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where $$x_n(t)= \begin{cases} nt & 0\leq t<\frac{1}{n}\\ \frac{1}{nt}&\frac{1}{n}<t\leq 1\\ \end{cases}$$

I don't think this is converging just by guess, but what is the procedure for determining convergence? This being a function makes it a little harder for me to understand.

What is $x_1$?

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It should really be $0\leq t\leq\frac{1}{n}$ and/or $\frac{1}{n}\leq t\leq 1$ to specify that $x_n(1/n)=1$ for all $n$. –  Jonas Meyer Feb 2 '12 at 5:57
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I suggest computing $\int_0^1 x_n(t)dt$. –  Jonas Meyer Feb 2 '12 at 5:58
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$x_1$ is not the constant function over $[0,1]$ - it's the identity function $x(t)=t$. –  rotten Feb 2 '12 at 6:27
    
Have you tried to draw a sketch of the function's graph? It is usually extremely helpful. –  Najib Idrissi Feb 2 '12 at 6:54
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1 Answer 1

up vote 2 down vote accepted

The graph of $f_n$ consists of a straight line segment from the origin to the point $[1/n,1]$ and a "hyberbolic arch" from $[1/n,1]$ to $[1,1/n]$.

Note that $\{f_n\}$ converges pointwise to 0: For $t=0$, we have $f_n(0)=0$ for all $n$, while for $t>0$: $$ \lim_{n\rightarrow\infty} f_n(t)=\lim_{n\rightarrow\infty} {1\over nt}=0. $$

Now note:

  • The distance from any $f_n$ to the 0 function in $C[0,1]$ is 1.

  • $\int_0^1| f_n(t)-0|\,dt= {1\over 2n}+ \int_{1/n}^1 {1\over n}{1\over t}\,dt= {1\over n}+ {\ln n\over n}$.

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