Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
You seem to be right. –  egreg Nov 27 '13 at 9:27
    
Thank you. But if I am right, then, I can't go further on this topic in the book. The book says, since the area under the graph of $f_n$ is $\frac1n$, $$||f_n||_{L^1}=\int_0^1|f_n(x)|dx\text{ ($x\in[0,1]$)}\longrightarrow0.$$ Again I am stuck! –  Sush Nov 27 '13 at 9:38
1  
As $\int_{0}^{\frac{1}{n}}2n^2-2n^3x dx$=n, the book seems to contain a typo. –  Peter Nov 27 '13 at 9:48
    
@Peter, So does this mean that $$||f_n||_{L^1}\longrightarrow\infty?$$ A bigger typo? –  Sush Nov 27 '13 at 9:51
    
No, this typo follows from the wrong calculated integral. –  Peter Nov 27 '13 at 9:53

1 Answer 1

up vote 1 down vote accepted

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 $$ \lim_{n\to\infty}f_n(0)=\infty. $$

share|improve this answer
    
Many many thanks, Sir. The author wants to show that the choice of norm makes a big difference in function spaces. And, as this example is false, will you please provide me an example (or any useful elementary link) to show that choice of norm really matters for function spaces? –  Sush Nov 27 '13 at 13:03
1  
@Sush I'll try supplying an example later. But the idea is to take a triangle with area $1/n$ so that the vertex on the $y$-axis goes far away. –  egreg Nov 27 '13 at 17:18
    
Many many thanks, Sir. Your correction to the example given is really amazing! –  Sush Nov 28 '13 at 2:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.