User:FUNG Cheok Yin/proof(i) A127793

Let $R=e_{1,1}+\sum _{i=2}^{\infty }\sum _{p=0}^{\infty }(e_{2^{p}i,i}\lceil i/2\rceil -e_{2^{p}(i+1),i}\lceil i/2\rceil )$

where $e_{i,j}$ is the matrix unit.

Let the matrix involved written as $A=\sum _{i}\sum _{j}e_{i,j}{\frac {1}{\lceil i/2\rceil }}\kappa _{i,j}$, where $\kappa _{i,j}=1$ if ${\frac {i}{2}}<j\leq i$, 0 if $j\leq {\frac {i}{2}}$ or $j>i$.

To prove that $R$ is the right inverse of $A$, we are going to show $\sum _{k}=a_{i,k}r_{k,j}=\delta _{i,j}$

Since $a_{1,k}=0$ for all $k>1$: for $i=1$, $j=1$, we have $a_{1,1}r_{1,1}=1\cdot 1=1$ ; for $i>1$, $j=1$, we have $a_{i,1}r_{1,1}=0\cdot 1=0$ .

For $i\geq 1$, $j>1$,

$\sum _{k}a_{i,k}r_{k,j}$

$=\sum _{k=1}^{i}{\frac {1}{\lceil i/2\rceil }}\kappa _{i,k}r_{k,j}$

$=\sum _{k=\lceil (i+1)/2\rceil }^{i}{\frac {1}{\lceil i/2\rceil }}r_{k,j}$

We list out the terms which will be OR MAY be involved in the above summation:

$\lceil j/2\rceil e_{j,j},-\lceil j/2\rceil e_{j+1,j},$

$\lceil j/2\rceil e_{2j,j},-\lceil j/2\rceil e_{2(j+1),j},$

$\lceil j/2\rceil e_{4j,j},-\lceil j/2\rceil e_{4(j+1),j},$

...

$\lceil j/2\rceil e_{2^{p-1}j,j}(A),-\lceil j/2\rceil e_{2^{p-1}(j+1),j}(C),$

$\lceil j/2\rceil e_{2^{p}j,j}(D),-\lceil j/2\rceil e_{2^{p}(j+1),j}(B)$

It is obvious that the first $p-2$ lines cancelled out.

It remains to determine when A, B, C and D will be involved in the summation.

Consider the inequality $2^{p-1}j\leq \lceil {\frac {i+1}{2}}\rceil <2^{p}j$ (*) :

If $2^{p-1}j=\lceil {\frac {i+1}{2}}\rceil$:

(that is, A is in the summation)

for $i$ even

$2^{p-1}j={\frac {i}{2}}+1$, $2^{p}j=i+2>i$ ;

for $i$ odd

$2^{p-1}j={\frac {i+1}{2}}$, $2^{p}j=i+1>i$.

We can see that C will be in the summation, but neither D nor the even larger B.

If $2^{p-1}<\lceil {\frac {i+1}{2}}\rceil$:

(that is, A is not in the summation)

we separate into two cases again. (I) $i<2^{p}j$ and (II) $2^{p}j\leq i$ .

(I):

Solve $i<2^{p}j$ and $2^{p-1}<\lceil {\frac {i+1}{2}}\rceil$ .

If $i$ is odd, $2^{p-1}j<{\frac {i+1}{2}}$

$2^{p}j<i+1$

$2^{p}j-1<i<2^{p}j$

Impossible for an integer $i$.

If $i$ is even,

$2^{p-1}j<{\frac {i}{2}}+1$

$2^{p}<i+2$

$2^{p}j-2<i<2^{p}j$

$i=2^{p}j-1$

but $i$ is even, so $p=0$, thus $j=i+1$ .

Then none of A, C, D, or B will be in the summation.

(II):

If $2^{p-1}(j+1)\geq \lceil {\frac {i+1}{2}}\rceil$,

(that is, C is in the summation)

then $2^{p}(j+1)\geq \lceil {\frac {i+1}{2}}\rceil \times 2>i$

$2^{p-1}(j+1)=2^{p-1}j+2^{p-1}<2^{p-1}j+2^{p-1}j\leq i$, since $j>1$.

Therefore we can conclude B is not in the summation.

D will be in the summation because $2^{p}j\leq i$ (II) and $\lceil {\frac {i+1}{2}}\rceil <2^{p}j$ (from (*) ).

Then C and D cancel out.

If $2^{p-1}(j+1)<\lceil {\frac {i+1}{2}}\rceil$,

(that is, C is not in the summation)

(i) Consider $p\neq 0$,

we will show $i>2^{p}(j+1)>2^{p}(j)>\lceil {\frac {i+1}{2}}\rceil$.

(That is, both D and B are in the summation.)

The middle greater than sign is obvious; the rightmost sign comes from (*); it remains to show the leftmost sign is valid.

If $i$ is even,

${\frac {i}{2}}=\lceil {\frac {i+1}{2}}\rceil >2^{p-1}(j+1)$

$i>2^{p}(j+1)$

If $i$ is odd,

${\frac {i-1}{2}}\geq 2^{p-1}(j+1)$

$i-1\geq 2^{p}(j+1)$

$i>2^{p}(j+1)$

(ii) Consider $p=0$,

D = $2^{p}j=1\cdot j=j$ .

Since ${\frac {j+1}{2}}<\lceil {\frac {i+1}{2}}\rceil$,

$i\geq j>\lceil {\frac {i+1}{2}}\rceil$,

D is in the summation.

B = $2^{p}(j+1)=j+1$.

Since ${\frac {j+1}{2}}<\lceil {\frac {i+1}{2}}\rceil$,

$j+1<\lceil {\frac {i+1}{2}}\rceil \cdot 2$:

if $i$ is even, $j+1<i+2$,

$j<i+1$

$j\leq i$

$j+1\leq i+1$

B will be in the summation (and cancel D out) except the edge case $i=j$;

if $i$ is odd,

$j+1<i+1$

B will be in the summation (and cancel D out) except the edge case $i=j$. $\square$

First Draft Finished on Sep 12 2022 (The first draft is neither well written nor concise, comments are welcome.)