Is the subset G totally bounded? - Proving or disproving using relevant theorems

[Eulogy]

Homework Statement

Let G = { f $\in$ C[0,1] : $^{0}_{1}$$\int$|f(x)|dx $\leq$ 1 }
Endowed with the metric d(f,h) = $^{0}_{1}$$\int$|f(x)-h(x)|dx. Is G totally bounded? Prove or provide counterexample

2. Relevant Theorems

Arzela-Ascoli Theorem, Theorems relating to compactness, equicontinuity etc and

Let M be a compact metric space. A subset of
C(M) is totally bounded iff it is bounded and equicontinuous. (in terms of the sup metric)

The Attempt at a Solution

I'm not to sure whether G is totally bounded or not to begin with. If I was working in the sup metric the subset is not totally bounded as I can find a family of functions which are not equicontinuous in G. However I'm not sure how or if this translates to the integral metric defined above. Any help would be much appreciated!

 

[Dick]

Try to find a sequence of functions $f_i(x)$ such that $d(f_i,f_j)=1$ for all $i \ne j$. Step functions will do. What would that tell you about total boundedness?

 

[Eulogy]

Such a sequence could not be totally bounded since if I took a finite cover of the space with balls radius say half then each function in the sequence must be in a separate ball and as the sequence is infinite this leads to a contradiction. But I can't think of a sequence functions that are continuous on [0,1] with this property?

 

[Dick]

Such a sequence could not be totally bounded since if I took a finite cover of the space with balls radius say half then each function in the sequence must be in a separate ball and as the sequence is infinite this leads to a contradiction. But I can't think of a sequence functions that are continuous on [0,1] with this property?

Can you think of a way to define f1 so that it's integral is 1/2, but it is only nonzero on [1/2,1]. Same thing for f2 but it is only nonzero on [1/4,1/2], put f3 on [1/8,1/4] etc.

 

[Eulogy]

Yeah just figured it out thanks for your help!