/n \leq 1/N < \delta$, we conclude that for all $x$ and for all $n \geq N$, $|f(x) - f_n(x)| < \varepsilon$. That's precisely the definition of uniform convergence.