# Maximum arrangements of nonattacking kings on the $2n\times 2n$ chessboard

Tricia Muldoon Brown
Georgia Southern University

###### Abstract

To count the number of maximum independent arrangements of $n^{2}$ kings on a
$2n\times 2n$ chessboard, we build a $2^{n}\times(n+1)$ matrix whose entries
are independent arrangements of $n$ kings on $2\times 2n$ rectangles.
Utilizing upper and lower bound functions dependent of the entries of the
matrix, we recursively construct independent solutions, and provide a
straight-forward formula and algorithm.

Keywords: chess, independence, kings, nonattacking
MSC: 05A15, 05-04

##  1 Introduction

The problem of finding and counting the number of independent, also called
nonattacking, arrangements of pieces on a chessboard is a long-established
problem in mathematics and computer science. Many variations have been studied
including modifications of traditional pieces and of board size and shape. In
particular, we will be concerned here with maximum independent arrangements on
a square chessboard where arrangements of pieces are always fixed, that is,
rotations and reflections are considered distinct. For some traditional chess
pieces, bishops, rooks, knights, and pawns, both the size of a maximum
independent set and the number of distinct maximal independent arrangements
are known. For the other pieces, kings and queens, only partial results are
known. For those interested in these kinds of questions, books by Dudeney [6],
Kraitchik [9], Madachy [11], Watkins [14], and Yaglom and Yaglom [16] provide
good references, and Table 1 illustrates the currently known enumerative
results.

| Kings | Queens | Bishops | Knights | Rooks | Pawns
---|---|---|---|---|---|---
Number of pieces |  |  |  |  |  |
in a maximum | $\big{\lceil}\frac{n}{2}\big{\rceil}^{2}$ | $n$ | $2n-2$ | $\big{\lceil}\frac{n}{2}\big{\rceil}^{2}+\big{\lfloor}\frac{n}{2}\big{\rfloor}^{2}$ | $n$ | $n\big{\lceil}\frac{n}{2}\big{\rceil}$
arrangement |  |  |  |  |  |
Number of | 1 if $n$ is odd, | unknown if |  | 1 if $n$ is odd, |  | 2 if $n$ is odd,
maximum | unknown if | $n>27$ | $2^{n}$ | 2 if $n$ is even | $n!$ |  ${n\choose n/2}^{2}$ if $n$ is even
arrangements |  $n>26$ is even |  |  |  |  |

Table 1: Enumerative results for maximum independent sets placed on an
$n\times n$ chessboard where $n>1$

For the still-open questions, it is known that the maximum number of
nonattacking queens on an $n\times n$ chessboard is $n$, but counting the
number of these arrangements is a famously difficult question as can be the
many variations on this problem. (See a survey paper by Bell and Stevens [2]
on this topic.) However, here we wish to consider questions of maximum
arrangements of nonattacking kings. First, the maximum size is easily found.
We observe that every $2\times 2$ square on the chessboard may contain at most
one king, so the upper bound on the number of kings is
$\lceil\frac{n}{2}\rceil^{2}$. But this number can be achieved for both odd-
and even-length boards by alternating kings in both rows and columns, as shown
in Figure 1.

Figure 1: Maximum arrangements of 16 kings on $7\times 7$ and $8\times 8$
chessboards

Determining the number of distinct maximum independent arrangements in the
case that $n$ is odd, is also straight-forward. The arrangement placing kings
in alternating positions and alternating rows as illustrated in Figure 1 is
unique. To see this, observe that every $2\times n$ strip and every $n\times
2$ strip must contain exactly $\lceil\frac{n}{2}\rceil$ kings in order to
achieve the maximum. Because the kings necessarily must at least be placed in
alternating rows or columns, the only ways to do this is to place a king in
the top row or leftmost column and alternate to the bottom row or rightmost
column. Thus kings must appear in all four corners and in alternating rows and
columns.

Counting maximum arrangements of independent kings on an even-length $2n\times
2n$ chessboard is not so simple as there is more latitude for king placement
in this case. Figure 2 illustrates some of these arrangements. As we will now
only be considering even-length boards, note that for the rest of the
discussion $n$ will always refer to the half-length of the even-length
chessboard.

Figure 2: Examples of maximum arrangements of nonattacking kings on
chessboards with even-lengths 6, 8, and 10

The question of enumerating maximum independent arrangements of kings on the
even-length $2n\times 2n$ chessboard has been studied by Knuth [7] in an
unpublished work and later by Wilf [15], and then Larsen [10] who gives an
asymptotic approximation. Entry A018807 in the Online Encyclopedia of Integer
Sequences [13] contributes counts up to $n=26$, and Kotěšovec [8] also
provides enumerative results in his extensive book on independent chessboard
arrangements. Variations have been studied by Calkin et. al. [4, 5] who
consider all non-maximum arrangements, while Abramson and Moser [1] count
arrangements of $n$ independent kings with exactly one king in each row and
column.

For the original question, Knuth’s technique was to partition the chessboard
into $2\times 2$ squares in such a way that would generate two vectors of
integers. He then sets up a graph where directed paths with certain
restrictions could be used to find maximum independent kings solutions. Wilf’s
strategy is to also partition the chessboard into $2n\times 2$ columns, but
then uses recursive matrix multiplication of the transfer matrix to identify
solutions. Algorithmically, this requires construction of the transfer matrix
and repeated matrix multiplication on the $2^{n}(n+1)\times 2^{n}(n+1)$
matrix. Here, we will use a smaller $2^{n}\times(n+1)$ matrix and apply
techniques similar to those used in an earlier work by the author [3] on
arrangements of independent pawns to find a formula to enumerate the number of
arrangements of $n^{2}$ kings. The complexity is still exponential, but
smaller with a worst-case estimate of $O(3^{n}n^{3})$.

##  2 Counting nonattacking arrangements

Let us begin similarly to Wilf [15] by constructing a matrix which we will
call $M_{2n}$ whose entries are all possible arrangements of $n$ independent
kings on a $2\times 2n$ rectangle. First, given such $2\times 2n$ rectangle,
we partition the rectangle into $n$ $2\times 2$ squares. From left to right,
index each of these squares with the integers $1,2,\ldots,n$. We note, each
$2\times 2$ square must contain exactly one king so that the set of kings
placed on the rectangle is both non-attacking and maximum. To determine the
placement of the king, we need to know if the king is in the top or bottom row
of the square and if the king is in the left or right column. Thus we
associate any $2\times 2n$ arrangement with a subset $A\subseteq[n]$ and an
index $1\leq k\leq n+1$ where $A$ equal the set of indices of the $2\times 2$
squares that have their king in the top row of the square and $k$ is the index
of the square such that all squares with index less than $k$ have a left king
and all squares with an index greater than or equal to $k$ have a right king.
We use the $2^{n}$ subsets to index the rows of the matrix $M_{2n}$ and the
values $1\leq k\leq n+1$ to index the columns. The $2\times 2n$ rectangles are
arranged from left to right in increasing order by the number of kings which
appear in the left column of their $2\times 2$ square in the partition of the
rectangle. Specifically, the column index is the index of the leftmost square
containing a king in the right column, and also is one plus the number of such
kings appearing in a left column. See Figure 3 for an example of this matrix
in the case $n=3$.

Figure 3: Entries in the matrix $M_{6}$

######  Lemma 1.

Every independent arrangement of $n$ kings on a $2\times 2n$ rectangle appears
exactly once as an entry in the matrix $M_{2n}$.

