Let us take a warm up quiz

Understand the problem

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

Source of the problem

TIFR PROBLEM 10

Topic
Continuous Function 
Difficulty Level
EASY
Suggested Book
REAL ANALYSIS BY S.K MAPA

Start with hints

Do you really need a hint? Try it first!

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?
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 \nrightarrow 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.
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 \implies \lim_{n \to \infty} \int_{0}^{1} f^{n} dx \leq 1 So, \int_{0}^{1} f^{n} dx \nrightarrow 2 as n \to \infty
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 \implies 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.

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