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# Problem Set I

In this problem set we present three questions.

(High School Students) Four squares, all of different areas, are cut from a rectangle, leaving a smaller rectangle of dimensions $1\times2$.  If the largest square has area 64, and the other three squares have side lengths that are whole numbers no larger than 7, what are their areas?

(Undergraduate Students) Given $f\in C([0,\infty ))$ such that $f(x)\rightarrow 0$ as $x\rightarrow \infty$ show that for any $\epsilon >0$ there is a polynomial $p$ such that

$\left\vert f(x)-e^{-x}p(x)\right\vert <\epsilon \,\,\, \forall x\in \lbrack 0,\infty ).$

(Masters Students and above) If $K$ is a compact subset of $\mathbb{R}^{n}$ show that the set

$A=\{x\in \mathbb{R}^{n}: d(x,K)=1\}$

has Lebesgue measure $0.$

## 19 thoughts on “Problem Set I”

1. Manjunath Krishnapur says:

For the UG problem: Consider $C_0$, the Banach space of continuous functions on $[0,\infty)$
that vanish at infinity (with sup-norm). Let $W$be the subspace of all
$p(x)e^{-x}$. If closure(W) is not the whole space, there is a non-zero
element in the dual space that vanishes on $W$. But the dual space is all
finite Borel measures on $[0,\infty)$ and vanishing on W means

$\int x^n e^{-x}d\mu(x)=0$

for all $n$. But the measure $d\nu(x)=e^{-x}d\mu(x)$ has finite
Laplace transform

$\int e^{ax}d\nu(x) < \infty$

for $a \leq 1$ and hence, if
all its moments vanish, then the measure is zero, implying that $\mu=0.$

• Neat solution even though the results used may not be typically covered in an undergraduate course. Attached below is perhaps a (slightly more) detailed proof of your idea.

Consider the Banach space of all continuous functions on $0,\infty )$ vanishing at $\infty$ with the supremum norm. We have to show that the
subspace $\{e^{-x}p(x):p$ is a polynomial$\}$ is dense in this space. If
not, then there is a continuous linear functional which vanishes on this
subspace but not everywhere. By Riesz Representation Theorem there is a real
measure $\mu$ such that $\int e^{-x}p(x)d\mu (x)=0$ for every $p$ but $\mu \neq 0.$ Writing $\mu$ as $\mu _{1}-\mu _{2}$ where $\mu _{1}$ and $\mu _{2}$ are positive finite measures we have $\int e^{-x}p(x)d\mu _{1}(x)=\int e^{-x}p(x)d\mu _{2}(x).$ Let $d\nu _{1}=e^{-x}d\mu _{1}$ and $d\nu _{2}=e^{-x}d\mu _{2}.$ Then $\int p(x)d\nu _{1}(x)=\int p(x)d\nu _{2}(x)$
for every polynomial $p$ but $\nu _{1}\neq \nu _{2}.$ Let $\phi _{j}(z)=\int e^{zt}d\nu _{j}(t),j=1,2.$ These functions are holomorphic in $\{\mbox{Re}z < 1\}.$ Also $\phi _{1}^{(n)}(0)=\phi _{2}^{(n)}(0)$ $\forall n\geq 0.$ From the power series expansion of these two function in $\{z:\left\vert z\right\vert <1\}$ it follows that they coincide on this
ball, hence on $\{\mbox{Re}z<1\}.$ In particular they coincide on the
imaginary axis which means $\int e^{ist}d\nu _{j}(t)=\int e^{ist}d\nu _{j}(t)$ $\forall s\in \mathbb{R}.$ This is a contradiction.

In our next post we present a solution based on undergraduate mathematical material.

2. Stotsi says:

Suppose $x_0\in A$. Let $B := \{ d(x,K) < 1\}$.

Observe that since $K$ is compact, there exists $y_0\in K$ such that $d(x_0,y_0) = 1$. By definition $B_1(y_0) = \{x: d(x,y_0) < 1\}$ is a subset of $B$.

This implies that for all $\epsilon \leq 1/2$, we have that

$\mu(B_\epsilon(x_0) \cap B) \geq \frac1{2^n} \mu(B_\epsilon(x_0))$

where $\mu$ is the Lebesgue measure. (The factor $1/2^n$ is very loose: Since a sphere tangent to $x_0$ is contained in $B$, locally an orthant centered at $x_0$ is contained in $B$.)

Hence we have that for every $x_0\in A$,

$\limsup_{\epsilon \to 0} \frac{\mu(B_\epsilon(x_0) \cap A)}{\mu(B_\epsilon(x_0))} \leq \frac{2^n-1}{2^n} < 1$

By the Lebesgue differentiation theorem, the set of $x_0\in A$, for $A$ measurable, such that the above condition holds must be measure zero. Hence $A$ has measure zero.

• The proof seems correct. With this only Undergraduate problem remains to be solved.

3. Sayan Das says:

A (hopefully correct) solution of problem 3:
Let S_{\alpha} ={x: d(x,K)= \alpha} where \alpha >0. Clearly each S_{\alpha} is just \alpha. S_{1}. So, these sets all have the same measure. Also, since each S_{\alpha} is compact (as K is compact), they have finite measure. If measure is nonzero, then we get a contradiction as S_{1-1/n} are all disjoint, and union of these sets should have finite measure, since the union is a bounded set (contained in the set whose dist. from K is less than or equal to 1). So S_{1} has measure 0.