###### Proof.

By construction, the set of matrix entries is contained in the set of
independent $n$-kings arrangements on $2\times 2n$ rectangles. Further the
matrix entries encompass all ways to places kings in the top or bottom row of
each rectangle and all the ways to place kings in left or right columns so
that all right kings appear to the left of all left kings. Any arrangement
that does not appear as a matrix entry would necessarily then have a right
king appearing to the right of a left king, but this contradicts independence.
Thus we have arrangement all possible arrangements of non-attacking kings on a
$2\times 2n$ rectangle appearing in exactly one entry of the
$2^{n}\times(n+1)$ matrix $M_{2n}$. ∎

Our goal now is to vertically concatenate these rectangular arrangements to
form allowable independent solutions. The first observation is that for such a
concatenation to be independent, the row index of the upper rectangle, must
contain the row index of the lower rectangle.

######  Lemma 2.

Given an independent arrangement of $n$ kings on the rectangle indexed by
$(A,k)$ where $A\subseteq[n]$ and $1\leq k\leq n+1$, any independent
arrangement on another $2\times 2n$ rectangle that can be concatenated below
to form an independent arrangement of size $4\times 2n$ must be indexed by $B$
where $B\subseteq A$.

###### Proof.

To the contrary, if $B\nsubseteq A$, then there exists an index $i$ that is in
$B$ and is not in $A$. This indicates a king in the top row of the strip
indexed by $B$ and a king in the bottom row of the strip indexed by $A$ in
square $i$. No matter the placement, left or right, for each king, the two
kings may attack. ∎

This condition is necessary but not sufficient. For example the strip in
position $(\\{2\\},2)$ may not be placed independently under the strip indexed
by $(\\{1,2\\},4)$ despite $\\{2\\}\subseteq\\{1,2\\}$. This is because the
king in the top row of strip $(\\{2\\},2)$ in square $2$ has been pushed to
the right, so it conflicts with the king in the bottom row of the strip
$(\\{1,2\\},4)$ in square $3$. We need a condition on the left and right
positions of the kings. To find this, we define two functions dependent on
$A$, $B$, and $k$.

######  Definition 1.

  1. 1.

Let $p(A,B,k)=\max\\{\\{i\in[n+1]:i<k,i\notin A,\mbox{ and }i-1\in
B\\}\cup\\{1\\}\\}$.

  2. 2.

Let $q(A,B,k)=\min\\{\\{i\in[n+1]:i>k,i\in B,\mbox{ and }i-1\notin
A\\}\cup\\{n+1\\}\\}$

######  Proposition 1.

Given a $2\times 2n$ arrangement of $n$ independent kings indexed by $(A,k)$,
the $2\times 2n$ independent arrangements that maybe independently
concatenated below $(A,k)$ are exactly those which are indexed by $(B,i)$
where $B\subseteq A$ and $p(A,B,k)\leq i\leq q(A,B,k)$.

###### Proof.

Lemma 2 forces $B\subseteq A$. Then, note that locally for all $1<j\leq n+1$
the concatenation of two strips indexed by $A$ and $B$ will never have a king
in square $j$ of $(A,k)$ and a king in square of $j$ of $(B,k)$ attacking each
other, because either both kings are in the top row, both are in the bottom
row, or the king in $(A,k)$ is in the top row while the king in $(B,i)$ is in
the bottom row. There are however two ways for a concatenation of rectangular
strips $(A,k)$ and $(B,i)$ to have attacking kings in squares indexed some
$j-1$ and $j$; either there is a right king in the top row of square $j-1$ in
$(B,i)$ and a left king in the bottom row of square $j$ in $(A,k)$, or there
is a right king in the bottom row of square $j-1$ in $(A,k)$ and a left king
in the top row of square $j$ in $(B,i)$. (See Figure 4.)

|  | i |  | k |  |  |
---|---|---|---|---|---|---|---
$\cdots$ | L | L | L | R | R | R | $\cdots$
$\cdots$ | L | R | R | R | R | R | $\cdots$
(a) Type 1 attacking kings

|  |  | k |  | i |  |
---|---|---|---|---|---|---|---
$\cdots$ | L | L | R | R | R | R | $\cdots$
$\cdots$ | L | L | L | L | R | R | $\cdots$
(b) Type 2 attacking kings

Figure 4: Possible attacks by arrangements of independent kings on two
$2\times 2n$ rectangles

Let us first consider how we can avoid the situation of an attack by a left
king in the fixed strip $(A,k)$ and a right king in strip $(B,i)$ for some
$i$. We note, the largest value for $i$ that could generate a pair of
attacking kings is $i=k-2$, as we can see from Figure 4(a), because for any
larger values there cannot be a left king in $(A,k)$ that is immediately to
the left of a right king in $(B,i)$. So we will have an attacking pair in this
case if the king in $(B,i)$ is in the top row and the king in $(A,k)$ is in
the bottom row, that is, if

  1. i.

$i<k-1$,

  2. ii.

$i\in B$, and

  3. iii.

$i+1\notin A$.

In particular, we look to the largest index over all values $i$ satisfying the
criteria above, that is, over all $i$ where there is a pair of attacking kings
of Type 1, because every index less than or to the left of this maximum would
still have the attacking pair of kings with one king in the square indexed by
that maximum and the other directly to the right. This is because moving to
the left in the matrix will only increase the number of right kings in the
arrangement indexed by $B$. We can then choose the index which is one greater
than the maximum as the starting value for which all strips indexed by $B$ and
any index greater than or equal to this maximum plus one will not have a Type
1 conflict. If no such value exists, then all arrangements in row $B$ will not
have a conflict with the arrangement $(A,k)$ and we set our lower bound to
$1$. Using a change of variable where we send all $i+1$ to $i$, we can say the
lower bound for non-attacking arrangements is:

$p(A,B,k)=\max\\{\\{i\in[n+1]:i<k,i\notin A,\mbox{ and }i-1\in
B\\}\cup\\{1\\}\\}.$

Similarly, we need to look at the situation of an attack by a right king in
$(A,k)$ by a left king in $(B,i)$ for some $i$. In Figure 4(b), we can see
that the smallest value for an attacking pair would be $i=k+2$. So to have an
attacking pair of the second type, we need

  1. i.

$i>k+1$,

  2. ii.

$i-1\in B$, and

  3. iii.

$i-2\notin A$.

After determining the minimum of these values $i$ where have pairs of
attacking kings of Type 2, we set the upper bound to the minimum minus one as
the last possible non-attacking arrangement when we move from left to the
right across a row of $M_{2n}$ increasing the number of queens in left
columns. If no such value exists, then all arrangements of $B$ have no
attacking pairs of the second type, and we can set the upper bound to $n+1$.
Again applying a change of variable sending all $i-1$ to $i$, we have

$q(A,B,k)=\min\\{\\{i\in[n+1]:i>k,i\in B,\mbox{ and }i-1\notin
A\\}\cup\\{n+1\\}\\}.$

∎

We illustrate these functions with the following example.

######  Example 1.

Let $n=7$. Let $A=\\{1,2,5,7,8\\}$, $B=\\{1,2,5,7\\}$ and $k=4$. We compute
the upper and lower bounds.

$\displaystyle p(A,B,4)$ $\displaystyle=\max\\{\\{i:i\notin A,i<4\mbox{ and
}i-1\in B\\}\cup\\{1\\}\\}$ $\displaystyle=\max\\{\\{3\\}\cup\\{1\\}\\}=3$
$\displaystyle q(A,B,4)$ $\displaystyle=\min\\{\\{i:i\in B,i>4\mbox{ and
}i-1\notin A\\}\cup\\{n+1\\}\\}$
$\displaystyle=\min\\{\\{5,7\\}\cup\\{8\\}\\}=5.$

Thus arrangements indexed by $B$ that may be concatenated below $(A,4)$ are
exactly $(B,3)$, $(B,4)$, and $(B,5)$. See Figure 5.

Figure 5: Example of independent arrangements $(\\{1,2,5,7\\},i)$ which may be
concatenated independently below the independent arrangement
$(\\{1,2,5,7,8\\},4)$

Now we are ready for the main result.

