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Here is the question from the book:

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Let [itex]f: \mathbb{R} \to \mathbb{R}[/itex] be a function. For any [itex]a\in \mathbb{R}[/itex], let [itex]f_a :\mathbb{R}\to \mathbb{R}[/itex] be the shifted function [itex]f_a(x):=f(x-a)[/itex].

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

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

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

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

Given [itex]\epsilon' > 0[/itex] take [itex]\epsilon = \epsilon'[/itex], and take [itex]N = N'[/itex].

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

Thus, [itex]|f(x_n) - f(x_0)| < \epsilon = \epsilon'[/itex].

But, [itex]f(x_n) = f(x_0-a_n) = f_{a_n}(x_0)[/itex] for all [itex]n>N[/itex].

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

Thus [itex]f_{a_n}[/itex] converges pointwise to [itex]f[/itex].

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

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

But, [itex]|f_{a_n}(x_0) - f(x_0)| = |f(x_0 - a_n) - f(x_0)| = |f(x_n) - f(x_0)|[/itex].

Hence, [itex]|f(x_n) - f(x_0)| < \epsilon[/itex] for all [itex]n>N[/itex].

That is, [itex]f(x_n)[/itex] converges to [itex]f(x_0)[/itex]. Thus, [itex]f[/itex] 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 [itex]'s. Is there an easy (fast and painless) way to do this?

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Let [itex]f: \mathbb{R} \to \mathbb{R}[/itex] be a function. For any [itex]a\in \mathbb{R}[/itex], let [itex]f_a :\mathbb{R}\to \mathbb{R}[/itex] be the shifted function [itex]f_a(x):=f(x-a)[/itex].

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

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

---------

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

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

Given [itex]\epsilon' > 0[/itex] take [itex]\epsilon = \epsilon'[/itex], and take [itex]N = N'[/itex].

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

Thus, [itex]|f(x_n) - f(x_0)| < \epsilon = \epsilon'[/itex].

But, [itex]f(x_n) = f(x_0-a_n) = f_{a_n}(x_0)[/itex] for all [itex]n>N[/itex].

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

Thus [itex]f_{a_n}[/itex] converges pointwise to [itex]f[/itex].

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

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

But, [itex]|f_{a_n}(x_0) - f(x_0)| = |f(x_0 - a_n) - f(x_0)| = |f(x_n) - f(x_0)|[/itex].

Hence, [itex]|f(x_n) - f(x_0)| < \epsilon[/itex] for all [itex]n>N[/itex].

That is, [itex]f(x_n)[/itex] converges to [itex]f(x_0)[/itex]. Thus, [itex]f[/itex] 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 [itex]'s. Is there an easy (fast and painless) way to do this?

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