# A sequence $f_n to f$ uniformly on $(0,1)$ where $f, f_n in C([0,1])$ implies $f_n(0) to f(0)$, $f_n(1) to f(1)$.

Mathematics Asked by CyCeez on January 5, 2022

This is a past problem for an Applied Math QR Exam:

Suppose $$f_n$$, for $$n in mathbb{N}$$, and $$f$$ are
continuous functions on the closed interval $$[0,1]$$. Assume that $$f_n to f$$
as $$n to infty$$ uniformly on the open interval $$(0,1)$$. Prove
that

$$f_n(0) to f(0) quad text{and} quad f_n(1) to f(1).$$

Am I wrong to think the proof is just an $$epsilon/3$$ argument, at least for $$f_n(0) to f(0)$$? If so, then I think that my proof is lacking subtleties to be precise. In particular, I don’t think I can give an exact $$N = N(epsilon)$$ in applying the definitions. I’m also not sure that proving $$f_n(1) to f(1)$$ is any different, except that maybe in the $$epsilon$$$$delta$$ definition for, say, $$f$$, we must say $$1 – x < delta$$ implies $$|f(x) – f(1)| < epsilon$$. Lastly, my proof isn’t too creative – just application of definitions. So, any more elegant proofs would be of great help, too.

Proof of $$f_n(0) to f(0)$$. Our goal is to show that $$forall epsilon > 0$$, $$exists N$$ such that $$n > N$$ implies $$|f_n(0) – f(0)| < epsilon$$.

By assumption, we know that $$f, f_n in mathcal{C}([0,1])$$ tells us

$$forall epsilon > 0, exists delta_0 > 0 text{ such that } |x| < delta_0 Longrightarrow |f(x) – f(0)| < frac{epsilon}{3},$$

$$forall epsilon > 0, exists delta_1 > 0 text{ such that } |x| < delta_1 Longrightarrow |f_n(x) – f_n(0)| < frac{epsilon}{3},$$

and that $$f_n to f$$ uniformly on $$(0,1)$$ tells us

$$forall epsilon > 0, exists M text{ such that } forall x in (0,1), n > M Longrightarrow |f_n(x) – f(x)| < frac{epsilon}{3}.$$

Now, for any $$x in (0,1)$$, we have

begin{align*}|f_n(0) – f(0)| &= |f_n(0) – f_n(x) + f_n(x) – f(x) + f(x) – f(0)| \ &leq |f_n(x) – f_n(0)| + |f_n(x) – f(x)| + |f(x) – f(0)| \ &< 3 cdot frac{epsilon}{3} = epsilon.end{align*}

So, choosing any $$N > M$$ and letting $$delta_0, delta_1 to 0$$ implies $$epsilon to 0$$, and therefore $$f_n(0) = f(0)$$.