Uniform Convergence: Showing Continuous & Uniformly Continuous

[mattmns]

Here is the question from the book:
----------
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 $f$.---------
$(\Rightarrow)$ 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.----------

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 [itex]'s. Is there an easy (fast and painless) way to do this?

 

[StatusX]

Copy into notepad, search and replace, using pre-spaces to distinguish opening from closing ones. If it's not too late.

 

[StatusX]

What is your definition of uniform continuity? (Please always post relevant definitions)

 

[mattmns]

Cool, that worked nicely! Any ideas about the exercise now

Sorry, I thought that was a standard definition. Here it is:

Uniform Continuity: Let $f:X\to Y$ 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 $x\in X$.

 

[StatusX]

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.