Sometimes I see in proofs they say "this function is pointwise convergent to limit a. Let's check if it's uniformly convergent to limit a." And then they do a uniform convergence check on a and prove that it cannot converge uniformly to limit a.

My question is, if we know a function converges pointwise to a limit, suppose it does converge uniformly, would it converge to the same limit?

  • $\begingroup$ The limit is always unique, so if the function converge and converge uniformly then yes both limits are the same! $\endgroup$
    – iiivooo
    Commented Apr 6, 2016 at 21:47
  • 1
    $\begingroup$ To further elaborate on @iiivooo (but without getting technical, as in the answers), uniform convergence is "convergence" plus "something extra". So if the sequence of functions converges (call the limit $f)$ and the sequence of functions converges uniformly (call the limit $g),$ then we have on one hand the functions converging to $f$ and on the other hand we have the functions converging (along with "something extra") to $g,$ so $f=g$ (because the functions can't converge to different limit functions). $\endgroup$ Commented Apr 7, 2016 at 20:38

3 Answers 3


Yes, it does. Let $(f_n)$ be a series of functions converging uniformly to $f$. Then $\forall x$ we have $|f_n(x) - f(x)| \leq \sup_x{|f_n(x) - f(x)|}$. Thus $\lim_{n \to \infty}f_n(x) = f(x)$ (pointwise convergence). This limit is unique.


I will explain this using definitions of both pointwise convergence and uniform convergence

suppose $ f_n : A\rightarrow\mathbb{R} $

Pointwise convergence definition $$ (\forall x \in A) (\forall \varepsilon>0) (\exists n_0=n_0(x,\varepsilon) \in \mathbb{N}) (\forall n>n_0) \left | f(x) - f_n(x) \right |<\varepsilon $$

Uniform convergence definition
$$ (\forall \varepsilon>0) (\exists n_0=n_0(\varepsilon) \in \mathbb{N}) (\forall x \in A) (\forall n>n_0) \left | f(x) - f_n(x) \right |<\varepsilon $$

Note that $ n_0(x,\varepsilon) $ means that $n_0$ is a function of $ x $ and $ \varepsilon $

And $ n_0(\varepsilon) $ is a function of $ \varepsilon $

Suppose $f_n\rightrightarrows f$ on $A$

So since in uniform convergence we can find such $ n_0=n_0(\varepsilon) $ that depends just on $ \varepsilon$ we can use that same $n_0$ for definition on pointwise convergence. Our chosen $n_0$ works for every $ x $ so for every $ x $ we can chose our $n_0=n_0(\varepsilon)$

That means functions $f_n$ converges poinwise to $f$ on $A$ when $f_n\rightrightarrows f$ on $A$

Reverse does not apply If a function converges it doesn't need to converge uniformly but if it does the limit is the same as pointwise convergence.


Suppose there exists some $\{f_n\}$ so that the sequence converges pointwise to $f$, but $\{f_n\}$ converges uniformly to $g \neq f$. Then there exists some $x \in \mathbb{R}$ so that $g(x) \neq f(x)$.

Let $\epsilon=\frac{|g(x)-f(x)|}{2}$...

This boils down to the uniqueness of a limit, since we a limit of a sequence of functions at some particular $x \in \mathbb{R}$ is more or less just a regular ole' sequence.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .