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The Problem: I am trying to prove the following assertion, which I believe, but am uncertain, is true:

For a function $f:I \rightarrow E$ defined on a nondegenerate interval $I$ to a Banach space $E$, if $\lim_{x \rightarrow p^+} f(x) = y$, there exists a sequence $(y_n)$ in $E$ that converges to $y$.

My thoughts: This is very similar, but not quite the same, as a function being continuous if and only if it is sequentially continuous. Here though, we don't exactly have continuity. Now, since the limit as $x$ approaches $p$ from the right exists, one can imagine contructing a sequence in the following manner: We have that for any $\varepsilon_i$ , we can find an $x_i$ such that $|f(x_i) - p|<\varepsilon_i$. So, set $y_i = f(x_i)$ and choose successively smaller epsilons and note that $y_i \rightarrow p$ as $i \rightarrow \infty$.

Now, I'm not so sure about this argument. Does it work? Is there a better way to approach this?

To provide context, as requested, I am supplying the theorem/proof where this was encountered. I have made some slight modifications due to the author's rather nonstandard notation and terminology:

Theorem: A regulated function $f:I\rightarrow E$ from a compact perfect interval $I$ to a Banach space $E$ is bounded.

Proof: If $f$ is not bounded, we find a sequence $(x_n) \in I$ such that $\|f(x_n)\| \geq n$ for all $n \in \mathbb{N}$. Because $I$ is compact, there is a subsequence $(x_{n_k})$ of $(x_n)$ and $x \in I$ such that $x_{n_k} \rightarrow x$ as $k \rightarrow \infty$. By choosing a suitable subsequence of $(x_{n_k})$, we find a sequence $(y_n)$ that converges monotonically to $x$. Since $f$ is regulated there is a $v \in E$ with $\lim f(y_n) = v$ and therefore $\lim \|f(y_n)\|=\|v\|$. Because every convergent sequence is bounded this contradicts the assumption that $\|f(x_n)\| \geq n$.

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I'm confused. Why doesn't $y_n=y$ for all $n$ work?\ – Alex Becker Feb 26 '12 at 3:42
@Alex Well, I guess technically it would work but, in this case I don't see how the hypothesis about the limit is used. I mean, for any point $x \in E$ the sequence $x_i = x$ for all $i$ converges to $x$ and this is independent of other considerations. So, is the hypothesis unneeded? – ItsNotObvious Feb 26 '12 at 3:54
Where did the statement come from? – Arturo Magidin Feb 26 '12 at 4:01
@Arturo Well, I'm trying to understand a proof and in the proof the author basically goes from the hypothesis about the limit and then infers that such a sequence must exist. So, if I'm reading it correctly, the theorem as I stated it is basically what the author is assuming. Specifically, the proof is showing that a regulated function is necessarily bounded. – ItsNotObvious Feb 26 '12 at 4:09
@3Sphere: And if you are not reading it correctly, we have no way of telling because you are reporting to us your understanding. So, how about providing a verbatim quote (with context)? – Arturo Magidin Feb 26 '12 at 4:15
up vote 1 down vote accepted

By definition, a function $f\colon [a,b]\to B$ from a compact interval to a Banach space is regulated if and only if for every $r\in (a,b)$, $\lim\limits_{t\to r^-}f(t)$ and $\lim\limits_{t\to r^+} f(t)$ both exist, as well as $\lim\limits_{t\to a^+}f(t)$ and $\lim\limits_{t\to b^-}f(t)$.

Since $y_n\to x$ monotonically, say from below, then $\lim\limits_{n\to\infty}f(y_n) = \lim\limits_{t\to x^-}f(t)$ exists; call the limit $v$. If it converges from above, use the fact that $\lim\limits_{t\to x^+}f(t)$ exists.

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