Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The more general quesion is under what conditions the folloing equality will hold (all functions are real valued): $$\lim_{x \rightarrow a} \ \lim_{j \rightarrow \infty} f_j(x) = \lim_{j \rightarrow \infty} \ \lim_{x \rightarrow a} f_j(x)$$

A more specific question is if it will hold for non-continuous functions $f_j$ that are uniformly convergent to a non-continuous limit function $f$ and all the single and iterated double limits exist.

Also some references would be useful.

share|cite|improve this question
Have you tried proving it yourself? – Nate Eldredge Jun 14 '11 at 16:24
See this previous question on sequences for some ideas. – Arturo Magidin Jun 14 '11 at 17:29
up vote 6 down vote accepted

The answer to the specific question is Yes.

Since $f_j$ converges (uniformly and hence) pointwise to $f$,
$$ \mathop {\lim }\limits_{x \to a} \mathop {\lim }\limits_{j \to \infty } f_j (x) = \mathop {\lim }\limits_{x \to a} f(x). $$ For any $\varepsilon > 0$, since $f_j$ converges uniformly to $f$, there exists $N = N(\varepsilon)$ such that $$ \sup _x |f_j (x) - f(x)| < \varepsilon $$ for any $j > N$. We assume that all limits exist. Hence, for any $j > N$, $$ |\lim _{x \to a} f_j (x) - \lim _{x \to a} f(x)| = |\lim _{x \to a} (f_j (x) - f(x))| \le \varepsilon . $$ Define $p_j = \lim _{x \to a} f_j (x)$ and $p = \lim _{x \to a} f(x)$. Then, $$ |\lim _{j \to \infty } p_j - p| = \lim _{j \to \infty } |p_j - p| \le \varepsilon , $$ since $|p_j - p| \leq \varepsilon$ for any $j > N$. Since $\varepsilon$ is arbitrary, $ \lim _{j \to \infty } p_j = p$, hence $$ \mathop {\lim }\limits_{j \to \infty } \mathop {\lim }\limits_{x \to a} f_j (x) = \mathop {\lim }\limits_{x \to a} f(x) = \mathop {\lim }\limits_{x \to a} \mathop {\lim }\limits_{j \to \infty } f_j (x). $$

share|cite|improve this answer

Uniform convergence means that for every $\varepsilon > 0$ there exists $j_0$ such that for all $j \geq j_0$ and for all $x \in D$(the domain of definition for the functions $(f_j),f$ you have $|f_j(x)-f(x)| <\varepsilon$.

By the way the question is defined, although $f,f_j$ are not continuous, the limits $L=\lim_{x \to a}f(x)$ and $L_j=\lim_{x \to a}f_j(x)$ exist. The question translates in proving or disproving the followint equailty:

$$\lim_{j \to \infty}L_j=L$$

Let $\varepsilon>0$, and from the definition of uniform continuity we know that there exists $j_0$ such that forall $j \geq j_0$ and forall $x \in D$ we have that $|f_j(x)-f(x)|<\varepsilon$. Taking $x \to a$ in the last inequality we get that $|L_j-L|<\varepsilon ,\ \forall j \geq j_0$. This proves the assertion.

As a remark, I think the question should be edited such that

  • the domain of definition of the functions $f_j,f$ is clear, i.e. $f_j,f: D \to \Bbb{R}$ where $D=\Bbb{R}$ or some other suitable set.

  • mention that $a$ is a limit point for $D$

  • although the functions $f_j,f$ are not continuous, for the question to be valid, we must assume the existence of the limits $\lim_{x \to a}f_j(x), \ \lim_{x \to a}f(x)$.

share|cite|improve this answer

Note that $$\left |\lim_{x\to a}f_j(x) - \lim_{x\to a} f(x)\right|\le \|f_j-f\|_\infty\to 0,\; j\to \infty$$

Thus $$\lim_{j\to\infty}\lim_{x\to a}f_j(x) = \lim_{x\to a} f(x) = \lim_{x\to a} \lim_{j\to\infty}f(x)$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.