Calculating the $\|f \|_{\max}$ and $\|f\| _{1} =\int_{a}^{b} |f|.$

Question

Calculating the maximum norm and $\|f\| _{1} =\int_{0}^{1} |f|$ of the following sequence of functions.

$f_n(x)=n(\frac 1 n-x) $ for $0 \leq x\leq \frac 1n$ and $f_n(x)=0$ for $x> \frac 1 n$

I was told that the maximum norm is n and $\|f\| _{1} $ is $1/2n$ but I do not understand how, Could anyone explain this for me please?

What if I changed the interval of integration to $[a,b],$ how can I adjust my sequence of functions and how the values of the norms may differ?

Answer 1

Consider $f_n(0)=1$ and the function decreases to reach $f_n(1/n)=0$ and stays there. So

$$\|f_n\|_\infty=1$$

For $\|.\|_1$ one has

$$\|f_n\|_1=n\int_0^{1\over n}\left({1\over n}-x\right)dx={1\over n}-{1\over 2n}={1\over 2n}$$