show that $f_{n} \rightarrow f$ uniformly on $\mathbb{R}$

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be uniformly continuous, and let $f_n(x) := f(x\ +\ \frac{1}{n} )\ \forall x\ \in\ \mathbb{R}$

Show that $f_{n} \rightarrow f$ uniformly on $\mathbb{R}$

I'm trying to prove this rigorously using the $\epsilon,\ N$ definition, but I don't know how to make use of the uniform continuity here.

Any help or insight is deeply appreciated.

Answer 1

Let $ e>0 $, there exists $ d>0 $ so that $ \forall x,y \in \mathbb{R} : |x-y| < d \implies |f(x) - f(y)| < e $.

Observe that there exists $n_0 \in \mathbb{N}$ so that $ \forall n\geq n_0 , ~ \frac{1}{n} < d$.

Consequently $ \forall n\geq n_0, ~ |x+\frac{1}{n} - x| <d \implies |f_n(x) - f(x)|< e$.

Answer 2

(I assume you meant to specify that $f(x) = x$.) Given an $\epsilon > 0$, let $N\in \Bbb N$ be so that $1/N < \epsilon$. I claim that for any $f_n$ with $n > N$ and any $x \in \Bbb R$, we have $|f_n(x) - f(x)| < \epsilon$. (This is the definition of uniform convergence applied to our sequence.)

To prove my claim, we insert and calculate: $$ |f_n(x) - f(x)| = \left|x + \frac1n - x\right| = \left|\frac1n\right| = \frac1n < \frac 1N < \epsilon $$

Answer 3

$f_n(x)=f(x+\frac{1}{n})$

For a fixed $x$ ;$\lim_{n\to \infty} f_n(x)=\lim_{n\to \infty }f(x+\frac{1}{n})=f(\lim_{n\to \infty }(x+\frac{1}{n}))=f(x) $ as $f$ is continuous.

So $f_n(x)\to f(x)$ pointwise.

Now $\sup_{x\in \Bbb R} |f_n(x)-f(x)|=\sup_{x\in \Bbb R}| f(x+\frac{1}{n})-f(x)|$

Since $f$ is uniformly continuous so $|(x+\frac{1}{n})-x|=\frac{1}{n}\to 0\implies |f(x+\frac{1}{n})-f(x)|\to 0$

Hence $M_n=\sup_{x\in \Bbb R} |f_n(x)-f(x)|=\sup_{x\in \Bbb R}| f(x+\frac{1}{n})-f(x)|\to 0$ as $n\to \infty$.

By Weirstress Test series converges uniformly.

Answer 4

You can use the definition of uniform convergence over $A$, that is, a sequence of functions $f_n$ uniformly converges to the limit function $f$ over $A$, if and only if: $$\lim_{n\to\infty}\sup_{x\in A}\left|f_n(x)-f(x) \right|=0$$ Now, $f$ is uniformly continuous that is, if a certain $\varepsilon>0$ is such that $$\left|f(x_1)-f(x_2) \right|<\varepsilon$$ thus there must exist at least a $\delta>0$ such that the difference between the corresponding arguments of $f$ is bounded by it, for any $x_1, x_2$: $$\left|x_1-x_2\right|<\delta$$

In your case, of course $x_1 :=x$ and $x_2:=x+\frac{1}{n}$ therefore: $$\forall \varepsilon>0:\left|f_n(x)-f(x) \right|=\left|f\left(x+\frac{1}{n}\right)-f(x) \right|<\varepsilon$$

$$\exists\delta>0:\left|\left(x+\frac{1}{n}\right)-x\right|=\left|\frac{1}{n}\right|<\delta$$ Calling $n_0 = \left \lfloor \frac{1}{\delta}\right \rfloor+1$ (the minimum greater integer), then, considering a certain $\bar x$, this is the definition of the limit of the sequence $f_n(\bar x)$ that is, $f_n(\bar x)$ has limit $f(\bar x)$, because, $$\forall \varepsilon>0, \exists n_0: \forall n\ge n_0 \Rightarrow \left|f_n(\bar x)-f(\bar x)\right|<\varepsilon$$ and we did found $n_0$ just before.

Since the continuity is uniform, the property above is valid for each $x_1,x_2$ (or equivalently, for each $x$) so that $\delta$ depends just on $\varepsilon$ and not on $\bar x$ and so this property holds also for the supremum over $\mathbb{R}$.