Experiences at a PhD Interview at TIFR-CAM


This is the record of my interview held at TIFR Bangalore on 20th March’14 for Integrated program in Mathematics. So this was my second time at TIFR Bangalore ,last year I could not get through! This time I prepared much harder, mainly I did Real Analysis and I had more experience this time since in this one year I went through many exams and had lots of idea about what kind of questions can be asked. So Interview was at 9 in the morning and I was the first to appear that day.

I went inside the room and there were 5 Professors sitting inside, panel consists of Prof K Sandeep, Prof Venky Krishnan, Prof Praveen, Prof AS Vasudeva Murthy( don’t remember the other one). I’ll use P for Professor and I for myself.

I-Good Morning to all of you!

P-Good morning Neeraj, Welcome have a seat

P-So Neeraj you completed your BSc last year ,since what have you been doing?

I- Sir I joined Int-PhD program at IISER Mohali last year!

P-So why do you want to join us?

I- Sir I’m particularly interested in doing Analysis and I think TIFR-CAM is a good option for me

P- Okay so what courses have you done in 2nd sem?

I- Measure Theory, Complex Analysis, Combinatorial Group Theory, Discrete Mathematics, Groups and Fields, but I’m not much comfortable with my sem courses!

P-Let me ask some Measure Theory first, can you give an example of Lebesgue integral function which is not Riemann Integrable?

I- Yes, consider $$ f(x) =left{begin{array}{ll}x & mbox{if } x in mathbb{Q} \-x & mbox{if } x in mathbb{Q}^cend{array}right.$$

P-Correct, now can you give an example of function which is Reimann Integrable but not Lebesgue integrable?

I- Consider the integral $ int^infty_0 frac{sin x}{x},dx$ , this integral is finite but $ int^infty_0 |frac{sin x}{x}|,dx$ is infinite.

P-Ok, find the Lebesgue integral of  $$f(x)=left{begin{array}{ll}exp{x} & mbox{if } xin mathbb{Q}cap[0,1] \exp{-x} & mbox{if } x in mathbb{Q}^ccap[0,1]end{array}right.$$

I-(I had no idea about this, I thought I should write something on board and I knew what I’m writing is wrong!) Is it $ e^xmu(mathbb{Q}cap[0,1])+e^{-x}mu(mathbb{Q}^ccap [0,1])$

P- That’s not correct you’re using the formula for simple function! Let’s leave it, Shall we start with Complex Analysis?

I-Sir i’m not much comfortable with that!

P-Okay give an example of sequence of continuous functions whose limit is not continuous.

I- Consider $$f_n(x) =left{begin{array}{lll}-1 & mbox{if } x < frac{-1}{n} \0 & mbox{if } frac{-1}{n} le x lefrac{1}{n} \1 & mbox{if} x>frac{1}{n}end{array}right.$$

This is continuous at $ x=0$ but limit function $ f(x)=sgn(x)$ not continuous at $ x=0$

P- I mean give an example of sequence of functions which are continuous on their domain but limit function is not continuous.

I- Then $ f_n(x)=x^n$ on $ [0,1]$ will work with limit function $$ f(x)=left{begin{array}{ll}0 & mbox{if } x in [0,1) \1 &mbox{if } x =1end{array}right.$$

P-Is it uniformly continuous on $ [0,1]$

I-No sir, since $ lim_{nto infty} sup_{xin [0,1]} |f_n(x)-f(x)|=1nrightarrow 0 $ as $ nto 0$

P-Okay, Is it uniform in interval $  [0,b]$ for some $$0<b<1$$

I- Yes, since $ lim_{nto infty} sup_{xin [0,b]} |f_n(x)-f(x)|=b^nto 0 $ as $ nto 0$

P-Yes, now consider a continuous function $ f(x):mathbb{R} to mathbb{R}$ satisfying $ |f(x)-f(y)|ge |x-y|$ and is bounded. Is it onto?

I-(I did this problem day before the interview but there $ f(x)$ was not bounded , and in excitement I didn’t noticed this change and starting proving that $ f(x)$ is onto! As a result I got stuck in between and then I realized my mistake) Sir it can’t be onto since it is bounded!

P-Can you give an example of such function.

I- $ f(x)= arctan x$ ( I thought he’s asking for bounded function which is not onto)

P-But does it satisfy the inequality property?

I-No because $ |f(x)-f(y)|ge |x-y| implies f’ge1 forall x in mathbb{R}$ which is not true here since $ f'(x)= frac{1}{1+x^2}le 1 forall x in mathbb{R}$

P- If a continuous function is mapping open intervals to open intervals will it map open sets to open sets( He asked me this because I used this somewhere while I was proving the result)

I-( I thought for a while)

P-(hints) every open set can be written as countable union of disjoint open intervals!

I- oh yes, then for any open set $ A=cup U_i$ for disjoint intervals $ U_i$, $ f(A)=cup f(U_i)$ and this union will be open!

P-Yeah that’s true, now consider the sequence of functions $$ f_n(x)= left{begin{array}{ll}sin nx & mbox{if } 0le x le frac{pi}{2n} \0 & mbox{if } otherwiseend{array}right.$$

Is it pointwise convergent?

I- Yes, if we let $ nto infty$ then $ [0,frac{pi}{2n}]$ becomes $ {0}$ and $ f_n to 0$

P- Make it more precise.

I-( I was little confused exactly how to show it formally)

P-Fix $ x$ and then let $ n to infty$

I- Okay, so if we fix $ x=c$ then c comes out of interval $ [0,frac{pi}{2n}]$ for large n and so $ f_n(c) to 0$ pointwise.

P-Is the sequence uniformly converging to 0?

I-No sir , since $ lim_{nto infty} sup_{x in [0,frac{pi}{2n}]} |sin x| = lim 1 =1 nrightarrow 0$

P- Okay Neeraj we’re done you may leave now!

I-Thank You very much.

[Neeraj is now an Integrated PhD student at the Centre for Applicable Mathematics, TIFR, Bangalore.]