ISI MMath Solutions and Discussion 2020 : PMB

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ISI MMath PMB 2020 Subjective Questions, solutions and discussions

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Problem 1.

(a) Let $\{f_n\}$ be a sequence of continuous real-valued functions on $[0,1]$ converging uniformly on $[0,1]$ to a function $f$ . Suppose for all $n\geq 1$ there exists $x_n\in [0,1]$ such that $f_n(x_n)=0$ . Show that there exists $x\in [0,1]$ such that $f(x)=0$ .

(b) Give an example of a sequence $\{f_n\}$ of continuous real-valued functions on $[0,\infty)$ converging uniformly on $[0,\infty)$ to a function $f$ , such that for each $n\geq 1$ there exists $x_n\in [0,\infty)$ satisfying $f_n(x_n)=0$ , but $f$ satisfies $f(x)\neq 0$ for all $x\in [0,\infty)$ .

Topic: Real Analysis, Difficulty level: Medium

Full Solution

(a)

The segment $[0,1]$ is compact. Hence, the sequence $(x_n)_{n\geq 1}$ has a convergent subsequence:

$\lim_{k\to \infty}x_{n_k}=x\in [0,1].$

The function $f$ is continuous, since it is a uniform limit of continuous functions.

Let $\epsilon>0.$ There exists $N\geq 1$ such that for all $n\geq N$

$\sup_{t\in [0,1]}|f_n(t)-f(t)|\leq \epsilon.$

There exists $K\geq 1$ such that for all $k\geq K$ $n_k\geq N.$ Then we estimate

$|f(x_{n_k})|\leq |f(x_{n_k})-f_{n_k}(x_{n_k})|+|f_{n_k}(x_{n_k})|\leq \sup_{t\in [0,1]}|f(t)-f_{n_k}(t)|\leq \epsilon.$

(b)

Let $f_n(x)=e^{-x}-\frac{1}{n},$ $f(x)=e^{-x}.$ We have

$\sup_{x\geq 0}|f_n(x)-f(x)|=\frac{1}{n}\to 0, n\to\infty.$

Also, $f_n(\log(n))=0.$ However, $f(x)>0$ for all $x\geq 0.$

It follows that $|f(x)|=\lim_{k\to\infty}|f(x_{n_k})|\leq \epsilon.$ Since $\epsilon>0$ is arbitrary, we deduce that $f(x)=0.$

Problem 2.

Let $f:[0,1]\to \mathbb{R}$ be continuous function. Show that

$\lim_{n\to\infty}\Pi^n_{k=1}\bigg(1+\frac{1}{n}f\bigg(\frac{k}{n}\bigg)\bigg)=e^{\int^1_0 f(x)}dx$

Topic: Real Analysis, Difficulty Level: Hard

Full Solution

The function $f$ is bounded on $[0,1]:$

$C=\sup_{t\in [0,1]}|f(t)|<\infty.$

Let $n>\frac{2}{C},$ so that for all $t\in [0,1]$ $\frac{|f(t)|}{n}<\frac{1}{2}.$

We will use elementary inequality

$|\log(1+x)-x|\leq 2x^2, \ |x|\leq \frac{1}{2}.$

Represent the product as

$\prod^n_{k=1}\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)=\exp\left(\sum^n_{k=1}\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)\right).$

We have an estimate

$\bigg|\sum^n_{k=1}\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)-\sum^n_{k=1}\frac{1}{n}f\left(\frac{k}{n}\right)\bigg|\leq$

$\leq \sum^n_{k=1}\bigg|\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)-\frac{1}{n}f\left(\frac{k}{n}\right)\bigg|\leq 2\sum^n_{k=1}\frac{f(\frac{k}{n})^2}{n^2}\leq \frac{2C^2}{n}.$

So,

$\sum^n_{k=1}\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)-\sum^n_{k=1}\frac{1}{n}f\left(\frac{k}{n}\right)\to 0, \ n\to\infty.$

Taking into account the convergence

$\sum^n_{k=1}\frac{1}{n}f\left(\frac{k}{n}\right)\to \int^1_0 f(x)dx, \ n\to \infty$

we deduce that

$\sum^n_{k=1}\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)\to \int^1_0 f(x)dx, \ n\to \infty,$

and finally

$\prod^n_{k=1}\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)=\exp\left(\sum^n_{k=1}\log\left(1+\frac{1}{n}f\left(\frac{k}{n}\right)\right)\right)\to \exp\left(\int^1_0 f(x)dx\right),\ n\to\infty.$

Problem 3.

