Does there exist a function satisfying this condition? 
Does there exist a non-negative continuous function $f:[0,1]\to \Bbb R$ such that
$\int _0^1 f^n \text{dx}\to 2$ as $n\to \infty $?

I am not getting anything from here. As I was googling I found the answer
Please help me and elaborate this explanation with proper detail. I would be more  thankful. Thanks in advance for your kindness.
 A: 
$f(x)$ needs to be strictly larger than $1$ on at least one point $x_+\in (0, 1)$ - otherwise $\int_0^1(f(x))^ndx\leq 1$ for all $n\geq 0$

First of all, why would this be a bad thing? It would be a bad thing because if all these integrals are at most $1$, then they can't converge to $2$, which is what we want.
So, why is it true? That's because $f$ being non-negative means we may interpret the integral as an area. This area has width $1$ (since the integral goes from $0$ to $1$), so if the area is ever going to be larger than $1$, then the graph must at some point be taller than  $1$. Otherwise the area is completely contained in a $1\times 1$ square, which has area $1$. And if the graph of $f$ is completely contained within the square, then any power of $f$ is completely contained in the square as well.

$x_+\in (\delta, 1-\delta)$

This is just to make sure that the $x_+$ we pick is sufficiently far from the edge of the interval $[0, 1]$ that $|x-x_+|<\delta$ implies that $x$ is inside $[0, 1]$ as well. This is done so we don't have to treat the edge with any special care: We just make sure we are far enough away from it.

$\int_0^1 (f(x))^ndx\geq 2\delta(1+\epsilon)^n$

Well, because of the continuity of $f$ we have declared that a rectangle of width $2\delta$ and height $1+\epsilon$ with bottom line centered at $x_+$ fits under the graph of $f$. Therefore, clearly the integral of $f$ must be at least as large as the area of this rectangle.
Generally, it also implies that a rectangle of height $(1+\epsilon)^n$ and width $2\delta$ centered at $x_+$ as before, fits under the graph of $(f(x))^n$. Therefore the integral of $(f(x))^n$ must be at least as large as the area of this rectangle. Since the area of the rectangle just keeps on increasing towards $\infty$ as $n$ increases, so too must the integral of $(f(x))^n$.
A: Let me write it out fully. Suppose for a contradiction that for some continuous and non-negative $f$ we have $\int f^n \rightarrow 2$. 
By definition of convergence of sequences, there is then some $N \in \mathbb{N}$ such that $n \geq N \Rightarrow \int f^n \geq 3/2$.
Suppose that $f \leq 1$ on $[0,1]$. Then $\int f^n \leq \int f$, because $f^n \leq f$. But if $f \leq 1$ we also know that $\int f \leq 1$ . 
Let $n \geq N$. Then we've just shown that $\int f^n \leq 1$. 
But in the previous paragraph we said that $\int f^n \geq 3/2$. This is a contradiction. 
So we must have $f(x_+)  > 1+2\epsilon$ for some $x_+ \in [0,1]$ and $\epsilon > 0$.
By continuity, there is some $\delta > 0$ such that $f \geq  1+\epsilon$ on 
$I_\delta:= (x_+ - \delta, x_+ + \delta) \cap [0,1]$.
Now, $$\int_{[0,1]} f^n(x)dx\geq \int_{I_\delta} f^n(x)dx \geq (1+\epsilon)^n\int_{I_\delta}dx \geq (1+\epsilon)^n \delta \rightarrow \infty,$$
as $n \rightarrow \infty$.
The first inequality uses the fact that $f$ is non-negative. 
The second uses the fact that $f$ is at least $(1+\epsilon)$ on $I_\delta$. 
The third uses that the length of $I_\delta$ is at least $\delta$.
We have arrived at a contradiction.
