Uniform convergence of piecewise sequence of functions

Question

I stumbled upon this problem and I'm having hard time understanding the solution , mainly the part about the supremum of the function.

Study the uniform convergence of the following sequence of functions on the interval [$0,1$].

$$f_n(x)= \begin{cases} n^2x, & \text{$0\le x\le \frac{1}{n}$}\\ n^2(\frac{2}{n}-x), & \text{$\frac{1}{n}< x< \frac{2}{n}$}\\ 0, & \text{$\frac{2}{n} \le x \le 1$} \end{cases}$$

The solution :

If $x=0$ then $f_n(x)=0$. This is obvious Let $x\neq0$. Then there exists $n_0 \in N$ such that $\frac{2}{n_0}<x$ and for every $n\ge n_0,\ f_n(x)=0$

Why do we only look at the case $x=0$ and $x>\frac{2}{n}$. What about $\frac{1}{n}<x<\frac{2}{n}$?

So $f_n(x)$ converges pointwise to $f(x)=0$.

Now we need to find $\sup \left \lvert{f_n(x)-f(x)}\right \rvert=\sup > f_n(x)\ge f_n(\frac{1}{n})=n \rightarrow \infty$

The last part is the one I'm at most confused about.

Similarly what is the supremum of the following sequence of functions?

$$f_n(x)= \begin{cases} nx, & \text{$0\le x\le \frac{1}{n}$} \\ 2-nx, & \text{$\frac{1}{n}<x<\frac{2}{n}$}\\ 0, & \text{$\frac{2}{n}\le x\le3$} \end{cases}$$

Answer 1

• For the first question the goal is to study point wise convergence. So $x$ is fixed and then $n$ goes to $+\infty$.

You have $\frac{1}{n}<x<\frac{2}{n}$ for only a finite number of $n$ so it does not matter when $n \to \infty$.

• As uniform convergence implies point wise convergence, if we suppose (by contradiction) that $f_n \to f$ uniformly then the convergence is also pointwise. But by uniqueness of the limit as we have shown the point wise convergence to $0$ we must have $f=0$.

At this point we have:

If $f_n \to f$ uniformly then $f=0$.

It remains to show that $f_n$ does not converge to $0$ i.e that: $$\|f_n-0\|\_\infty \not \to0$$ as $$\sup_{x} |f_n(x)-0| \geq \left|f \left( \frac{1}{n} \right)-0\right| = n$$ you can conclude that $f_n \not \to 0$ and so $f_n$ does not converge uniformly to any function.

---

For the second example similarly you can show that $f_n$ converge pointwise to $0$ but that: $$\sup |f_n-0| \geq 1$$ so the sequence does not converge uniformly.

---

Edit

(If you want to find the supremum of such function the first thing is to plot the functions. You can then observe see it is a moving triangle with base $(0/n,2/n)$ and height $n$.)

To show that the supremum is $n$ for the first case you can show both points:

• There exist $x_0$ such that $f_n(x_0)=n$ (more generally, if your function is less "nice" there exist a sequence $x_k$ such that $f_n(x_k) \to n$ when $k \to \infty$).
• For all $x$, $f_n(x)\leq n$.

Answer 2

For every fixed $x$, then, unless $x=0$, $x>\frac2n$ if $n$ is large enough. But then $f_n(x)=0$, and this proves that $\lim_{n\to\infty}f_n(x)=0$.

On the other hand, $\sup f_n\geqslant f_n\left(\frac1n\right)=n$. Therefore, your sequence is cannot converge uniformly to the null function (or to any bounded function).