ISI MMath Solutions and Discussion 2020 : PMB

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.

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3 Comments

  • 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).Hencef”(x)=0for every real numberx. This impliesf(x)=Ax+BwhereA,Bare real constants. Sincefis bounded, we must haveA=0. Hencef$ is a constant function.

  • 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?

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