Show that the closed unit ball $B[0,1]$ in $C[0,1]$ is not compact

Solution 1:

Use functions that are $n^2$ on $[0,1/n]$ and zero on $[1/n+\epsilon,1]$ and on $[1/n,1/n+\epsilon]$ connect them to be continuous. Then the integral of any one of them is close to $n$ (choose an appropriate $\epsilon$ that will depend on $n$) and I believe the integral of the difference of any two of them is at least one.

Solution 2:

The uniform case is classic: use $f_n(x)=x^n$. There is a lot to be learned from this example.

That example won't work in the integral case, because this sequence converges to the zero function in the integral sense. Instead, try making a function which is a continuous approximation to a function with a jump: for instance it is $1$ on $[1/4,3/4]$, zero outside $[1/4-1/(4n),3/4+1/(4n)]$, and continuous (say, linear) in between. These look like trapezoids which are approaching a rectangle.

Another possibility is $\sin(nx)$, but I think the proof might be more difficult in this case.

Solution 3:

For 1. take $f_n$ to be the function whose graph is given by connecting the points $(0,0), ({1 \over n+1}, 0), ({1\over 2} ({1 \over n+1} + {1 \over n}),1),({1 \over n}, 0), (0,0) $. Then $d(0,f_n) = 1$, and $d(f_n,f_m) = 1$ for all $n \neq m$ hence there are no convergent subsequences.

$f_n$ is a 'tooth' of height $1$ with support in $[{1 \over n+1},{1 \over n}]$.

For 2. in a similar fashion, take $f_n$ to be the function whose graph is given by connecting the points $(0,0), ({1 \over n+1}, 0), ({1\over 2} ({1 \over n+1} + {1 \over n}), 2n (n+1)), ({1 \over n}, 0)), (0,0) $. Then , as above, $d(0,f_n) = 1$, and $d(f_n,f_m) = 2$ for all $n \neq m$ hence there are no convergent subsequences.

$f_n$ is a 'tooth' of height $2n(n+1)$ with support in $[{1 \over n+1},{1 \over n}]$.