If $f$ is uniformly continuous and $a_{n+1}-a_n\to0$ then $f(a_{n+1})-f(a_n)\to 0$

Question

$a_n$ is a sequence, $f$ is a uniformly continuous function in $R$.

I need to prove that $lim_{n\to\infty}(a_{n+1}-a_n) = 0 \implies lim_{n\to\infty}(f(a_{n+1})-f(a_n)) = 0$.

Here's what I thought of doing:

1. $lim_{n\to\infty}(a_{n+1}-a_n) = 0 \implies lim_{n\to\infty}(a_{n+1}) = lim_{n\to\infty}(a_{n}) = L \in R$
2. Since $f$ is continuous than for every sequence $(X_{n=1}^{\infty})$ that converges to $L \in R$ the following holds: $f(X_n)_{n\to \infty} \to f(L)$, and so $f(a_n)_{n\to \infty} \to f(L), f(a_{n+1})_{n\to \infty} \to f(L)$. This leads to $lim_{n\to\infty}(f(a_{n+1})-f(a_n))= lim_{n\to\infty}f(a_{n+1})-lim_{n\to\infty}f(a_{n}) = L - L = 0$

I guess this solution is wrong somehow, because I didn't use the fact that $f$ is uniformly continuous.

Is this solution wrong? If it is, what's wrong about it? And could you provide a valid solution?

Moreover, If $f$ was continuous in R but not Uniformly continuous in R, would that statement still be true?

Answer 1

Consider $\epsilon>0$. By uniform continuity, there exist $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$

Now $|a_{n+1}-a_n|\rightarrow 0$ so there exists $N$ such that $|a_{n+1}-a_n|<\delta$ for $n>N$

So $|f(a_{n+1})-f(a_n)|<\epsilon$ for $n>N$, which proves your result.

Your mistake is that you assumed that $a_n$ is itself convergent, which it doesn't have to be.

EDIT:

For a counterexample of your assumption, think about $a_n=1+\frac{1}{2}+...+\frac{1}{n}$, or if you want it to be bounded you may take $a_n=\sin\left(1+\frac{1}{2}+...+\frac{1}{n}\right)$. Then $a_{n+1}-a_n\rightarrow 0$, but $a_n$ is not convergent.

And for counterexample with $f$ continuous, but not uniformly continuous, take $a_n=\sqrt n$ like in the comments, and $f(x)=x^2$. Then $a_{n+1}-a_n\rightarrow 0$, but $f(a_{n+1})-f(a_n)=1$

Answer 2

In principle, it is possible that $\lim_{n\to\infty} (a_{n+1}-a_n)=0$ but still the sequence $(a_n)$ might not be convergent. So, assuming that $\lim_{n\to\infty}a_n=L\in\mathbb{R}$ would be a mistake.

It is better to think about the problem in this way: since $f$ is uniformly continuous, given $\epsilon>0$ there is $\delta>0$ depending only on $\epsilon$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$.

By the definition of convergent sequences, given $\delta>0$ there is $N\in\mathbb{N}$ such that whenever $n\geq N$, $|a_{n+1}-a_n|<\delta$.

Putting these facts together, it is easy to prove that for every $\epsilon>0$ there is $N\in\mathbb{N}$ such that for $n\geq N$, $|f(a_{n+1})-f(a_n)|<\epsilon$, which is the definition of $\displaystyle{\lim_{n\to\infty}f(a_{n+1})-f(a_n)=0}$.

Answer 3

If $\lim_{n\to\infty}(a_{n+1}-a_{n})=0$, this means that for all $\delta>0$, there exists $N_{\delta}$ such that for any $n\geq N_{\delta}$, we have $\vert a_{n+1}-a_{n}\vert<\delta$.

Now, since $f$ is uniformly continuous, we know that for any $\epsilon>0$, there exists $\delta>0$ such that $\vert x-y\vert<\delta$ implies $\vert f(x)-f(y)\vert<\epsilon$

With the initial observation, we know that when $n\geq N_{\delta}$, we have $\vert a_{n+1}-a_{n}\vert<\delta$, which implies that $\vert f(a_{n+1})-f(a_{n})\vert<\epsilon$, which means that for any $\epsilon>0$, we can take $N_{\epsilon}=N_{\delta}$ such that for all $n\geq N_{\epsilon}=N_{\delta}$, we have $\vert f(a_{n+1})-f(a_{n})\vert<\epsilon$, i.e. $\lim_{n\to\infty}(f(a_{n+1})-f(a_{n}))=0$