######  Theorem 1.

Define a sequence of matrices $M^{\ell}$ with entries $\\{m_{A,k}^{\ell}\\}$
for $1\leq\ell\leq n$, $A\subseteq[n]$, and $1\leq k\leq n+1$ such that
$m^{1}_{A,k}=1$ for all pairs $(A,k)$ and for $1<\ell\leq n$

$m^{\ell}_{A,k}=\sum_{B\subseteq
A}\sum_{i=p_{(}A,B,k)}^{q_{(}A,B,k)}m^{\ell-1}_{B,i}.$

Then $K(n)$, the number of maximum arrangements of nonattacking kings on the
$2n\times 2n$ chessboard, is

$K(n)=\sum_{A\subseteq[n]}\sum_{k=1}^{n+1}m^{n}_{A,k}.$

###### Proof.

The sum of the entries of the matrix $M^{1}$ gives the number of nonattacking
arrangements of $n$ kings on a $2\times 2n$ strip as each of the entries
corresponds bijectively with a specific strip indexed by a subset
$A\subseteq[n]$ and a value $1\leq k\leq n+1$. Because $p(A,B,k)$ and
$q(A,B,k)$ provide bounds for all valid arrangement which may be independently
concatenated below $(A,k)$, by Proposition 1 the double sum

$m^{2}_{A,k}=\sum_{B\subseteq A}\sum_{i=p_{(}A,B,k)}^{q_{(}A,B,k)}m^{1}_{B,i}$

counts all valid $4\times 2n$ maximum arrangements of nonattacking kings where
the top $2\times 2n$ arrangement is $(A,k)$, that is, the entries
$m_{A,k}^{2}$ correspond to the number of independent arrangements on $2\times
2n$ rectangles which may be concatenated below the arrangement $(A,k)$ while
preserving independence. Their sum thus is the total number of nonattacking
arrangements of $2n$ kings on a $4\times 2n$ strip. As this process is
repeated, we see that a $2\ell\times 2n$ independent arrangement is found by
concatenating a single independent arrangement on a $2\times 2n$ rectangle
$(A,k)$ above and existing independent arrangement of the $2(\ell-1)\times 2n$
strip so that independence is also preserved along this concatenation.
Inductively, we know the number of $2(\ell-1)\times 2n$ arrangements with
arrangement $(B,i)$ at the top is $m_{B,i}^{\ell-1}$, and as in the case where
$\ell=2$ we utilize the upper and lower bound functions $p$ and $q$ to sum
over precisely the arrangements which can be concatenated independently ∎

######  Example 2.

Let $n=3$.

$m^{1}=\begin{bmatrix}1&1&1&1\\\ 1&1&1&1\\\ 1&1&1&1\\\ 1&1&1&1\\\ 1&1&1&1\\\
1&1&1&1\\\ 1&1&1&1\\\ 1&1&1&1\\\ \end{bmatrix}\quad
m^{2}=\begin{bmatrix}32&32&32&32\\\ 12&16&16&16\\\ 14&14&14&14\\\
16&16&16&12\\\ 7&7&8&8&\\\ 6&8&8&6\\\ 8&8&7&7\\\ 4&4&4&4\\\ \end{bmatrix}\quad
m^{3}=\begin{bmatrix}408&408&408&408\\\ 88&134&134&134\\\ 110&110&110&110\\\
134&134&134&88\\\ 38&38&46&46\\\ 30&44&44&30\\\ 46&46&38&38\\\ 16&16&16&16\\\
\end{bmatrix}$ $K(3)=\sum_{A\subseteq[3]}\sum_{k=1}^{4}m^{3}_{A,k}=3600.$

Of course, if we want to enumerate arrangements on a $2m\times 2n$ rectangular
chessboard with $m\leq n$, we can simply utilize matrices $M^{m}$. We have the
following corollary:

######  Corollary 1.

The number of maximum arrangements of nonattacking kings on a $2m\times 2n$
rectangle for $m\leq n$ is given by the sum

$K(n,m)=\sum_{A\subseteq[n]}\sum_{k=1}^{n+1}m^{m}_{A,k}.$

Finally, Theorem 1 provides the following algorithm for computing the values
of $K(n)$.

######  Algorithm 1.

  1. 1.

For every triple $(A,B,k)$ such that $B\subseteq A\subseteq[n]$ and $1\leq
k\leq n+1$, compute $p(A,B,k)$ and $q(A,B,k)$.

  2. 2.

Define the matrix $M^{1}$ with rows indexed by subsets of $[n]$ and columns
indexed by $1\leq k\leq n+1$ where each entry is set to 1.

  3. 3.

For $2\leq\ell\leq n$, determine the entries in the matrix $M_{\ell}$ by
summing over entries $m_{B,i}^{\ell-1}$ in the matrix $M^{\ell-1}$ where
$B\subseteq A$ and $p(A,B,k)\leq i\leq q(A,B,k)$.

  4. 4.

Sum the entries of $M^{n}$.

We can estimate the computational complexity of this process. It is
computationally cheap to determine the values of $p(A,B,k)$ and $q(A,B,k)$.
Each of these functions run in polynomial $\mathcal{O}(n^{2})$, that is,
$2n+1$ ways to check the requirements multiplied by $n$ for all possible
values of $i$, plus $n$ to determine the minimum. However, the difficulty is
that the algorithm sums over all pairs of subsets $B\subseteq A\subseteq[n]$
in $\sum_{i=0}^{n}{n\choose i}2^{i}=3^{n}$ ways, providing an overall upper
bound on the order of $\mathcal{O}(3^{n}n^{3})$. We note, many of the entries
in each matrix $M$ are repeated; for example columns indexed by $i$ and
$n+2-i$ are equal, so some futher modifications to this bound can be made, but
the algorithm already an improvement on previous methods. A Mathematica [12]
program that uses Algorithm 1 to calculate the values of $K(n)$ is available
upon request.

## References

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  * [2] J. Bell and B. Stevens, _A survey of known results and research areas for $n$-queens_, Discrete Math. 309 (2009) no. 1, pp. 1–31.
  * [3] T. M. Brown, _The problem of pawns_ , Elec. J. Combin. 22(3) (2019) P3.21.
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  * [5] N. J. Calkin, K. James, S. Purvis, S. Race, K. Schneider and M. Yancey, _Counting kings: explicit formulas, recurrence relations, and generating functions! Oh my!_ , Proceedings of the Thirty-Seventh Southeastern International Conference on Combinatorics, Graph Theory and Computing. Congr. Numer. 182 (2006) pp. 41-51.
  * [6] H. E. Dudeney, Amusements in Mathematics, Edinburgh: Thomas Nelson & Sons, Limited (1917).
  * [7] D. Knuth, Non-attacking Kings on a Chessboard, unpublished, https://www-cs-faculty.stanford.edu/~knuth/preprints.html#unpub (1994).
  * [8] V. Kotěšovec, Non-attacking chess pieces, 6ed, self pub. http://www.kotesovec.cz (2013).
  * [9] M. Kraitchik, Mathematical Recreations, New York: W. W. Norton & Company Inc. (1942).
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  * [11] J. S. Madachy, Mathematical Recreations, New York: Dover (1979).
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  * [13] N. J. A. Sloane, editor, The On-Line Encyclopedia of Integer Sequences, published electronically at https://oeis.org (2020).
  * [14] J. J. Watkins, Across the Board; The Mathematics of Chessboard Problems, Princeton: Princeton University Press (2004).
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  * [16] A. Yaglom and I. Yaglom, Challenging Mathematical Problems with Elementary Solutions. Vol I: Combinatorial Analysis and Probability Theory (Trans. James McCawley, Jr.), San Francisco, CA: Holden-Day Inc. (1964, Revised and edited by Basil Gordon).


