I was self-reading Mathematics for Economists by Simon and Blume. On page 815, Section 29.4, he has discussed "Norms on Function Space". And here I am stuck:
Let $$f_n = \begin{cases} 2n^2-2n^3x, & \text{$0\leq x\leq\frac1n$,} \\ 0, & \text{$\frac1n\leq x\leq1.$} \\ \end{cases}$$ The graph of $f_n$ is a line segment of slope $-2n^3$ from $(0,2n^2)$ to $(\frac1n,0)$ and then runs along $x$-axis from $(\frac1n,0)$ to $(1,0)$. The area under the graph of $f_n$ is $\frac1n$ and thus $$||f_n||_{L^1}=\int_0^1|f_n(x)|dx\text{ ($x\in[0,1]$)}\longrightarrow0.$$
But I think the corresponding area should be $$\frac12\times \text{Base}\times\text{Height}=\frac12\times\frac1n\times2n^2=n.$$ Please let me know where I am wrong.
You're right. The consequence derived from this wrong computation is of course false too. Indeed $$ \lim_{n\to\infty}\|f_n\|_{L^1}=\infty. $$
A “correct” example might be $$ f(x)=\begin{cases} n-n^3x & \text{for $0\le x\le\frac{1}{n^2}$}\\ 0 & \text{for $\frac{1}{n^2}<x\le 1$} \end{cases} $$ Then $$ \int_0^1 f(x)\,dx=\int_0^{1/n^2}(n-n^3x)\,dx= n\frac{1}{n^2}-\frac{1}{2}n^3\frac{1}{n^4}=\frac{1}{2n} $$ so $$ \lim_{n\to\infty}\|f_n\|_{L^1}=\lim_{n\to\infty}\frac{1}{2n}=0 $$ but the sequence of functions $(f_n)$ is not pointwise convergent, because $$ \lim_{n\to\infty}f_n(0)=\infty. $$