(a) Let $f: \mathbb{R}\to\mathbb{R}$ be a twice continuously differentiable function. Show that

$\lim_{h\to 0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}=f''(x)$

for all $x\in \mathbb{R}$ .

(b) Show that if $f$ further satisfies

$\frac{1}{2y}\int^{x+y}_{x-y}f(t)dt=f(x)$

for all $x\in \mathbb{R},y>0$ , then there exist $a,b\in \mathbb{R}$ such that $f(x)=ax+b$ for all $x\in\mathbb{R}$ .

Topic: Real Analysis, Difficulty Level: Medium

Full Solution

(a)

We use Taylor’s expansion of $f$ with the remainder in the integral form:

$f(x+h)=f(x)+hf'(x)+\int^{x+h}_x f''(y)(x+h-y)dy.$

Then

$\frac{f(x+h)+f(x-h)-2f(x)}{h^2}=$

$=\frac{1}{h^2}\bigg(f(x)+hf'(x)+\int^{x+h}_x f''(y)(x+h-y)dy+$

$+f(x)-hf'(x)+\int^{x-h}_x f''(y)(x-h-y)dy-2f(x)\bigg)=$

$=\frac{1}{h^2}\int^{x+h}_x f''(y)(x+h-y)dy+\frac{1}{h^2}\int^{x-h}_x f''(y)(x-h-y)dy=$

$=\frac{1}{h^2}\int^{x+h}_x (f''(y)-f''(x))(x+h-y)dy+\frac{1}{h^2}\int^{x-h}_x (f''(y)-f''(x))(x-h-y)dy+f''(x)$

We use continuity of second derivative: given $\epsilon>0$ there exists $\delta>0$ such that

$|z-x|\leq \delta \Rightarrow |f''(z)-f''(x)|\leq \epsilon.$

Let $|h|\leq \delta.$ Then

$\bigg|\frac{f(x+h)+f(x-h)-2f(x)}{h^2}-f''(x)\bigg|\leq$

$\leq \frac{1}{h^2}\bigg|\int^{x+h}_x (f''(y)-f''(x))(x+h-y)dy\bigg|+\frac{1}{h^2}\bigg|\int^{x-h}_x (f''(y)-f''(x))(x-h-y)dy\bigg|\leq$

since $|x+h-y|\leq |h|$ and $|x-h-y|\leq |h|$

$\leq \frac{2}{h^2}\int^{x+|h|}_x \epsilon |h|dy=2\epsilon.$

The convergence

$\lim_{h\to 0}\frac{f(x+h)+f(x-h)-2f(x)}{h^2}=f''(x)$

is proved.

(b)

Differentiate the identity

$\frac{1}{2y}\int^{x+y}_{x-y}f(t)dt=f(x)$

with respect to $y:$

$\frac{f(x+y)+f(x-y)}{2y}-\frac{1}{2y^2}\int^{x+y}_{x-y}f(t)dt=0.$

It follows that

$f(x+y)+f(x-y)=\frac{1}{y}\int^{x+y}_{x-y}f(t)dt=2f(x)$

and

$\frac{f(x+y)+f(x-y)-2f(x)}{y^2}=0.$

Letting $y\to 0$ we deduce $f''(x)=0.$ So $f(x)=ax+b.$

Problem 4.

Let $f:\mathbb{R}\to\mathbb{R}$ be twice continuously differentiable function. Show that if $f$ is bounded and $f''(x)\geq 0$ for all $x\in \mathbb{R}$ then $f$ must be constant.

