TIFR 2014 Problem 4 Solution -Checking for Uniform continuity


TIFR 2014 Problem 4 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:

Let (f:[0,\infty)\to \mathbb{R}) be defined by

( f(x)=x^{2/3}logx)  for (x>0 ) (f(x)=0) if (x=0)

Then

A. f is discontinuous at x=0

B. f is continuous on ([0,\infty)) but not uniformly continuous on ([0,\infty))

C. f is uniformly continuous on ([0,\infty))

D. f is not uniformly continuous of ([0,\infty) but uniformly continuous of ((0,\infty))

Discussion:

First, we find whether (f) is continuous or not. (f) is "clearly" continuous on ((0,\infty)).

Question is, what happens near (x=0)?

Take limit.

(lim_{x\to 0}\frac{logx}{x^{-2/3}}) is in (\frac{-\infty}{\infty}) form.

We use the L'Hospital Rule.

(lim_{x\to 0}\frac{logx}{x^{-2/3}})=(lim_{x\to 0}\frac{1/x}{-\frac{2}{3}x^{-2/3-1}})

=(lim_{x\to 0}-3/2{x^{2/3}})

=(0).

Okay! So we now know that (f) is at-least continuous.

Now we present the HINT s:

For (x>0), (f'(x)=\frac{2}{3}x^{-1/3}logx + x^{2/3}x^{-1}) (By the formula for calculating derivatives of product).

After simplification,

(f'(x)=\frac{2logx+3}{3x^{1/3}}).

We want the boundedness of (f') and we don't care about what happens close to zero as much. (Because we are going to use the first hint).

(lim_{x\to \infty}\frac{2logx+3}{3x^{1/3}}) is in (\frac{\infty}{\infty}) form.

Once again, we use the L'Hospital rule.

(lim_{x\to \infty}\frac{2logx+3}{3x^{1/3}}=lim_{x\to \infty}\frac{2/x}{3/3 x^{-2/3}})

(= lim_{x\to \infty} 2x^{-1/3}=0)

What does this tell us? This tells us that (f') is bounded on (say) ([1,\infty))

So (f) is uniformly continuous on ([1,\infty)). Also, (f) is uniformly continuous on ([0,1]).

So given (\epsilon>0) we have two "delta's" for these two intervals. (which satisfies the condition for uniform continuous in these intervals respectively). Take the delta which is less. And this gives the uniform continuity for (f).

Helpdesk

TIFR 2013 problem 19 | Non-uniformly continuous function

This problem from TIFR 2013, Problem 19 discusses the example of a non-uniformly continuous function.

Question: TIFR 2013 problem 19

True/False?

Every differentiable function $f:(0,1) \to [0,1]$ is uniformly continuous.

Hint: 

$\sin(1/x)$ is not uniformly continuous. However, range is not $[0,1]$. But its a simple matter of scaling.

Discussion:

Let $f(x)=\sin(\frac{1}{x})$ for all $x\in(0,1)$. For simplicity, we first prove that this $f$ is not uniformly continuous. Then we will scale things down, which won't change the non-uniform continuity of $f$.

Note that $f$ is differentiable. To show that $f$ is not uniformly continuous, we first note that as $x$ approaches $0$, $\frac{1}{x}$ goes through an odd multiple of  $\frac{\pi}{2}$ to an even multiple of  $\frac{\pi}{2}$ real fast. So in a very small interval close to $0$, I can find two such points which gives value $1$ and $0$.

Let $x=\frac{2}{(2n+1)\pi}$ and $y=\frac{2}{2n\pi}$.

Then $|x-y|=\frac{2}{2n(2n+1)\pi}$.

Since the right hand side of above goes to zero as $n$ increases, given any $\delta > 0$ we can find $n$ large enough so that $|x-y|<\delta$. For these $x$ and $y$, $|f(x)-f(y)|=1$.

Of course, choosing $\epsilon$ as any positive number less than $1$ shows that $f$ is not uniformly continuous.

We have with us a function $f$ which is bounded, differentiable and not uniformly continuous.

To match with the questions requirements, notice $-1\le f(x)\le 1$.

So $0\le 1+f(x) \le 2$ And $0\le \frac{1+f(x)}{2} \le 1$.

Define $g(x)= \frac{1+f(x)}{2}$ for all $x\in(0,1)$.

