TIFR 2013 problem 35 | Sequence Bounded / Unbounded

Let's discuss a problem and find out whether a sequence is bounded or unbounded from TIFR 2013 Problem 35. Try before reading the solution.

Question: TIFR 2013 problem 35

True/False?

Let \( \{a_n\} \) be any non-constant sequence in \(\mathbb{R}\) such that \( a_{n+1}=\frac{a_n + a_{n+2} }{2} \) for all \(n \ge 1 \). Then \(\{a_n\} \) is unbounded.

Hint:

The given expression is same as \( a_{n+1}-a_n = a_{n+2} -a_{n+1} \).

Discussion:

The distance between two successive terms in the given sequence is constant. It is given by \( |a_{n+1}- a_n| = |a_n - a_{n-1}| = ... = |a_1-a_0| \).

So for the sequence to be non-constant, \( a_1 \ne a_0 \). Because otherwise, the sequence will have distance between any two successive terms zero, which is just another way of saying that the sequence is constant.

There are two cases:

Case 1: \(a_1 > a_0\).

Then \(a_{n+1} > a_n\), that is the sequence is increasing, and not only that, it is an arithmetic progression with common difference \(a_1-a_0 (> 0)\). Therefore, the sequence is unbounded above.

Case 2: \(a_1<a_0\).

Then as in the previous case the sequence this time will become a decreasing sequence, not only that, it is an arithmetic progression with common difference \( < 0 \). Therefore, the sequence is unbounded below.

Remark:

We don't really need to take the two cases. The key point is that the given recurrence relation is that of an arithmetic progression whose common difference is non-zero. Hence the sequence has to be unbounded.

 

Useful links:

TIFR 2013 Problem 27 Solution -Bounded derivatives and Compactness


TIFR 2013 Problem 27 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.

Also Visit: College Mathematics Program of Cheenta

Problem:True/False?

If every differentiable function on a subset \(X\subset \mathbb{R}^n \) (i.e, restriction of a differentiable function on a neighbourhood of \(X\) ) is bounded, then \(X\) is compact.

Hint:

Recall that  \(X\subset \mathbb{R}^n \) is compact if and only if it is closed and bounded. Find suitable functions to show bounded and closed respectively.

Discussion:

Let \(f(x)=(||x||^2,||x||^2,...,||x||^2)\), then \(f\) is differentiable. Hence \(||x||\) must be bounded on \(X\), which is the same as saying \(X\) is bounded.

Note, \(||x||\) is the standard euclidean norm. And \(f\) is differentiable because each component of \(f\) is differentiable.

Now all we need is a function which will show that \(X\) is closed as well.

Let \(p\in \mathbb{R}^n \) be a point which is a limit point of \(X\). Suppose that \(p\notin X \). Then \(||x-p|| \ne 0 \) for any \(x\in X\). Hence the function \(g(x)=(\frac{1}{||x-p||},\frac{1}{||x-p||},...,\frac{1}{||x-p||} )  \) is a well-defined and differentiable function. Again, the differentiation part follows from the fact that it is differentiable in each component. And that it is differentiable in each component is easy to check. (Hint: Take the partial derivatives).

Now that we have \(g\), a differentiable function, we know that \(g\) must be bounded on \(X\). But as \(x \to p \) we see that \(||g(x)|| \to \infty \), i.e, g is not bounded. The only assumption we have made so far is that \(p\) is not in \(X\). This must be false, so \(p\in X\). So in fact, every limit point of \(X\) is in \(X\). This shows that \(X\) is closed.

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