# Uniform Convergence

Here is the question from the book:
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Let $f: \mathbb{R} \to \mathbb{R}$ be a function. For any $a\in \mathbb{R}$, let $f_a :\mathbb{R}\to \mathbb{R}$ be the shifted function $f_a(x):=f(x-a)$.

(a) Show that $f$ is continuous if and only if, whenever $(a_n)_{n=0}^{\infty}$ 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 if and only if, whenever $(a_n)_{n=0}^{\infty}$ is a sequence of real numbers which converges to zero, the shifted functions $f_{a_n}$ converge uniformly to $f$.

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$(\Rightarrow)$ Let $x_0 \in \mathbb{R}$. Suppose $f$ is continuous. That is, given $\epsilon > 0$ there exists $\delta > 0$ such that if $|x-x_0| < \delta$ then $|f(x)-f(x_0)| < \epsilon$.

Let $x_n = x_0 - a_n$. So given $\delta > 0$ there exists $N' > 0$ such that $|x_n - x_0|< \delta$ for all $n > N'$.

Given $\epsilon' > 0$ take $\epsilon = \epsilon'$, and take $N = N'$.

So by the continuity of $f$ we get that $|x_n - x_0| < \delta$ for all $n>N' = N$.

Thus, $|f(x_n) - f(x_0)| < \epsilon = \epsilon'$.

But, $f(x_n) = f(x_0-a_n) = f_{a_n}(x_0)$ for all $n>N$.

So we have $|f_{a_n}(x_0) - f(x_0)| < \epsilon$ for all $n>N$.

Thus $f_{a_n}$ converges pointwise to $f$.

$(\Leftarrow)$ Suppose given $a_n \to 0$ that $f_{a_n}$ converges pointwise to $f$. Given $x_0$ take a sequence $x_n$ which converges to $x_0$. That is, $a_n = x_0 - x_n \to 0$.

Since $f_n$ converges pointwise to $f$, given $\epsilon > 0$ there is some $N>0$ such that $|f_{a_n}(x_0) - f(x_0)| < \epsilon$ for all $n>N$.

But, $|f_{a_n}(x_0) - f(x_0)| = |f(x_0 - a_n) - f(x_0)| = |f(x_n) - f(x_0)|$.

Hence, $|f(x_n) - f(x_0)| < \epsilon$ for all $n>N$.

That is, $f(x_n)$ converges to $f(x_0)$. Thus, $f$ is continuous.

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I don't think there is anything wrong with any of that (if there is, or you have any comments please say something).

However, I am not too sure about part (b).

The first direction is the same as in part (a), but the other direction I am not sure about.

I have been trying just the basic epsilon-delta definition way (similar to what I did in part (a) with this direction) and not really getting anywhere (which I guess could be expected since we don't have a uniform continuity definition with sequences [at least in our book]).

Any ideas? Thanks!

edit... I am converting the \$ signs to $'s. Is there an easy (fast and painless) way to do this? Last edited: ## Answers and Replies Related Calculus and Beyond Homework Help News on Phys.org StatusX Homework Helper Copy into notepad, search and replace, using pre-spaces to distinguish opening from closing ones. If it's not too late. StatusX Homework Helper What is your definition of uniform continuity? (Please always post relevant definitions) Cool, that worked nicely! Any ideas about the exercise now Sorry, I thought that was a standard definition. Here it is: Uniform Continuity: Let [itex]f:X\to Y$ be a map from one metric space $(X,d_X)$ to another $(Y,d_Y).$ Then $f$ is uniformly continuous if, for every $\epsilon > 0$, there exists a $\delta > 0$ such that $d_Y(f(x),f(x')) < \epsilon$ whenever $x,x'\in X$ are such that $d_X(x,x')<\delta$.

Here is our book's definition of uniform convergence:

Let $(f^{(n)})_{n=1}^{\infty}$ be a sequence of functions from one metric space $(X,d_X)$ to another $(Y,d_Y)$, and let $f: X\to Y$ be another function. We say that $(f^{(n)})_{n=1}^{\infty}$ converges uniformly to $f$ on $X$ if for every $\epsilon > 0$ there exists $N>0$ such that $d_Y(f^{(n)}(x),f(x))<\epsilon$ for every $n>N$ and $x\in X$.

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StatusX
Homework Helper
It is standard, it's just a lot easier to refer to a definition that's here. Remeber that the definition of continuity in terms of sequences was just a result of the more general definition of limits in terms of sequences, and you should still be able to apply that.