# On Constructing Functions, part 2

/This post is the second example in an ongoing list of various sequences of functions which converge to different things in different ways.

*Also in this series:*

Example 1: converges almost everywhere but not in $L^1$

Example 3: converges in $L^1$ but not uniformly

Example 4: $f_n$ are integrable and converge uniformly to $f$, yet $f$ is not integrable

Example 5: converges pointwise but not in $L^1$

Example 6: converges in $L^1$ but does not converge anywhere

## Example 2

**A sequence of functions $\{f_n:\mathbb{R}\to\mathbb{R}\}$ which converges to 0 uniformly but does not converge to 0 in $L^1$.**

**This works because:** The sequence tends to 0 as $n\to \infty$ since the height of each function tends to 0 and the the region where $f_n$ is taking on this decreasing height is tending towards all of $\mathbb{R}^+$ ($(0,n)$ as $n\to \infty$) (and it's already 0 on $\mathbb{R}^-\cup\{0\}$). The convergence is uniform because the number of times we have to keep "squishing" the rectangles until their height is less than $\epsilon$ does not depend on $x$.

**The details:** Let $\epsilon>0$ and choose $N\in \mathbb{N}$ so that $N>\frac{1}{\epsilon}$ and let $n>N$. Fix $x\in \mathbb{R}$.

- Case 1 ($x\leq 0$ or $x\geq n$) Then $f_n(x)=0$ and so $|f_n(x)-0|=0< \epsilon$.
- Case 2 ($0< x < n$ ) Then $f_n(x)=\frac{1}{n}$ and so $|f_n(x)-0|=\frac{1}{n}< \frac{1}{N}<\epsilon$

**Remark:**Here's a question you could ask: wouldn't $f_n=n\chi_{(0,\frac{1}{n})}$ work here too? Both are tending to 0 everywhere and both involve rectangles of area 1. The answer is "kinda." The problem is that the convergence of $n\chi_{(0,\frac{1}{n})}$ is pointwise. BUT Egoroff's Theorem gives us a way to actually "make" it uniform! We've seen this before in a previous example.

For a measurable set $X\subset \mathbb{R}$, denote the set of all Lebesgue integrable functions $f:X\to\mathbb{R}$ by $L^1(X)$. Then a sequence of functions $\{f_n\}$ is said to

*converge in $L^1$*to a function $f$ if $\displaystyle{\lim_{n\to\infty}}\int|f_n-f|=0$.