Topic: Real Analysis, Difficulty Level: Medium

Full Solution

By assumption, the derivative of $f$ is increasing. Assume there exists $x_0\in \mathbb{R},$ such that $f'(x_0)>0.$ Then for all $x>x_0$

$f(x)=f(x_0)+\int^x_{x_0}f'(y)dy\geq f(x_0)+\int^x_{x_0}f'(x_0)dy=f(x_0)+(x-x_0)f'(x_0)\to \infty, \ x\to\infty.$

It contradicts the fact that $f$ is bounded.

Assume there exists $x_0\in \mathbb{R},$ such that $f'(x_0)<0.$ Then for all $x<x_0$

$f(x)=f(x_0)-\int^{x_0}_{x}f'(y)dy\geq f(x_0)-\int^{x_0}_xf'(x_0)dy=f(x_0)+(x-x_0)f'(x_0)\to \infty, \ x\to-\infty.$

Again it contradicts the fact that $f$ is bounded. We deduce that $f'(x)=0$ and $f$ is constant.

Problem 5.

Let $J$ be a $2\times 2$ real matrix such that $J^2=-I$ , where $I$ is the identity matrix

(a) Show that if $v\in\mathbb{R}^2$ and $v\neq 0$ , then the vectors $v,Jv\in\mathbb{R}^2$ are linearly independent

(b) Show that there exists an invertible $2\times 2$ real matrix $U$ such that

$U JU^{-1}=\begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}$

Topic: Multivariable Calculus, Difficulty Level: Medium

Full Solution

(a)

Assume there are scalars $\alpha,\beta$ such that

$\alpha v+\beta Jv=0.$

Apply $J$ to this equality:

$\alpha Jv-\beta v=0.$

From these two identities we get

$\alpha(\alpha v+\beta Jv)-\beta (\alpha Jv-\beta v)=(\alpha^2+\beta^2)v=0.$

(b)

Let $V$ be the matrix with the first column given by the vector $v$ and the second column given by the vector $Jv:$

$V=\begin{pmatrix} v & Jv \end{pmatrix}.$

$V$ is invertible, since $\{v,Jv\}$ is a basis of $\mathbb{R}^2.$ Let $U=V^{-1}.$ Then

$JV=\begin{pmatrix} Jv & J^2v\end{pmatrix} =\begin{pmatrix} Jv & -v\end{pmatrix}=\begin{pmatrix} v & Jv\end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}=V\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}$

So,

$UJU^{-1}=V^{-1}JV=\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}.$

Since $v\ne 0$ we get $\alpha^2+\beta^2=0,$ and $\alpha=\beta=0.$ Linear independence of $v$ and $Jv$ are proved.

Problem 6.

Suppose $V$ is a $3$ -dimensional real vector space and $T:V\to V$ is a linear map such that $T^3=0$ and $T^2\neq 0$ .

(a) Show that there exists a vector $v\in V$ such that the set $\{v,T(v),T^2(v)\}$ is a basis of $V$ .

(b) Suppose $S:V\to V$ is another linear map such that $S^3=0$ and $S^2\neq 0$ . Show that there exists an invertible linear map $U:V\to V$ such that $S=UTU^{-1}$ .

Topic: Linear Algebra, Difficulty Level: Medium

Full Solution

(a)

Let $v\in V$ be such that $T^2v\ne 0.$ Assume the system $\{v,T(v),T^2(v)\}$ is linearly dependent. Then there are scalars $a,b,c$ such that

$av+bT(v)+cT^2(v)=0.$

Apply $T$ to this identity and use the fact $T^3=0:$

$aT(v)+bT^2(v)=0.$

Apply $T$ to this identity and use the fact $T^3=0:$

$aT^2(v)=0.$

(b)

Let $v\in V$ be such that $\{v,T(v),T^2(v)\}$ is a basis of $V.$ Let

$W=\begin{pmatrix}v & T(v) & T^2(v) \end{pmatrix}$

be the matrix whose columns are given by vectors $v,T(v),T^2(v).$ The matrix $W$ is invertible. We have

$TW=\begin{pmatrix}T(v) & T^2(v) & 0 \end{pmatrix} = \begin{pmatrix}v & T(v) & T^2(v) \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0\\ 0 & 1 &0 \end{pmatrix} =W \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0\\ 0 & 1 &0 \end{pmatrix}$

It follows that

$W^{-1}TW=\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0\\ 0 & 1 &0 \end{pmatrix}$

for some invertible matrix $W$ .

