# Continuous functions that pointwise converge to zero, but the integral tending towards infinity.

I wish to find a continuous function $f_n:[0,1]\rightarrow\mathbb{R}$ such that $f$ converges pointwise to 0, but also $\int^1_0f_n\rightarrow\infty$

Then another $f_n:[0,\infty)\rightarrow\mathbb{R}$ continuous satisfying $||f_n||_\infty\rightarrow 0$ but with $\int_0^\infty f_n=1$

For the first one every point in 0 to 1 must converge to 0, but the area underneath them must not (which really confuses me because $f$ is continuous, it can't jump)

For the second one, the sup norm tending towards zero means the largest value $f_n$ takes must tend towards zero, but the area must be 1. I'm thinking of some sort of function that spreads out might work but keeping it continuous is the difficult part.

I've been thinking for over an hour (I hate to admit) and really quite stumped, this is a past exam question (no solutions available) and I have no ideas left about how to find such a function.

I've been looking for a theorem to negate or abuse so I can get a definition I must get a function to satisfy, which is really hard with the continuous constraint.

For the first question, consider on the interval $[0, \frac{2}{n}]$ the isosceles triangle with height $n^{3}$. Then I leave to you to calculate the function (there are two linear parts and then zero) and to check it converges to zero.

For the second problem you can rearrange the functions $\frac{1}{n}\chi_{[0,n]}$ in a way it is continuous and has integral $1$ (you can do again with a linear part to join the two constant parts).

• I've ticked this because you didn't just give me a magical answer, thanks. It would have been nice to have used say.... n*(the other person's example) and shown that was 1 (I've enjoyed playing with this on paper) but imagining triangles helped, thanks! – Alec Teal Apr 27 '14 at 22:14
• Basic geometric examples often are very good. Anyway, notice that the other example starts with a triangle and the first deformation leads to my first one. The second is simply a smart use of change of variable in integration (so another tool you can keep in mind when looking for counterexamples). – Stefano Apr 28 '14 at 7:07

Let $$g(x)=\begin{cases}x&\text{if }0\le x\le 1\\ 2-x&\text{if }1\le x\le 2\\ 0&\text{if }x\ge 2\end{cases}$$ and then $f_n(x)=n^2g(nx)$ for the first problem and $f_n(x)=\frac1ng(x/n)$ for the second problem.

• How did you come to such magic? What did your picture look like! (I'd really like to be able to do this myself) also for the first one how does that pointwise converge to 0? it'll always have a peek, or does shifting it mean it doesn't? – Alec Teal Apr 27 '14 at 21:40
• For $x=0$, $f_n(x)=0$ for all $n$. For $x>0$, $f_n(x)=0$ as soon as $n\ge\frac2x$. – Hagen von Eitzen Apr 27 '14 at 21:48
• So that peek sort of ... vanishes (it can't be at 0 but it can't be beyond something tending towards zero....)? – Alec Teal Apr 27 '14 at 21:58
• This is odd, I see exactly what you mean, for any non zero x there is an n where it becomes zero, which is... not a lesser value of x's problem as that'll also eventually become zero.... – Alec Teal Apr 27 '14 at 22:02