For the interval $[0,1],$ define $f_n(x)=n(\frac 1 n-x) $ for $0 \leq x\leq \frac 1n$ and $f_n(x)=0$ for $x> \frac 1 n$. Then the maximum norm tends to 1 but the $\|{f}\|_{1} = (1/2n)$ norm tends to zero.
Now, Transforming this example to be on the interval $[a,b].$ define $g_n(x)=f_n(\frac {x-a} {b-a})$ for $0 \leq \frac {x-a} {b-a}\leq \frac 1n$ and $g_n(x) = f_n(x)=0$ for $\frac {x-a} {b-a} > \frac {1}{n}.$
But I was told that the maximum norm of the new function $g_{n}$ is still 1 (which I do not understand how )and $\|{g}\|_{1}$ norm equals $((b - a)/2n)$ (which also I do not understand how ) , could anyone show me the details of the calculations please?
Note: $\| f\| _{1} =\int_{a}^{b} | f|$
$\{f_n(x): 0 \leq x \leq 1\}=\{g_n(y): a \leq y \leq b\}$ because $a \leq y \leq b$ iff $0 \leq \frac {y-a} {b-a} \leq 1$. Take sup on both sides to to get the first result.
$\int_a^{b} |g_n(y)|dy=\int_a^{b} |f_n(\frac {y-a} {b-a})|dy$. Make the the change of variables $x =\frac {y-a} {b-a}$. [ Note that $dy=(b-a)dx$]. Hence $\int_a^{b} |g_n(y)|dy=\frac {b-a} {2n}$.