Let $E$ a normed vector space and $F$ a Banach space. Let $A \subset E$, $x_0$ an accumulation point and $f : A \to F$ a uniformly continuous function. Show that $\lim_{x \to x_0} f(x)$ exist.

Proof (partial) :

Since $(a_n)$ is a convergent sequence, then its Cauchy. $f$ is a uniform continuous function; hence, if $(a_n)$ is Cauchy, $(f(a_n))$ is also Cauchy. Again, because $(f(a_n))$, then its convergent and hence the limit exists.

Here, I have to show then the limit is independent of the convergent sequence.

Given $x_n\to x_0,y_n\to x_0$, we know $\lim f(x_n),\lim f(y_n)$ exists and aim to show $\lim f(x_n)=\lim f(y_n)$. Consider sequence $z_n$ defined as $(x_1,y_1,x_2,y_2,\cdots)$ which is a combination of $(x_n),(y_n)$. Then we know $z_n\to x_0$ and by similar argument, we know $\lim f(z_n)=l\in \mathbb{R}$. Since $f(x_n),f(y_n)$ are two subsequence of a convergence sequence $\lim f(z_n)$, we know they must convergent to the same limit, hence $\lim f(x_n)=\lim f(y_n)$.

Question :

($1$) Why the convergence of the sequence $(f(a_n))$ assures us that the limit exists?

($2$) Why do we need the uniform continuity in the second part of the proof?

($3$) Why is it necessary to show that the limit is independent of the convergent sequence?

  • $\begingroup$ (3) Because the limit is always unique (in a Hausdorff space). Hence you must be sure that different sequences do not lead you to different limits. $\endgroup$
    – Siminore
    Feb 2 '16 at 14:18
  • (1) tells you that, for any sequence $a=(a_n)_n$ such that $a_n \xrightarrow[n\to\infty]{} a_0$, then the sequence $(f(a_n))_n)$ converges to some limit $\ell_a$.

  • So now, for the limit $\lim_{x\to 1_0}f(x)$ to exist, you need to show that $\ell_a$ is independent of the sequence $a$ -- i.e., that actually this limit is always the same. (Think otherwise of the sequence $\frac{(-1)^n}{n}$ with the function $f$ (not continuous) defined on $\mathbb{R}$ by $$f(x) = \begin{cases} 1&\text{ if } x\geq 0\\ -1 &\text{ otherwise.}\end{cases}$$ The sequences $a=(\frac{-1}{n})_n$ and $b=(\frac{1}{n})_n$ both converge to $0$, and both $(f(a_n))_n$ and $(f(b_n))_n$ converge, but... not to the same limit. This is what (3) is for.

  • As for (2), you only need uniform continuity in the first part: namely, to say that the image by $f$ of a Cauchy sequence is also Cauchy. (This is not true if you only assume $f$ to be continuous.)


(1) $f(a_n)$ is a Cauchy sequence in a Banach space, which by definition is complete, i.e. every Cauchy sequence converges. To say that any sequence converges is just saying that its limit exists.

(2) I do not think one actually needs uniform continuity. However, since $E$ can be incomplete, $x_0 \notin A$ or even $x_0 \notin E$ is possible (it is only said to be an accumulation point). Usually continuity is defined by $\lim_{x \to x_0} f(x) = f(x_0)$ but there is no way to check this if $x_0 \notin A$, i.e. $f$ is not defined for $x_0$. Hence, I guess that the uniform continuity is just a fancy way to get the most general possible statement.

(3) $\lim_{x \to x_0} f(x) = z$ means by definition that for every $x_n \to x_0$ we have $\lim_{n \to \infty} f(x_n) = z$.


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