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TIFR 2014 Problem 3 Solution -Function bounds from derivative limits

TIFR 2014 Problem 3 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

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Let (f:\mathbb{R}\to \mathbb{R}) be a differentiable function such that (\lim_{x\to \infty} f'(x)=1) then

A. f is bounded

B. f is increasing

C. f is unbounded

D. f' is bounded


The easiest function to look at is the function which has derivative 1 at all points. That is (f(x)=x). This function is not bounded. So option A is wrong.

Now, the function (f) is increasing. But let's say we disturb the same (f) around 0 a little bit, so that the function is not increasing any more. We can do this disturbing so that the function still remains differentiable, but it loses the increasing part of it. This tells us that B is wrong.

We can disprove D in a similar manner, for even though (f') can be bounded in the positive side of x-axis, what if (f'\to \infty) as (x\to -\infty)? And this case in absolutely possible, because we have no information about the nature of (f) for -ve values of x. We can construct such an example easily by the help of drawing graphs, or constructing functions explicitly. (Use (x^2) when x is negative and some linear function/translation of x when x is positive, and make sure the function transition is smooth in between).

Now, we have eliminated all but option C. We prove option C.

What does the expression (\lim_{x\to \infty} f'(x)=1) mean?

It means that for large values of (x) (f'(x)) is close enough to 1.

That means, for a sequence going to (\infty) we have (f') at the sequence-points going to 1.

Take the sequence as (n\in \mathbb{N}). Then

(\lim_{n\to \infty} lim_{h \to 0} \frac{f(n+h)-f(n)}{h} =1).

Let's forget about (h\to 0). Let's say, h=1.

Then (f(n+1)-f(n)=f'(y_n)(n+1-n)) by the mean-value theorem. Where (y_n\in[n,n+1]). As (n\to \infty), (y_n \to \lim_{x\to \infty} f'(x)=1).

That means, given (\epsilon >0); for large values of (n)

(|f(n+1)-f(n)-1|=|f'(y_n)-1|< \epsilon ). This means that (f) increases approximately by 1 for an increment of 1 in the value of (x). This means that (f) behaves like (x) for large values of (x). Which proves that (f) must be unbounded.


  • What is this topic: Real Analysis
  • What are some of the associated concept: Increasing function, Differentiability
  • Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

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