If $S$ is another linear map of $V$ such that $S^2\ne 0$ and $S^3=0,$ then for some invertible matrix $Q$ we have

$Q^{-1}SQ=\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0\\ 0 & 1 &0 \end{pmatrix}$

So,

$Q^{-1}SQ=W^{-1}TW$

and

$S=UTU^{-1}$

for $U=QW^{-1}.$

But $T^2(v)\ne 0.$ So, $a=0$ and consequently $b=0, c=0.$ Linear independence of $\{v,T(v),T^2(v)\}$ is proved.

Problem 7.

Let $K$ be a field, and let $R$ be the ring $K[x]$ . Let $I\subset R$ be the ideal generated by $(x-1)(x-2)$ . Find all maximal ideals of the ring $R/I$ .

Topic: Ring Theory, Difficulty Level: Hard

Full Solution

Consider the mapping $f:K[x]\to K^2$ given by

$f(P(x))=(P(1),P(2)).$

$f$ is a ring homomorphism. The kernel of $f$ coincides with $I.$ Indeed,

$f(P(x))=0 \Leftrightarrow P(1)=P(2)=0 \Leftrightarrow P(x)=Q(x)(x-1)(x-2)\Leftrightarrow P\in I.$

Given $a,b\in K$ consider $P(x)=(b-a)x+(2a-b).$ Then

$f(P(x))=(P(1),P(2))=(a,b).$

So, $f$ maps $K[x]$ onto $K^2.$ By the isomorphism theorem, $P/I$ is isomorphic to $K^2$ . Consider a maximal ideal $J$ of $K^2.$ $J\ne \{(0,0)\}.$ Consider arbitrary nonzero $(a,b)\in J.$ Then

$(a,0)=(1,0)(a,b)\in J, \ (0,b)=(0,1)(a,b)\in J.$

If $a\ne 0$ and $b\ne 0,$ then $(1,0),(0,1)\in J$ and $J=K^2.$ In this case $J$ is not a maximal ideal. So, either

$J=\{0\}\times K$ or $J=K\times \{0\}.$

Conclusion. There are two maximal ideals in $R/I.$ One is generated by $(x-1),$ another is generated by $(x-2).$

Problem 8.

Let $G$ be a finite group, and let $H$ be a normal subgroup of $G$ , Let $P$ be a Sylow $p$ -subgroup of $H$

(a) Show that for all $g\in G$ , there exists $h\in H$ such that $gPg^{-1}=hPh^{-1}$ .

(b) Let $N=\{g\in G \mid gPg^{-1}=P\}$ . Let $HN$ be the set $HN=\{hn \mid h\in H,n\in N\}$ . Show that $G=HN$ .

Topic: Group Theory, Difficulty Level: Easy

Full Solution

(a)

Let $g\in G.$ $H$ is a normal subgroup of $G$ , so

$gPg^{-1}\subset gHg^{-1}\subset H.$

It follows that $gPg^{-1}$ is a subgroup of $H$ .

But $|gPg^{-1}|=|P|$ and $gPg^{-1}$ is a Sylow $p-$ subgroup of $H.$ By Sylow theorem, $P$ and $gPg^{-1}$ are conjugate in $H$ , i.e. there exists $h\in H$ such that

$gPg^{-1}=hPh^{-1}.$

(b)

Let $g\in G.$ There exists $h\in H$ such that

$gPg^{-1}=hPh^{-1}.$

Then

$(h^{-1}g)P(h^{-1}g)^{-1}=P.$

So,

$n=h^{-1}g\in N$

and

$g=hn\in HN.$

Equality $G=HN$ is proved.

