Continuous Function: TIFR 2017 Problem 10

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

Start with hints

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.

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

Start with hints

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.