# Equivalent formulation of uniform continuity (uniform convergence of shifted functions)

I'm working on the following exercise (from T.Tao's Analysis 2 book):

"Let $f\colon \mathbb{R}\to\mathbb{R}$ be a function. For any $a\in\mathbb{R}$, let $f_a\colon\mathbb{R}\to\mathbb{R}$ be the shifted function $f_a(x):=f(x-a)$.

a) Show that $f$ is continuous iff, whenever $(a_n)_{n=1}^\infty$ is a sequence of real numbers which converges to zero, the shifted functions $f_{a_n}$ converge pointwise to $f$.

b) Show that $f$ is uniformly continuous iff, whenever $(a_n)_{n=1}^\infty$ is a sequence of real numbers which converges to zero, the shifted functions $f_{a_n}$ converge uniformly to $f$."

Now, I proved (a) and (b) (the rightward implication) but I'm stuck on the remaining one, namely the leftward implication of (b).

EDIT: I'm writing the proof below, I need just an hint to finish it.

(Proof)

Let $\varepsilon>0$ and $x_0\in X$ be arbitrary, let also $(x_n)_{n=1}^\infty$ be a sequence converging to $x_0$ and consider the sequence $(a_n)_{n=1}^\infty$ defined by $a_n:=x_0-x_n$; then by hypothesis there exists $N>0$ such that $|f_{a_n}(x_0)-f(x_0)|=|f(x_n)-f(x_0)|<\varepsilon\ \forall n\geq N$ and since $(f_{a_n})_{n=1}^\infty$ converges uniformly to $f$ and $x_0$ and $(x_n)_{n=1}^\infty$ were chosen arbitrarily we have that this $N>0$ is such that $|f(x_n)-f(x_0)|<\varepsilon$ for all $x_0\in\mathbb{R}$ and $x_n \to x_0$. (TODO)

Now, how can I translate what I've found into $\exists \delta>0: |f(x)-f(x_0)|<\varepsilon\ \forall x,x_0\in X$ such that $|x-x_0|<\delta$?

• There is a direct proof showing that the sequential definition of uniformly continuous is satisfied. This proof by contradiction is just applying the negation of 'f uniformly continuous' n times to get our hands on a sequence. – Matt A Pelto Oct 1 '17 at 6:25

Suppose that the sequence of shifted functions $\{f_{a_n} \}_{n=1}^\infty$ converges uniformly to $f$ whenever $\{a_n\}_{n=1}^\infty$ is a sequence of real numbers that converges to zero.
We shall assume that $f$ is not uniformly continuous and reach a contradiction. Since $f$ is not uniformly continuous we know there is a $\hat{\varepsilon}>0$ so that for each $n=1,2,\ldots$ there are points $x_n,y_n \in \mathbb{R}$ such that $|x_n-y_n|<\frac{1}{n}$ but $|f(x_n)-f(y_n)| \geq \hat{\varepsilon}$. For $n=1, 2, \ldots$ we set $a_n:=x_n-y_n$. Since $a_n \to 0$ as $n \to \infty$, we know that the sequence of functions $\{f_{a_n} \}_{n=1}^\infty$ converges uniformly to $f$. Hence we may find a positive integer $N$ such that $\displaystyle \sup_{x \in \mathbb{R}}|f_{a_n}(x)-f(x)|<\hat{\varepsilon}$ whenever $n \geq N$. In particular we have that $|f_{a_N}(x_N)-f(x_N)|<\hat{\varepsilon}$, but $f_{a_N}(x_N)=f(y_N)$. This is a contradiction, and so our assumption must be false. Therefore $f$ is uniformly continuous.