A new norm on $BLO$, matters of approximability and duality

FRANCESCA ANGRISANI 111 Department of Information Engineering, Computer
Science and Mathematics (DISIM),
Via Vetoio, Coppito 1, L’Aquila.
<EMAIL_ADDRESS>
Present address: Viale di Augusto n.122 -80125 Napoli - since 15th July to
31st August: only via mail

Abstract –  In this paper, we obtain an alternative expression for the
distance of a function in $BLO$ from the subspace $L^{\infty}$. The distance
is the one induced by choosing a new "norm" on $BLO$, equivalent to the usual
one and that has the advantage to make the distance to $L^{\infty}$
explicitely and exactly computable. We also prove that this new norm has got
the norm attaining property on $BLO$.
We address the issue of approximability by truncation as it is not obvious
even in the closure of $L^{\infty}$ in $BLO$ that functions can be
approximated by truncation. As a matter of fact, a few examples of this are
presented. The easier issue of approximability by convolution is also
addressed.
In the last section we finally prove that the "dual" of $VLO$ is isometric to
$VMO^{\ast}$.

_Key words:_ $BLO$, $L^{\infty}$, distance, truncation, approximability,
duality.

_Mathematics Subject Classification:_ $46E30$

##  1 Introduction

As it is well known, Lebesgue spaces $L^{p}(\Omega,\mu)$ and many of their
generalizations (Grand Lebesgue, Lorentz, Orlicz, Marcinkiewicz spaces) are
defined by conditions that, in some intuitive sense, limit the size of the
function, but in a way such that the distribution of the values of the
function is irrelevant.
In other words, one measures the mass of the function irrespective of where it
is allocated. This is formalized by saying that these spaces are
_rearrangement invariant_ , that is, given two functions $f$ and $g$ such that

$|\\{x\in\Omega:|f(x)|\geq\lambda\\}|=|\\{x\in\Omega:|g(x)|\geq\lambda\\}|,\quad\forall\lambda\geq
0,$

then $f$ belongs to the space if and only if $g$ does; two functions
satisfying the above property are called equimeasurable. In the aforementioned
Lebesgue spaces, for example, this is made clear by the so-called Cavalieri
Principle.
If a function $f$ belongs to a given rearrangement invariant function space
$X$, then the same can be said of its _non-increasing rearrangement_

$f^{*}:t\in[0,|\Omega|]\mapsto\inf\Big{\\{}\lambda:|\\{x\in\Omega:|f(x)|\geq\lambda\\}|\leq
t\Big{\\}}\in\mathbb{R}^{+},$

which is a non-increasing function that is equimeasurable to the starting
function $f$ (see [22]).
Not all function spaces, though, are rearrangement invariant.
An interesting aspect of functions is the way they oscillate, which is
neglected in rearrangement invariant spaces, is crucial to the definition of
other interesting function spaces, as spaces defined by mean of oscillation.
In the mathematical literature, a plethora of different meanings and formal
definitions have been associated to the word "oscillation".
In reality in many cases spaces defined by means of oscillation, in many
different sense of the word, fit into two modes: spaces in which oscillation
is bounded and spaces in which oscillation is vanishing, i.e. arbitrarily
small when measured on a sufficiently small set. This structure, called _o-O
type structure_ , fits into a very general and abstract framework for couple
of non-reflexive Banach function spaces, introduced by K.M. Perfect in 2013
(see [24], [25], [26]). He shows that an example of $(o,O)$ pair is
$(BMO,VMO)$.
Other examples of $(o,O)$ pair of functional spaces are: $(B,B_{0})$ ([12]) (B
is a very large space of functions, that generalize $BMO$ introduced by
Brezis, Bourgain and Mironescu) [8] (see also [17] and [18] ) and $B_{0}$ an
important subspace of $B$), the well known $(Lip_{\alpha},lip_{\alpha})$ (see
[5]), Orlicz space with $\Delta_{0}$ condition and their Morse subspace
$(M^{\psi},L^{\psi})$ (see [6]).
The framework introduced by Perfect often is also useful to obtain some
information about dual spaces, distance formulas and result of atomic
decomposition of the predual of the $O$-space (see for example [4], [13], [7],
[12], [5]).
However space defined by means of oscillation has got also many application.
In fact, for example, $BMO$ was introduced by John and Nirenberg ([21]) to
approach a problem of elasticity theory, in particular to describe the elastic
strain.
In this paper we focus on two subcones of $BMO$ and $VMO$ called respectively
$BLO$ and $VLO$.
Also this two cones are defined by mean of oscillation and it is natural to
imagine the couple $(BLO,VLO)$ as an $o-O$ pair.
Unfortunately the theory introduced by Perfect can not be applied to these
classes of functions, as they are only cones and not vectorial spaces.
$BMO$ and $BLO$ which are function classes defined by imposing a bound on two
different kinds of oscillation, the so called mean oscillation of $f$ over
$I$, i.e.

$\frac{1}{|I|}\displaystyle\int_{I}\left|f(x)-\frac{1}{|I|}\displaystyle\int_{I}f(y)\,dy\right|\,dx.$

and the so called lower oscillation of $f$ over $I$, i.e.

$\frac{1}{|I|}\displaystyle\int_{I}f(x)\,dx-\inf\limits_{x\in I}f(x)$

respectively, where $I$ is an interval if we are dealing with functions of a
single variable, while it is a hypercube with sides parallel to the coordinate
hyperplanes if we wish to define the function class in $\mathbb{R}^{n}$.
In particular, in this paper we introduce a new equivalent norm in $BLO$,
inspired by some characterizations of $BLO$ by Coifman and Rochberg in 1980
(see [11]), that is much more natural when addressing the aforementioned
distance problem.
We also prove the our new norm has got the _norm attaining property_ , i.e.
our norm is attained on the closed unit balls of $BLO$. This property does not
hold for classical norm on $BLO$ (it holds only on $VLO$ functions [3]).
In the second part of paper we discuss the behaviour of $BLO$ functions with
respect to truncation.
As the norm on $BLO$ is defined by means of oscillation, we may expect it to
have strange behaviour with respect to truncation, which is a procedure acting
on the size of the function.
That is exactly what happens: we show that bounded functions, functions that
can be $BLO$-approximated by their truncations and functions that can be
$BLO$-approximated by bounded functions are three completely different
concepts, forming subspaces $L^{\infty},T,\overline{L^{\infty}}^{BLO}$ in
strict consecutive inclusion:

$L^{\infty}\subsetneq T\subsetneq\overline{L^{\infty}}^{BLO}.$

Examples are provided in the section 5.
Then we also deal the easier issue of approximation by convolution.
Another reason for which $BMO$ is a very important space is that it plays the
same role in the theory of Hardy spaces $H^{p}$ that the space $L^{\infty}$ of
essentially bounded functions plays in the theory of $L^{p}$ spaces.
In fact C. Fefferman and E.M. Stein in 1972 proved that $H^{1}$ is the predual
of $BMO$ (see [15]).
Instead it is still unknown what is the dual space of $BMO$.
So, finally we characterize the "dual" of $VLO$ (cleary we will use a weak
definition of dual of $VLO$): in particular we prove that it is isometric to
$VMO^{*}$.

##  2 Notation

######  Definition 2.1.

We say that a real valued locally integrable function $f(x)\in
L^{1}_{loc}(\mathbb{R})$ is of Bounded Mean Oscillation ($f(x)\in
BMO(\mathbb{R})$) if:

$\sup\limits_{I}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}|f(x)-f_{I}|\,dx=\|f\|_{BMO}<\infty$
(1)

where $f_{I}$ denotes
$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}f(x)\,dx=\frac{1}{|I|}\int_{I}f(x)\,dx$
and $I\subset\subset\mathbb{R}$ is the generic compact interval.