Since sum of two uniformly continuous functions is uniformly continuous and a scalar multiple of uniformly continuous function is uniformly continuous, if  $g(x)$ was uniformly continuous, then $f(x)=2g(x)-1$ would also be uniformly continuous.

This proves that g is in fact not uniformly continuous. It is still differentiable, and range is $[0,1]$, which shows that the given statement is actually false.

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TIFR 2013 Problem 14 Solution -Uniform Continuous or Not?


TIFR 2013 Problem 14 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 program 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

Let \(f:\mathbb{R}\to\mathbb{R}\) be defined by \(f(x)=sin(x^3)\). Then f is continuous but not uniformly continuous.

Hint: Can you find a sequence whose terms can get arbitrarily close to each other but the function gives distant values?

Discussion:

The function sin takes the value 1 at \(4n+1\) multiples of \(\pi/2\) and it is -1 at \(4n-1\) multiples of \(\pi/2\) .

Let \(x_n=(n\pi+\pi/2)^{1/3}\) That is, the function takes the value +1 when n is even and it is -1 when n is odd.

Now \(x_{n+1}-x_n\) \(=\) \(\frac{( n+1 )\pi+\pi/2-(n\pi+\pi/2)}{\text{terms involving n}}\)

This gives that the two terms \(x_{n+1}\) and \(x_n\) are close to each other. Because, the limiting value of the difference is zero. (So if you give me any positive real number \(\delta\) I can find an n such that the difference of two consecutive terms is less than that (\delta\) )

And what happens to \(f(x_n)\)? It is +1 and -1 for two consecutive terms (or -1 and +1). Therefore, the difference \(|f(x_{n+1})-f(x_n)|\) is always 2.

In particular, if I give \(\epsilon=1\) then whatever \(\delta\) you produce I will select two consecutive terms in the above sequence \(x_n\) which has distance less than \(\delta\) and the difference of values of \(f\) would not be less than 1.

This proves that \(f\) is not uniformly continuous.

Remark: \(f\) is continuous because it is a composition of two continuous functions ( the sine function applied to the polynomial function \(x\to x^3\) ).

Helpdesk

Differentiability and Uniform Continuity

Problem: Every differentiable function f:  (0, 1) --> [0, 1] is uniformly continuous.

Discussion;

False

Note that every differentiable function f: [0,1] --> (0, 1) is uniformly continuous by virtue of uniform continuity theorem which says every continuous map from closed bounded interval to R is uniformly continuous. However in this case the domain is an open interval.

We can easily find counter example such as $latex f(x) = \sin ( \frac {1}{x} ) $. Intuitively speaking the function oscillates (between -1 and 1) faster and faster as we get close to x = 0. Hence we can get two arbitrarily close values of x such that their functional value's difference equals a particular number (say 1) therefore exceeding any $latex \epsilon < 1 $

An interesting discussion:

differentiability and uniform continuity

Uniform Continuity

Problem: Let f: R --> R be defined by $latex f(x) = sin (x^3) $. Then f is continuous but not uniformly continuous.

Discussion:

True

It is sufficient to show that there exists an $latex epsilon > 0 $ such that for all $latex \delta > 0 $ there exist $latex x_1 , x_2 \in R $ such that $latex | x_1 - x_2 |< \delta $ implies  $latex | f(x_1) - f(x_2) | > \epsilon $ .

Assume $latex epsilon = 0.99 $ . Let $latex x_1 = (k \pi )^{\frac{1}{3}} $ and $latex x_2 = (k \pi + \frac{\pi}{2} )^{\frac{1}{3}} $.

Hence $latex |x_1 - x_2 | = (k \pi + \frac{\pi}{2})^{\frac{1}{3}} - (k \pi)^{\frac{1}{3}} < \frac {1}{k^{2/3}} $ . (This is achieved by some simple algebra like rationalization )

Now if we take $latex k > \frac {1}{\delta ^{3/2}} $ then $latex |x_1 - x_2| < \frac {1}{k^{2/3}} < \delta $ but $latex |f(x_1) - f(x_2) | = 1 > 0.99 = \epsilon $

Hence there exists an $latex \epsilon $ (= 0.99) such that for any value of $latex \delta >0 $ we will get $latex x_1 , x_2 $ such that $latex | x_1 - x_2 |< \delta $ implies  $latex | f(x_1) - f(x_2) | > \epsilon $ .