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 
 A: 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.
A: The first thing I did was connect the points in the plane $(-1,0),(-1+1/n,1), (1-1/n,1),(1,0)$ with line segments, here thinking of $n$ as large. You then have the graph of an $f_n$ on $[-1,1]$ for which computations are easy. Note that $|f_n'|=n$ on the intervals to the left and right, with $f_n'=0$ in between. Thus
$$\int_{-1}^1|f_n'|^p = 2 \cdot n^p\cdot \frac{1}{n}= 2n^{p-1} \to 0\,\, \text { as } n\to \infty.$$
The sequence $f_n\to 1$ pointwise on $(-1,1),$ so we have our example. Except that these $f_n$ are not smooth.
It turns out you don't have to do much to find smooth $f_n$ like the above. Suppose $f$ is smooth on $[a,b],$ and $f$ increases from $0$ to $1$ on $[a,b].$ Then
$$\tag 1  \int_a^b(f')^p \le (b-a)^{1-p}.$$
Proof: By Holder and the FTC,
$$\int_a^b (f')^p =  \int_a^b(f')^p\cdot 1 \le (\int_a^b  f')^{p}(\int_a^b 1^{1/(1-p)})^{1-p} = 1\cdot (b-a)^{1-p}.$$
Of course there is also a version of $(1)$ for decreasing $f.$
This implies that every sequence of smooth versions of the $f_n$ above will work. Thus for $n=2,3,\dots$ let $f_n\in C^\infty(\mathbb R)$ be such that $f_n(x)=0$ for $|x|\ge 1,$ $f_n$ increases from $0$ to $1$ on $[-1,-1+1/n],$ $f_n=1$ on $[-1+1/n,1-1/n],$ and $f_n$ decreases from $1$ to $0$ on $[1-1/n,1].$ Then $f_n\to 0$ pointwise on $(-1,1),$ and $(1)$ (and its decreasing cousin) imply
$$\int_{-1}^1 |(f_n)'| \le 2n^{1-p} \to 0.$$