$BMO$, first introduced by John and Niremberg in 1961, is a vector space and
$\|\cdot\|_{BMO}$ is a seminorm whose kernel is the set of all functions which
are almost everywhere equal to a constant.
If we consider $BMO$ modulo this subspace, we get a Banach space.

######  Definition 2.2.

We say that a $BMO(\mathbb{R})$ function $f(x)$ is of Vanishing Mean
Oscillation ($f\in VMO(\mathbb{R})$) if it also satisfies:

$V(f)=\limsup\limits_{|I|\to
0}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}|f(x)-f_{I}|\,dx=0.$
(2)

######  Definition 2.3.

We say that a real valued locally integrable function $f(x)\in
L^{1}_{loc}(\mathbb{R})$ has Bounded Lower Oscillation ($f(x)\in
BLO(\mathbb{R})$) if

$\sup\limits_{I}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}[f(x)-\inf_{I}f]\,dx=\sup\limits_{I}[f_{I}-\inf_{I}f]=\|f\|_{BLO}<\infty.$
(3)

As for $BMO$, we will think of the class of $BLO(\mathbb{R})$ functions modulo
constants.

######  Definition 2.4.

We say that a $BLO(\mathbb{R})$ function has Vanishing Lower Oscillation
($f\in VLO(\mathbb{R})$) if it also satisfies:

$W(f)=\limsup\limits_{|I|\to 0}[f_{I}-\inf_{I}f]=0.$ (4)

Of course $\|\cdot\|_{BLO}$ vanishes only on the set of almost everywhere
constant functions, is sub-additive and homogenous for positive scalars but
$BLO\not=-BLO$222An example of a function $f(x)=-\log(|x|)\in BLO$ but not in
$-BLO$ was pointed out by Korey [23]. For this reason and this reason only,
$BLO$ is not a vector space (but only a cone) and $\|\cdot\|_{BLO}$ is not a
norm on it in the classical sense. Even if it lacks homogeneity for negative
constants, following other authors (see [23] and [11]), we will call this
quantity a norm on $BLO(\mathbb{R})$ anyways.
Also the class of all $VLO$ functions is not a vector space for the same
reason.
Sometimes we will use the notation $BMO$, $VMO$ $BLO$ and $VLO$: we always
intend $BMO(\mathbb{R})$, $VMO(\mathbb{R})$ $BLO(\mathbb{R})$ and
$VLO(\mathbb{R})$.
It is obvious, by the definitions, that:

$\|f\|_{BMO}\leq 2\|f\|_{BLO},\quad\|f\|_{BMO},\|f\|_{BLO}\leq
2\|f\|_{L^{\infty}}\quad\forall f\in L^{1}_{loc}.$ (5)

Let us know introduce the $A_{1}$ class of Muckenhoupt.

######  Definition 2.5.

A weight $w:\mathbb{R}\rightarrow[0,+\infty[$ belongs to the $A_{1}$ class of
Muckenhoupt if:

$A_{1}(w)=\sup_{I\subset\subset\mathbb{R}}\frac{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}w(x)dx}{inf_{I}w}<\infty$

or equivalently:

Inspired by a work of Coifman and Rochberg on a decomposition of $A_{1}$
weights through the Hardy-Littlewood maximal operator and by its consequences
on $BLO$, we define a new, equivalent norm on BLO.
This new norm $\|\cdot\|_{BLO}^{\prime}$ will have the advantage of making

$dist_{BLO^{\prime}}(f,L^{\infty}):=\inf\limits_{g\in
L^{\infty}}\|f-g\|_{BLO}^{\prime}$

explicitely and exactly computable.
As a matter of fact, we will prove the equality:

$dist_{BLO^{\prime}}(f,L^{\infty})=\sigma(f):=\inf\\{\mu>0:\
e^{\frac{f}{\mu}}\in A_{1}\\}$ (6)

showing that, in this new norm, the distance from $L^{\infty}$ is exactly the
reciprocal of the critical exponent for which the function $e^{f}$ is in
$A_{1}$. The right hand side in (6) is well defined for a known duality
between $A_{1}$ and $BLO$ which is presented in the following section. In a
previous paper, (see [2]), we proved that

######  Theorem 1.

There exist two absolute constants $d_{1},d_{2}>0$ such that for every real
valued function $f\in BLO(\mathbb{R})$ the following inequalities hold:

$d_{1}\sigma(f)\leq dist_{BLO}(f,L^{\infty})\leq d_{2}\sigma(f)$ (7)

where:

$\sigma(f)=\inf\\{\mu>0:\ e^{\frac{f}{\mu}}\in A_{1}\\}$ (8)

In other contests similar results have been obtained also in [20], [27], [9],
[10], [14] and [16].
Our new result will be a quantitative improvement of Theorem 1 of [2].

##  3 Preliminaries

We begin by quoting an important result concerning $BMO$.

######  Lemma 2 (John-Nirenberg inequality, [21]).

There exist two absolute constants $c_{1},c_{2}>0$ such that, for every $f\in
BMO$, $I\subset\mathbb{R}$ and $\lambda>0$ we have:

$|\\{t\in I:\ |f(t)-f_{I}|>\lambda\\}|\leq
c_{1}e^{-c_{2}\lambda/\|f\|_{BMO}}|I|.$ (9)

There is a strong and useful connection between $BMO$ and $A_{2}$ (see [11]
and also [1]).
This next lemma, due to Coifman and Rochberg, shows, instead, a less known
connection between the Muckenhoupt class $A_{1}$ and $BLO$.

######  Lemma 3 (R.R. Coifman, R. Rochberg).

Let $w\in L^{1}_{loc}(\mathbb{R})$. Then:

  * 1.

$w\in A_{1}\Rightarrow\log(w)\in BLO$ and
$\|\log(w)\|_{BLO}\leq\log(A_{1}(w))$

  * 2.

$w\in BLO\Rightarrow\ e^{w/\mu}\in
A_{1},\quad\forall\mu>\frac{\|w\|_{BMO}}{c_{2}}$

where $c_{2}$ is the constant from (9) in Lemma 2.

###### Proof.

For the proof we refer to [29],p. 157. ∎

Let now introduce another important tool that we will use:

######  Definition 3.1.

The operator $M:L_{loc}^{1}\rightarrow\mathbb{R}$ such that:

$Mw(x)=\sup\limits_{I\ni
x}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}|w(t)|dt$

is called Hardy Littlewood maximal operator.

We quote another result Coifman and Rochberg, concerning powers of $Mw$ for
$w\in L^{1}_{loc}(\mathbb{R})$.

######  Lemma 4 (R.R. Coifman, R. Rochberg, [11]).

Let $w\in L^{1}_{loc}$ be a weight and $\varepsilon\in[0,1)$. Then:

$Mw(x)^{\varepsilon}\in A_{1}$

with $A_{1}$ constant depending on $\varepsilon$ but not on $w$.

The two authors also proved that, modulo $L^{\infty}$ functions, all $A_{1}$
functions arise as some power of $Mw$ for a suitable $w\in
L^{1}_{loc}(\mathbb{R})$, i.e:

######  Theorem 5 (R.R. Coifman, R. Rochberg, [11]).

Assume $w\in A_{1}$.
There are functions $0<A<b(x)<1\in L^{\infty}$, $g\in L^{1}_{loc}$ and a
number $\varepsilon\in[0,1)$ such that:

$w(x)=b(x)Mg(x)^{\varepsilon}$

For the convenience of the reader we will repeat the proof of this theorem
paying particular attention to the value of $A$ and how it depends on
$A_{1}(w)$.
To do so, the so called reverse Holder inequality is needed:

######  Theorem 6.

If $w\in A_{1}$, then there is a sufficiently small value of $\eta>0$ and a
constant $c_{\eta}$ depending on $\eta$ but not on $I\subset\mathbb{R}$ such
that:

$\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}w^{1+\eta}(x)dx\right)^{\frac{1}{1+\eta}}\leq
c_{\eta}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}w(x)dx.$

In particular, we have $w^{1+\eta}\in A_{1}$ and $A_{1}(w^{1+\eta})\leq
c_{\eta}^{1+\eta}A_{1}(w)^{1+\eta}$.
Furthermore, $c_{\eta}<c_{3}:=2e,\ \forall\eta<1$.

