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.

Similar Problems

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

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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

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.

Similar Problems

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

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