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# Understand the problem

There exists a non-negative continuous function $f:[0,1] \rightarrow \mathbb{R}$ such that $\int_{0}^{1} f^{n} d x \rightarrow 2$ as $n \rightarrow \infty$ (a) TRUE (b) FALSE

Do you really need a hint? Try it first!

"Hint 1" : Rather I want to say that it is a comment on the question that here $f^{n}$ does not mean $f \circ f \circ . . . . . . \circ f$ ($n$ times). Here $f^{n}= f \bullet f . . . . . \bullet f$ ($n$ times). Now do you want to think again with this disclosure? "Hint 2" : Make two cases Case 1:  $0<f(x) \leq 1\ \forall x \in [0,1]$ [Observe that $f$ is non negative function] Case 2:  $f(x) > 1$ for some $x \in [0,1]$ Now prove for each case that $\int_{0}^{1} f^{n} dx \rightarrow 2$ as $n \to \infty$. So, by the last statement you have guessed the validity of the statement . It is a false statement!! In the next two cases, we will basically prove two cases. "Hint 3" : Case 1:  If $f(x) \leq 1\ \forall\ x \in [0,1]$ then $f^{n}(x) \leq 1\ \forall\ x \in [0,1]$ So, $\int_{0}^{1} f^{n} dx \leq\ \int_{0}^{1} 1 dx = 1$ $\rightarrow \lim_{n \to \infty} \int_{0}^{1} f^{n} dx \leq 1$ So, $\int_{0}^{1} f^{n} dx \rightarrow 2$ as $n \to \infty$ "Hint 4" : Case 2:  Suppose $f(x) > 1$ for some  $y \in [0,1]$ Now, as $f(x)$ is continuous function We have $f(x) > 1+\epsilon\ \forall\ x \in (y - \delta, y + \delta)$ for some $\epsilon , \delta > 0$ $\rightarrow f^{n}(x) > (1+ \epsilon)^{n}$ then we have $\int_{0}^{1} f^{n}\ dx > \int_{y - \delta} ^{y + \delta} (1+ \epsilon)^{n}\ dx$ [If $y \in [0,1]]$ or $\int_{1 - \delta}^{1} (1+ \epsilon)^{n}\ dx$ [If $y=1$] or $\int_{0}^{0 + \delta} (1+ \epsilon)^{n}\ dx$ [If $y=0$] In either case , $\int_{0}^{1} f^{n}\ dx > (1 + \epsilon)^{n}\ \delta \to \infty$   So, the statement is false.

# Similar Problems

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