Thank to this result we obtain that for every $w\in A_{1}$ there is always a
$\eta\in(0,1)$ such that $A_{1}(w^{1+\eta})\leq
c_{3}^{1+\eta}A_{1}(w)^{1+\eta}$.

###### Proof of Theorem 5.

Choose a sufficiently small $\eta<1$ as in the reverse Holder inequality and
define

$b(x)=\frac{w(x)}{[M(w^{1+\eta}(x))]^{\frac{1}{1+\eta}}}\text{ and
}g(x)=w(x)^{1+\eta},$ (10)

noticing that $g(x)\in L^{1}_{loc}(\mathbb{R})$.
We have that $b(x)\leq 1\iff w^{1+\eta}(x)\leq M(w^{1+\eta}(x))$ but this
latter inequality holds a.e.since almost every point $x\in\mathbb{R}$ is a
Lebesgue point for $w^{1+\eta}(x)\in L^{1}_{loc}(\mathbb{R})$.
A lower bound on $b(x)$ can be obtained by observing that:

$b(x)=\frac{w(x)}{[M(w^{1+\eta}(x))]^{\frac{1}{1+\eta}}}=\left[\frac{w(x)^{1+\eta}}{M(w^{1+\eta}(x))}\right]^{\frac{1}{1+\eta}}\geq\left[\frac{1}{A_{1}(w^{1+\eta})}\right]^{\frac{1}{1+\eta}}\geq\frac{1}{c_{3}A_{1}(w)}>0.$
(11)

Now let us take $\varepsilon\in(0,1)$ to be $\frac{1}{1+\eta}$.
We have that $w(x)=b(x)\left[Mg(x)\right]^{\varepsilon}$ so that the theorem
is proven and we can choose A as $\frac{1}{c_{3}A_{1}(w)}$. ∎

By combining theorems 4 and 5, in light of the duality between $BLO$ and
$A_{1}$ expressed by lemma 3 we get that:

$f(x)\in BLO\iff\exists\ \alpha>0,\ b(x)\in L^{\infty},\ g(x)\in
L^{1}_{loc},\quad f(x)=\alpha\log(Mg(x))+b(x)$ (12)

As a matter of fact, if we have $f\in BLO$, then by lemma 3 there is a $\mu>0$
such that $e^{f/\mu}\in A_{1}$ can be decomposed as in Theorem 5 and the
conclusion follows by taking logarithms of both sides.
The inverse implication follows from the fact that $(Mg)^{\frac{1}{2}}\in
A_{1}$ by lemma 4, which in turn implies
$\alpha\log(Mg(x))=2\alpha\log((Mg)^{\frac{1}{2}})$ is in $BLO$ by lemma 3.
The obvious inclusion $L^{\infty}\subset BLO$ completes this line of
reasoning.
The logical equivalence in (12) leads us to define:

$\|f\|_{BLO}^{\prime}=\inf\\{\alpha+\|b\|_{\infty}:\ \alpha>0,\ b\in
L^{\infty}\ \text{ s.t. }\exists g\in L^{1}_{loc},\ f=\alpha\log(Mg)+b\\}$
(13)

which has, of course, the same properties of $\|\cdot\|_{BLO}$. In the next
section we prove the equivalence of the two.

##  4 Equivalence of the norms

######  Theorem 7.

There exist two absolute constants $d_{1},d_{2}>0$ such that, for every $f\in
BLO$:

$d_{1}\|f\|_{BLO}\leq\|f\|_{BLO}^{\prime}\leq d_{2}\|f\|_{BLO}$ (14)

where $||f||_{BLO}^{\prime}$ has been defined in (LABEL:newnorm).

###### Proof.

During this proof, $c_{1}$ and $c_{2}$ will still denote the constants in
equation (9) of lemma 2.
We will divide the proof in steps, addressing the latter inequality first.
Step 1: Showing that $\exists c_{4}>0$ such that $\forall f\in BLO,\
A_{1}\left(e^{\frac{fc_{2}}{4\|f\|_{BLO}}}\right)\leq c_{4}$:
Let us take $\lambda=\frac{4\|f\|_{BLO}}{c_{2}}\log(\zeta)$ in John-Niremberg
inequality (9) from lemma 2.
We have:

$\Bigg{|}\,\Bigg{\\{}t\in I:\
|f(t)-f_{I}|>\frac{4\|f\|_{BLO}}{c_{2}}\log(\zeta)\Bigg{\\}}\,\Bigg{|}\leq
c_{1}\zeta^{-\frac{4\|f\|_{BLO}}{\|f\|_{BMO}}}|I|.$ (15)

Notice that:

$|f(t)-f_{I}|>\frac{4\|f\|_{BLO}}{c_{2}}\log(\zeta)\iff
e^{\frac{c_{2}|f(t)-f_{I}|}{4\|f\|_{BLO}}}>\zeta.$ (16)

By Cavalieri’s principle we have:

$\int_{I}e^{\frac{c_{2}|f(t)-f_{I}|}{4\|f\|_{BLO}}}dt=\int_{0}^{\infty}|\\{t\in
I:\ e^{\frac{c_{2}|f(t)-f_{I}|}{4\|f\|_{BLO}}}>\zeta\\}|d\zeta\leq\\\
\leq|I|+\int_{1}^{\infty}|\\{t\in I:\
e^{\frac{c_{2}|f(t)-f_{I}|}{4\|f\|_{BLO}}}>\zeta\\}|d\zeta.$ (17)

Thank to (16) we are able to bound the last integral so that (17) becomes:

$\int_{I}e^{\frac{c_{2}|f(t)-f_{I}|}{4\|f\|_{BLO}}}dt\leq|I|+|I|\int_{1}^{\infty}c_{1}\zeta^{-\frac{4\|f\|_{BLO}}{\|f\|_{BMO}}}\leq|I|+|I|\int_{1}^{\infty}c_{1}\zeta^{-2}d\zeta=|I|(1+c_{1}).$
(18)

where $\|\cdot\|_{BMO}\leq 2\|\cdot\|_{BLO}$ was also used.
By dividing both sides of (18) by $|I|$ and using $x\leq|x|$, we get to

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}e^{\frac{c_{2}(f(t)-f_{I})}{4\|f\|_{BLO}}}dt\leq
1+c_{1}$ (19)

so that:

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}e^{\frac{c_{2}}{4\|f\|_{BLO}}f(t)}dt\leq(1+c_{1})e^{\frac{c_{2}f_{I}}{4\|f\|_{BLO}}}\leq(1+c_{1})e^{\frac{c_{2}(\inf_{I}f+\|f\|_{BLO})}{4\|f\|_{BLO}}}$
(20)

where we used $f\in BLO$. Lastly:

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}e^{\frac{c_{2}}{4\|f\|_{BLO}}f(t)}dt\leq(1+c_{1})e^{\frac{c_{2}}{4}}\inf_{I}e^{\frac{c_{2}}{4\|f\|_{BLO}}f(t)}$
(21)

and this first step is concluded by taking
$c_{4}=(1+c_{1})e^{\frac{c_{2}}{4}}$.
Step 2: Showing that $\|f\|_{BLO}^{\prime}\leq d_{2}\|f\|_{BLO}$ Take $f(x)\in
BLO$. By the first step we have $w(x)=e^{\frac{f(x)c_{2}}{4\|f\|_{BLO}}}\in
A_{1}$.
By theorem 5 there exist $b\in L^{\infty}$, $0<A<b(x)<1$,
$\varepsilon\in[0,1)$, $g\in L^{1}_{loc}$ such that

$w(x)=b(x)[Mg(x)]^{\varepsilon}$ (22)

and so

$f(x)=\frac{4\|f\|_{BLO}}{c_{2}}\log(b(x))+\frac{4\|f\|_{BLO}}{c_{2}}\varepsilon\log(Mg(x)).$
(23)

In particular, define

$B(x)=\frac{4\|f\|_{BLO}}{c_{2}}\log(b(x))$ and
$\alpha=\frac{4\|f\|_{BLO}}{c_{2}}\varepsilon$.

