Proving two metrics are not equivalent

[Lily@pie]

Homework Statement

The vector space of continuous functions on [0,1] is given the following metrics

$d_{∞} (f,g) = sup_{x\in [0,1]} |f(x)-g(x)|$

$d_{2} (f,g) = (∫^{1}_{0} |f(x)-g(x)|^{2} dx)^{1/2}$

Are these two metrics equivalent?

Homework Equations

If $d_{∞}$ and $d_{2}$ are equivalent, then for every $f \in C[0,1]$ and every $\epsilon>0$, there exist a $\delta>0$ such that $B^{d_{∞}}_{\delta}(f)\subset B^{d_{2}}_{\epsilon}(f)$ and $B^{d_{2}}_{\delta}(f)\subset B^{d_{∞}}_{\epsilon}(f)$

The Attempt at a Solution

I know that these two metrics are not equivalent, but i can't find a function at which i can find an $\epsilon>0$ for every $\delta>0$ where they are not the subset of each other.

I know that $d_{∞}$ is like the highest point of the g(x) if I let f(x)=0 and g(x) be some other function. $d_{2}$ is like the area under the graph between [0,1].

 

[tiny-tim]

Hi Lily@pie!

I know that $d_{∞}$ is like the highest point of the g(x) if I let f(x)=0 and g(x) be some other function. $d_{2}$ is like the area under the graph between [0,1].

Yes, you're practically there …

you need a sequence g_n(x) with constant maximum but decreasing area ​

 

[Lily@pie]

I'm clueless

any hints?

 

[tiny-tim]

how about a curve which starts at (0,1) and curves gracefully down to (1,0) ?

then reduce the area ​

 

[Lily@pie]

Would this be a function g(x)=(x-1)² where 0≤x≤1??

And f(x) = 0 for all x

For all δ>0 such that $d_{2} (f,g)$ will be undefined! oh my god... what did i do...

 

[tiny-tim]

Would this be a function g(x)=(x-1)² where 0≤x≤1??

i was thinking of somthing that gets flatter a lot quicker,

like e^-x (but adjusted to fit into [0,1])