Exercise 6.2.7

Question

Let $f$ be uniformly continuous on all of $\mathbf{R}$, and define a sequence of functions by $f_{n}(x)=f\left(x+\frac{1}{n}\right)$. Show that $f_{n} \rightarrow f$ uniformly. Give an example to show that this proposition fails if $f$ is only assumed to be continuous and not uniformly continuous on $\mathbf{R}$.

Ulisse Mini · Accepted Answer

Given $\epsilon>0$ set $\delta>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$. Then set $N > 1/\delta$ so that $n \ge N$ implies (since $1/n<\delta$) $$ |f(x) - f_n(x)| = |f(x) - f(x+1/n)| < \epsilon $$ Which shows $(f_n) \to f$ uniformly. To see this doesn't work if $f$ is only continuous, consider $f(x) = x^2$. We have $$ |f(x)-f_n(x)| = \left|x^2 - \left(x + \frac{1}{n}\right)^2\right| = \left|\frac{2x}{n} + \frac{1}{n^2}\right| $$ which given a fixed $n$, becomes arbitrarily big as $x$ goes to infinity. Hence $(f_n)$ does not converge uniformly.

Exercise 6.2.7

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Exercise 6.2.7

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