Since $1\geq b(x)\geq\frac{1}{c_{3}A_{1}(w)}$, taking logarithms and using the
first step we have that

$0<|B(x)|<[\log(c_{3})+\log(c_{4})]\frac{4\|f\|_{BLO}}{c_{2}}=:c_{5}\frac{4\|f\|_{BLO}}{c_{2}}.$
(24)

Also notice $\alpha<\frac{4\|f\|_{BLO}}{c_{2}}$ since $\varepsilon<1$.
By combining this with (24) we get
$\|f\|_{BLO}^{\prime}\leq\alpha+\|B\|_{\infty}\leq\left[(1+c_{5})\frac{4}{c_{2}}\right]\|f\|_{BLO}$,
so that by denoting $d_{2}=\left[(1+c_{5})\frac{4}{c_{2}}\right]$ we conclude
the second step.
Step 3: Showing $d_{1}\|f\|_{BLO}\leq\|f\|_{BLO}^{\prime}$.
Let us choose any $\alpha>0$, $g\in L^{1}_{loc}$ and $b\in L^{\infty}$ such
that $f=\alpha\log(Mg)+b$.
We have:

$\|f\|_{BLO}\leq\alpha\|\log(Mg)\|_{BLO}+\|b\|_{BLO}\leq
2\alpha\|\log([Mg]^{\frac{1}{2}})\|_{BLO}+2\|b\|_{\infty}.$

Thank to the first proposition in Lemma 3 we have:

$\|f\|_{BLO}\leq
2\alpha\log\left[A_{1}([Mg]^{\frac{1}{2}})\right]+2\|b\|_{\infty}.$

By Theorem 4 we have that $A_{1}([Mg]^{\varepsilon})$ does not depend on $g$,
so that $A_{1}([Mg]^{\frac{1}{2}})\leq c_{6}$, with $c_{6}$ absolute constant.
For this reason we have

$\|f\|_{BLO}\leq 2\alpha\log(c_{6})+2\|b\|_{\infty},$ (25)

for each possible decomposition of $f$, so that by taking the infimum and
denoting $\frac{1}{d_{1}}=\max\\{2,2\log(c_{6})\\}$ we conclude the last step
of the proof. ∎

######  Theorem 8.

For every $f\in BLO$ define:

$\sigma(f)=\inf\Bigg{\\{}\mu>0:\ e^{\frac{f}{\mu}}\in A_{1}\Bigg{\\}}$

we have that:

$dist_{BLO^{\prime}}(f,L^{\infty})=\inf\limits_{h\in
L^{\infty}}\|f-h\|_{BLO}^{\prime}=\sigma(f)$

###### Proof.

First notice that:

$dist_{BLO^{\prime}}(f,L^{\infty})=\inf\limits_{h\in
L^{\infty}}\inf\\{\mu+\|b\|_{\infty}:\ f-h=\mu\log(Mg)+b\\}$

so that, for any $\mu$ and $g$, by a proper choice of $h$, $b$ can always be
chosen to be $0$.
This means that

$dist_{BLO^{\prime}}(f,L^{\infty})=\inf\\{\mu>0:\ \exists g(x)\in
L^{1}_{loc},\ f(x)-\mu\log[Mg(x)]\in L^{\infty}\\}.$ (26)

On the other hand, since a weight $e^{\frac{f}{\mu}}\in L^{1}_{loc}$ is in the
Muckenhoupt class $A_{1}$ if and only if it can be written as
$b(x)[Mg(x)]^{\varepsilon}$ for suitable $0<A<b(x)\in L^{\infty}$, $g\in
L^{1}_{loc}$ and $\varepsilon\in[0,1)$, we have:

$\sigma(f)=\inf\\{\mu>0:\ \exists b\in L^{\infty},\ \varepsilon<1,\ g\in
L^{1}_{loc},\ e^{\frac{f(x)}{\mu}}=b(x)[Mg(x)]^{\varepsilon}\\}$ (27)

or, by taking logarithms:

$\sigma(f)=\inf\Bigg{\\{}\mu>0:\ \exists b(x)\in L^{\infty},\ \varepsilon<1,\
g(x)\in L^{1}_{loc},\
\frac{f(x)}{\mu}=b^{\prime}(x)+\varepsilon\log[Mg(x)]\Bigg{\\}},$ (28)

where we have denoted $b^{\prime}(x)=\log(b(x))\in L^{\infty}$.
Equivalently:

$\sigma(f)=\inf\\{\mu>0:\ \exists\varepsilon<1,\ g(x)\in L^{1}_{loc},\
f(x)-\mu\varepsilon\log[Mg(x)]\in L^{\infty}\\}$ (29)

is the same as:

$\sigma(f)=\inf\\{\mu>0:\ \exists g(x)\in L^{1}_{loc},\ f(x)-\mu\log[Mg(x)]\in
L^{\infty}\\}.$ (30)

Let us better explain this last logical step, between equation (29) and (30).
A real number $a$ satisfies

$a<\inf\\{\mu>0:\ \exists\varepsilon<1,\ g(x)\in L^{1}_{loc},\
f(x)-\mu\varepsilon\log[Mg(x)]\in L^{\infty}\\}$ (31)

if and only if there is no $\varepsilon<1$ and $g\in L^{1}_{loc}$ such that
$f-a\varepsilon\log(Mg(x))\in L^{\infty}$, which is equivalent to saying that:

$\nexists a^{\prime}<a:\ f-a^{\prime}\log(Mg(x))\in L^{\infty}$ (32)

which happens if and only if:

$a\leq\inf\\{\mu>0:\ \exists g(x)\in L^{1}_{loc},\ f(x)-\mu\log[Mg(x)]\in
L^{\infty}\\}.$

The conclusion follows by comparing (26) and (30).
∎

###  4.1 Norm attaining property

In this section, we will show that our new norm has the norm attaining
property on $BLO$, i.e. this norm is attained on the closed unit balls of
$BLO$.
The classical norm on $BLO$ instead has got norm attaining property only on
function in $VLO$. Actually, there are examples of $BLO$ functions for whom
the norm attaining property does not hold (see [3]).

######  Definition 4.1.

Let $\Omega\subset\mathbb{R}$. We say that a measurable function $f$ belongs
to $EXP(\Omega)$ if there exists a number $\mu\in[0,\infty)$ such that

$\int_{\Omega}e^{\frac{f}{\mu}}<\infty.$ (33)

######  Theorem 9.

If $f\in BLO$, then $\sigma(f)$ is a minimum, i.e.

