Functional Analysis integral problem Im attempting to solve an example from my course, 

Prove that for any $0<p<1$ there is a sequence of functions $(f_{i})_{i=1}^{\infty}$ in $C^{\infty}_{0}$($[-1,1]$) such that, 
$$ lim_{i \to \infty}=\int_{\mathbb{R}} |\frac{df_{i}}{dx}(y)|^{p}dy=0 $$
and for any $x \in (-1,1) $ we have that the following holds 
$$lim_{i \to \infty}f_{i}(x)=1 $$

To solve it I first thought about DCT but clearly it is not helpful, 
So now Im thinking 
-maybe i have to use that $C^{\infty}_{0}$ is dense in $L_{p}$ spaces
-or use a common test function like $e$ to some power?
Help greatly appreciated im really stuck 

Let $\phi: \mathbb R^+ \to [0,1]$, smooth, $\epsilon>0$ 


*

*$\phi(x) = 0$ for all x $\leq \epsilon$,

*$\phi(x) = 1$ for all $x \geq 1$ and 

*$\phi'(x) \leq 2$ for all $x\in \mathbb{R}^+$. 
For $x\in [-1,1]$ we now define 
$$
f_i(x) := \phi(i (1-x^2)).
$$
Now, we only need to check the requested properties. Clearly, if $x \in (-1,1)$, $i \to \infty$, then $\phi(i (1-x^2))$ goes to 1 by the second property of $\phi$. Moreover, $f_i$ is smooth and has compact support by construction.
Now let us have a look at the integrals. Here it is essential that $p< 1$.
$$
\int_{[-1,1]} |f_i'(y)|^p dy = \int_{[-1,1]} |\phi'(i (1-y^2)) 2 y i |^p dy.
$$
By assumption, we have that $\phi'(x)= 0 $ for all $x \geq 1$. This implies that 
$$
\int_{[-1,1]} |\phi'(i (1-y^2)) 2 y i |^p dy = \int_{(1-y^2)\leq \frac{1}{i}, |y|\leq 1} |\phi'(i( 1-y^2)) 2 y i |^p dy \leq (4 i)^p \int_{(1-y^2)\leq \frac{1}{i}, |y|\leq 1} dy \leq (4 i)^p \int_{(1-|y|)\leq \frac{1}{i}, |y|\leq 1} dy \leq 2 (4 i)^p /i.
$$
As a consequence, we have that 
$$
\int_{[-1,1]} |f_i'(y)|^p dy \leq C i^p/i,
$$
for some constant $C>0$ which finishes the proof.
