# Pointwise convergence and uniform convergence converge to same limit?

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?

• The limit is always unique, so if the function converge and converge uniformly then yes both limits are the same! – iiivooo Apr 6 '16 at 21:47
• 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). – Dave L. Renfro Apr 7 '16 at 20:38

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.