Does convergence on finite intervals imply uniform convergence?

Question

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?

Answer 1

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 $0$).

Answer 2

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.

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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 $1/2^{n+1}$.

Answer 3

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.

Answer 4

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 $n$.

Answer 5

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 $(2n,n)$.