# Does convergence on finite intervals imply uniform convergence?

Let $f_n(x) \rightarrow 0$, $n\rightarrow \infty$ for all $x \in \mathbb{R}$. Does this imply $f_n(x) \rightarrow 0$ uniformly on finite intervals?

I could think it could be proofen like this maybe:

1. Let the interval $I$ be compact without loss of generalization
2. Choose a finite subcover for $I$
3. take $N := \sup N_j$ for the $\epsilon$-$\delta$-proof

But I am not sure if it is okay this way, or if it can be done more easy. Is it maybe a known lemma?

-
What is a finite interval? Finite length? What is the domain of the $f, f_n$? – Jonas Teuwen Sep 6 '11 at 12:44
$f_n, f: \mathbb{R} \rightarrow \mathbb{R}$. I thought it would be finite length, so on $R$ just any interval except the ones "involving" $\infty$ $(-\infty, b]$ and so on – Johannes L Sep 6 '11 at 12:57

I may have misunderstood the question, but there are plenty of examples of function sequences on the finite interval $[0,1]$ that converge to $0$ pointwise but not uniformly. In finding one, it may be easier to draw pictures of graphs instead of trying equations. I'm thinking of a sequence which is zero on the interval except for a triangular spike of width $1/2^n$ that reaches height $1$. We can see that for any fixed $x$, eventually it's values is the sequence will be all zeros, but $||f_n-0||_{\infty} = 1$ and in particular, does not tend to 0 so the convergence is not uniform.

Here is an explicit example:

Let $f_n:[0,1] \to [0,1]$ be defined as such: $$f_n(x) = 2^{n+1}x$$ for $0\leq x\leq 1/2^{n+1}$ , $$f_n(x) = -2^{n+1} \left( x- \frac{1}{2^n} \right)$$ for $1/2^{n+1}<x\leq 1/2^n$ and $$f_n(x)=0$$ for $1/2^n < x \leq 1$.

Essentially, this is a flat line except for a triangular spike from $0$ to $1/2^n$, with it's peak height of $1$ reached half way, at $1/2^{n+1}$.

-
Where is the spike of height $1$ centered? If it is fixed, then that point does not tend to $0$. If it moves around, where is it for $f_n$? – robjohn Sep 6 '11 at 13:09
Sorry, I guess I'll just write out the equation explicitly so people can draw it out themselves. Give me a moment. – Ragib Zaman Sep 6 '11 at 13:12
+1 for a counterexample with continuous functions. @robjohn, if the left edge of the spike stays fixed and the spike itself becomes thinner and thinner, then any point to the right of the edge will eventually be to the right of the entire spike. – Henning Makholm Sep 6 '11 at 13:18
@Ragib And even if I add "Let $f$ be continous and almost everywhere differentiable" your counter-example would still work. Ok. Actually I also saw this kind of counter-example in Calculus I but with the case where I want to apply it to I didn't think of them. – Johannes L Sep 6 '11 at 13:26
I suppose you meant $f_n(x)=x\;2^{n+1}$ for $0\le x\le1/2^{n+1}$. In any case, this is much better. It is best to give a more concrete counterexample than a vague, qualitative description (which could just as easily describe something wrong). – robjohn Sep 6 '11 at 14:22

Just to give another example:

Define $$f_n: \mathbb{R} \rightarrow \mathbb{R}$$ $$x \mapsto n^2 (1-x) \left(|(x-1) x|-x^2+x\right) x^n$$

If you look at the plot (or the formula) you will see that $f_n \rightarrow 0$ pointwise everywhere but it doesn't converge uniformly. (Outside of $[0,1]$ $f_n$ evaluates to $0$).

-
Nice example! +1. – Jonas Teuwen Sep 6 '11 at 13:22

No, it does not.

Imagine we had a sequence of functions such that $f_n = x^n$ on the interval $[0,1)$ and $f_n \equiv 0$ everywhere else. We see that $\lim f_n = 0$. But it does not converge uniformly in a neighborhood around 1 (it does everywhere else, though, i.e. if you take out any neighborhood of 1, the sequence converges uniformly).

Why is this, despite your proof? Because you let an interval be compact WLOG. Well, there is a loss of generalization.

-
I had a continous and almost everywhere differentiable function in mind (I forget to state this condition, now I let it as it is because it might be helpful to others this way) but in this case it's not so clear to me like in your example why I "lose generality" with changing the interval.. – Johannes L Sep 6 '11 at 13:30
@Johannes: If you append the reflection of my function from the left of 1 to the right of 1 (you can smooth out the lower points, too, if you want, with some bump function) then we have a continuous function, differentiable everywhere except at 1, that is not uniformly continuous. – mixedmath Sep 6 '11 at 13:34
I was going to comment that $f_n$ was not continuous at $1$, but upon re-reading the question, I see that continuity was not required. Even with continuous $f_n$, the statement is false. Equicontinuity is needed. – robjohn Sep 6 '11 at 17:33
@robjohn: indeed - at first, I thought I would have to strain harder to think of a counterexample. But then I wimped out and came up with something very simple. In fact, it was the function on the cover of my first real analysis book (with lots of 0 on both sides). – mixedmath Sep 6 '11 at 17:38

The counterexample I usually use, and I like it aesthetically since it doesn't need a piecewise definition, is $$f_n(x)=4x^n(1-x^n)$$ for $x\in[0,1]$. $f_n(1)=0$ for all $n$, and $f_n(x)\le4x^n$ for all other $x\in[0,1]$. However, $\|f_n\|_{L^\infty}=1$ for all $n$.

-

It is in fact very easy to construct a sequence $(f_n)$ in $C_b(\mathbb{R})$ that is unbounded (w.r.t. $\lVert\cdot\rVert_\infty$) yet still converges uniformly to $0$ on compact subsets (and in fact it converges to $0$ pointwise everywhere):

$$f_n(x) = \begin{cases} 0 & \text{if } 0 \leq x \leq n, \\\\ x-n & \text{if } n \lt x \leq 2n, \\\\ n & \text{if } x > 2n, \\\\ f_n(-x) & \text{if } x < 0. \end{cases}$$

On the interval from $n$ to $2n$ the graph of $f_n$ is merely the straight line segment joining $(n,0)$ and $(2n,n)$.

-