• Thanks for submitting and trying the problems. The above solution is not correct as well.

4. Sayan Das says:

Solution of problem 3: Let S= {x: d(x,K)=1}. S is clearly measurable. Let x \in S. Then, there exists k \in K such that d(x,k)=1. k has to be in boundary of K. (Consider the line joining k and x. This line touches the boundary at b, then b is closer to x than k). So, S is really a “translate” of the boundary of K. But boundary of K has measure zero, as Lebesgue measure is regular. Also, Lebesgue measure is translation invariant, which implies S has measure 0.

5. Mihir Vilas Deo says:

The solution for 1st question
The big squre is of side 8u therefore for the original rectangle the small side of rectangle is 8u
then let the 8+x be the bigger side
then after removal of big squre 3 squres and 1 rectangle of dim 1 by 2
hence (8*x)-2=sum of three squares {here * means multiplication}
and sides are less than 7
and 8*5-2=38
but 38=25+4+9 ie sum of squares of 2 3 and 5
hence the areas of rest 3 squares are 25squ.9squ and 4squ.
The total area of origianl rectangle is 104 squ

• Nishith Mohan says:

By Maclaurins Theorem f(x)=f(0)+ xf^’ (0)+x^2/2! f^” (0)+ ………….
e^(-x)=1-x+x^2/2!-x^3/3!+⋯
-ε≤f(x)- e^(-x) p(x)≤ε
p(x)≤(f(0)+ xf^’ (0)+x^2/2! f^” (0)+ ………….+ ε)/(1-x+x^2/2!-x^3/3!+⋯)

As ε is a constant So,
p(x)≤(f(0)+ ε+ xf^’ (0)+x^2/2! f^” (0)+ ………….+ )/(1-x+x^2/2!-x^3/3!+⋯) = p(x)≤ (c+xf^’ (0)+x^2/2! f^” (0)+ …………)/(1-x+x^2/2!-x^3/3!+⋯)
Similarly
p(x)≥(f(0)+ xf^’ (0)+x^2/2! f^” (0)+ ………….- ε)/(1-x+x^2/2!-x^3/3!+⋯)
p(x)≥(c^’+xf^’ (0)+x^2/2! f^” (0)+ ………….)/(1-x+x^2/2!-x^3/3!+⋯)
As both are in polynomial form then p(x) exists.

• f is just a continuous function vanishing at infinity. It need have a Taylor series expansion even in a small neighbourhood of 0. The emphasis in this problem is on unform approximation on teh infinite interval [0,infinity)

• Perhaps another way to answer this question is to think of what are the possible areas of the three squares that are cut-out. Then use graph paper- to check out the various possibilities; after which you can proceed to write a methodical proof.

• One way to start is to identify the possible answers. They are 1,4,9,16, and three other numbers. Once you have the list then you can come to conclusion.

6. Soumya Sinha Babu says:

Oh , Ok . So I need to show that the functions ‘ $e ^{( - k x ) }$‘ , $k \in N$ ( natural numbers ) , are in the closure of A . But that can be shown by taking , ( say for k = 2 )

$g_n ( x )$ = The n – th Order Taylor Expansion of $e^{( - x )}$,

and showing that

$| e^{( - 2 x )} - e^{( - x )} g_n( x ) | \rightarrow 0$

as $n\rightarrow \infty$.

Thus , we can complete the argument above .

Sorry for some typo – s in reply 1 , sometimes I have written ‘ $e^{( x )}$ ‘ in place of ‘$e^{( - x )}$ ‘.

• The above will not lead to a solution to Problem 2. Sorry.

• Kavi Rama Murthy says:

You have come to an interesting point in the proof. Try to prove that the approximation of e^(-2x) works uniformly on [0,infinity). PS: I still don’t understand what algebra you are considering.

7. I propose the following solution to Problem 2 .

I use ‘ l ‘ to denote ” infinity ” , and $X = C [ 0 , l )$. ” E ” means ” belongs to ” .

Let

$A = \{ \frac{p ( x )}{ e ^ {( - x ) }} : ' p ' \mbox{ is a polynomial } \}.$

We have to show that A is dense in X . Since X is a locally compact Hausdorff space , we use the Stone – Weierstrass Approximation Theorem for X . Obviously , A is a sub-algebra of X . We only need to check whether A vanishes nowhere and A separates points .

Given any ‘ a ‘ E $latex[ 0 , l )$ , we can always define ,

$f ( x ) = \frac{( x + a )}{ e^{(- x )}}$

So that ‘ f ( a ) ‘ does not equal 0 . Therefore , there does not exist any ‘ a ‘ E [ 0 , l ) such that for every ‘ f ‘ E X , f ( a ) = 0 . From this observation , we ascertain that A vanishes nowhere .

Define ,

$g ( x ) = \frac{1}{e^{( - x )}} = e^{( x )} .$

Clearly , g E X .

But given two unequal numbers ‘ x ‘ and ‘ y ‘ , we always have ,

$| g ( x ) - g ( y ) | \geq 0 .$

Hence , A separates points also . Now , by Stone – Weierstrass Approximation Theorem , A is dense in X ,

QED .

• Dear Soumya,
Your set A is not an algebra. It is not closed under multiplication. Anyway it is nice that you are trying to solve this interesting problem. Good luck!
Rama Murthy