To view ISI MMATH 2019 solutions, hints, and discussions: Click Here

To view ISI MMATH 2018 solutions, hints, and discussions: Click Here

To view ISI MMATH 2017 solutions, hints, and discussions: Click Here

To view ISI MMATH 2016 solutions, hints, and discussions: Click Here

To view previous year’s question papers: Click Here

5 Comments

Dhanraj Shrivastav

February 4, 2021 at 5:15 pm Reply

Here is a less elegant solution to Problem 4 which does not require integration:

We first show that $f''$ is identically zero. Suppose $f''(c)>0$ for some real number $c$ . Since $f''$ is continuous, there is an open interval $J$ containing $c$ such that $f''(x)>0$ for all $x \in J$ . Thus, $f'$ is strictly increasing on $J$ . Let $x_1,x_2 \in J$ with $x_1<x_2$ . Then $f'(x_1)<f'(x_2)$ .
By Taylor's Theorem, there is $d \in (x_1,x_2)$ such that

$f(x_2)-f(x_1)=(x_2-x_1)f'(x_1)+\frac{(x_2-x_1)^2}{2}f''(d).$

Noting that $f'(x_1)<f'(d)<f'(x_2)$ , \frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(x_1)+\frac{(x_2-x_1)}{2}f''(d)f'(d). $Hence$ f”(x)=0 $for every real number$ x $. This implies$ f(x)=Ax+B $where$ A,B $are real constants. Since$ f $is bounded, we must have$ A=0 $. Hence$ f$ is a constant function.
Dhanraj Shrivastav

February 18, 2021 at 8:41 pm Reply

I am a little confused about Problem 3 part (a). It can be solved using only L-Hopital’s Rule without using continuity of second derivative. Is there any reason that hypothesis was included?
- Annonymous
  
  May 2, 2021 at 7:47 pm Reply
  
  You are using $\lim_{h\to0} f''(x+h)=f''(x)$ which can be guranteed by continuity of $f''$
Anshika

May 2, 2022 at 7:12 pm Reply

Please solve CMI msc math previous year paper
- FractionsHub
  
  May 5, 2022 at 3:24 pm Reply
  
  Official Solutions has already been published by CMI.

ISI MMath Solutions and Discussion 2020 : PMB

ISI MMath PMB 2020 Subjective Questions, solutions and discussions

Sorry, did not get time to post the hints.

Question Paper ISI M.Math 2020 PMA Objective Questions Solutions and Discussions: Click Here

To view ISI MMATH 2019 solutions, hints, and discussions: Click Here

To view ISI MMATH 2018 solutions, hints, and discussions: Click Here

To view ISI MMATH 2017 solutions, hints, and discussions: Click Here

To view ISI MMATH 2016 solutions, hints, and discussions: Click Here

To view previous year’s question papers: Click Here

Problem 1.

Topic: Real Analysis, Difficulty level: Medium

Full Solution

Problem 2.

Topic: Real Analysis, Difficulty Level: Hard

Full Solution

Problem 3.

Topic: Real Analysis, Difficulty Level: Medium

Full Solution

Problem 4.

Topic: Real Analysis, Difficulty Level: Medium

Full Solution

Problem 5.

Topic: Multivariable Calculus, Difficulty Level: Medium

Full Solution

Problem 6.

Topic: Linear Algebra, Difficulty Level: Medium

Full Solution

Problem 7.

Topic: Ring Theory, Difficulty Level: Hard

Full Solution

Problem 8.

Topic: Group Theory, Difficulty Level: Easy

Full Solution

To view ISI MMATH 2019 solutions, hints, and discussions: Click Here

To view ISI MMATH 2018 solutions, hints, and discussions: Click Here

To view ISI MMATH 2017 solutions, hints, and discussions: Click Here

To view ISI MMATH 2016 solutions, hints, and discussions: Click Here

To view previous year’s question papers: Click Here

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