$\sigma(f)=\min\\{\mu>0:\ e^{\frac{f}{\mu}}\in A_{1}\\}$

###### Proof.

Since $f$ belongs to $BLO$, we have that $e^{\frac{f}{\mu}}\in A_{1}$, and so:

$\sup_{I\subset\subset\mathbb{R}}\frac{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}e^{\frac{f}{\mu}}dx}{\inf_{I}e^{\frac{f}{\mu}}}<\infty$

i.e. there exists a constant $C$ such that:

$\sup_{I\subset\subset\mathbb{R}}{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}e^{f-\inf_{I}f}dx}\leq
C.$

So, to obtain the thesis, we just have to prove that:

$\inf\Bigg{\\{}\mu>0:\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int
e^{\frac{h}{\mu}}\leq C\Bigg{\\}},$ (34)

is a minimum, where $h$ is a $L^{\infty}$ function.
Throughout this proof we will use $\lambda=1/\mu$ instead of $\mu$. Let us
consider $g=e^{h}$.
Let us suppose that $g$ is not almost everywhere equal to $1$ and define
$M=\\{\sup{\lambda>0:g\in L^{\lambda}}\\}$.
Let us first suppose that $M<+\infty$.
It is well known that $||g||_{\lambda}$ and hence $||g||_{\lambda}^{\lambda}$
is continuous with respect to $\lambda$ in $[0,M).$
We also notice that it is strictly increasing because $g$ is not almost
everywhere equal to $1$.
So, there are two possibilities: either $||g||_{M}^{M}$ is at most $C$ (and
$\frac{1}{M}$ is the minimum we were looking for in the theorem) or
$||g||_{M}^{M}\in(C,+\infty]$ but there is unique $M^{\prime}<M$ such that
$||g||_{M^{\prime}}^{M^{\prime}}=C$, by the definition of limit of
$||g||_{\lambda}^{\lambda}\to||g||_{M}^{M}$ as $\lambda$ goes to $M$,
monotonicity and the intermediate value theorem.
In either case, the infimum (34) is attained.
Now let us consider the case that $M=+\infty$.
Of course $\lim_{\lambda\to+\infty}||g||_{\lambda}=||g||_{\infty}>1$ (it is
possible that $||g||_{\infty}$ is infinite but that is not a problem) because
$h$ is not almost everywhere equal to $0$ but this means that
$\lim_{\lambda\to\infty}||g||_{\lambda}^{\lambda}$. So there exists
$M^{\prime}<\infty$ such that $||g||_{M^{\prime}}^{M^{\prime}}=C$, again by
definition of limit, monotonicity and intermediate value theorem.
If $g$ is almost everywhere equal to $1$, then $h=0$ a.e., so the minimum in
(34) is attained for every $\mu>0$.

∎

######  Corollary 10.

If $h\in EXP$, then $\sigma(h)$ is a minimum, i.e.
The quantity

$\inf\Bigg{\\{}\mu>0:\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{[a,b]}e^{\frac{h}{\mu}}\leq
C\Bigg{\\}}$ (35)

is a minimum.

###### Proof.

It suffices to use the fact that $f\in L^{\infty}$ in the proof of Theorem 9,
so the thesis follows trivially from the proof of Theorem 9 ∎

######  Remark: 1.

The quantity in (35) is a norm on $EXP$. By corollary 10, $EXP$ with this norm
is norm attaining.

##  5 Approximability

###  5.1 Example of an unbounded function in $\overline{L^{\infty}}^{BLO}$

It follows from our new distance formula that the closure of $L^{\infty}$ in
$BLO$ can be characterized as:

$f\in\overline{L^{\infty}}^{BLO}\iff\forall\mu>0,\ e^{\frac{f}{\mu}}\in
A_{1}.$ (36)

Using the characterization in (36), let us start by providing an example of an
unbounded function in $\overline{L^{\infty}}^{BLO}$, proving the strict
inclusion $L^{\infty}\subset\overline{L^{\infty}}^{BLO}$ is strict.
Throughout this example we restrict the attention to the interval
$J=(0,\frac{1}{e})$ and consider the function $f(x)=\log(-\log(x))$, that
clearly does not belong to $L^{\infty}(J)$.
However, we will prove $e^{rf(x)}=[-log(x)]^{r}$ is $A_{1}$ on
$(0,\frac{1}{e})$ for arbitrarily large $r$, concluding through (36) that it
is in the closure of $L^{\infty}$ in $BLO$.
To prove this, we observe that $g_{r}(x)=[-log(x)]^{r}$ is decreasing and
always greater than $1$ in $J$. This helps with the extimate of
$A_{1}(g_{r})$.
Indeed

$A_{1}(g_{r})=\sup\limits_{I\subset\subset
J}\frac{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}g_{r}(x)dx}{\inf_{I}g_{r}}=\sup\limits_{[0,b]\subset\subset
J}\frac{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{0}^{b}g_{r}(x)dx}{\inf_{[0,b]}g_{r}}=\sup\limits_{b\leq
1/e}\frac{1}{g_{r}(b)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{0}^{b}g_{r}(x)dx.$
(37)

This is because $g_{r}(x)$ is decreasing and for every interval $I=[a,b]$.
By considering the interval $I^{\prime}=[0,b]$ we have the obvious
inequalities:

  * 1.

$(g_{r})_{I^{\prime}}\geq(g_{r})_{I}$

  * 2.

$\inf_{I^{\prime}}g_{r}=g_{r}(b)=\inf_{I}g_{r}$.

By taking $b=e^{-t}$ we have

$A_{1}(g_{r})=\sup\limits_{t\geq
1}\frac{e^{t}}{t^{r}}\int_{0}^{e^{-t}}[-log(x)]^{r}dx.$ (38)

We will now show via the induction principle on the integer _r_ that the
quantity of which we are taking the supremum in (37) is decreasing with
respect to $t$ for every positive integer $r$.
As a matter of fact, the right hand side equals $1+\frac{1}{t}$ for $r=1$ so
the base case of the induction principle is satisfied.
We will now proceed by induction and assume it is decreasing for $r-1$ and let
us use this notation:

$f_{r}(t)=\int_{0}^{e^{-t}}[-log(x)]^{r}dx.$

Using integration by parts we get the recursive formula

$f_{r}(t)=\frac{t^{r}}{e^{t}}+rf_{r-1}(t)$

and so we obtain

$\frac{e^{t}}{t^{r}}f_{r}(t)=1+\frac{r}{t}\left[\frac{e^{t}}{t^{r-1}}f_{r-1}(t)\right].$

As the product of two positive decreasing functions is still decreasing, we
proved that the right hand side in equation (38) is decreasing in $t$ for
every $n$ and this implies:

$A_{1}(g_{r})=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{0}^{1/e}[-\log(x)]^{r}dx$

because the supremum in (37) is obtained when $t=1$.
One could compute $A_{1}(g_{r})$ exactly with this strategy, but for us it is
enough to observe that it is finite for every $r$, i.e. $A_{1}(e^{rf})<\infty$
for every $r\in\mathbb{N}$ and $f(x)\in\overline{L^{\infty}}^{BLO}$ by the
aforementioned characterization.

###  5.2 Approximability by truncation

In the previous section, we explicitely showed that the inclusion
$L^{\infty}\subset\overline{L^{\infty}}^{BLO}$ is strict.

######  Definition 5.1.

Let us call $T$ the set of $BLO$ functions $f$ such that, if we define the
truncated functions,

$T_{k}f=\begin{cases}k&\text{if }f(x)\geq k\\\ f(x)&\text{if }-k\leq f(x)\leq
k\\\ -k&\text{if }f(x)\leq-k\end{cases}$

we get $\lim\limits_{k\to\infty}\|f-T_{k}f|_{BLO}=0$.

We will now also show that both inclusions in

$L^{\infty}\subset T\subset\overline{L^{\infty}}^{BLO}$ (39)

are strict.
Our example of an unbounded function that is approximable by truncation is
exactly the one from the previous section. In fact we will show that
$f(x)=\log(-\log(x))\in T$ even if it is unbounded, concluding
$L^{\infty}\subset T$ is strict.
To do so, let us notice that:

$f-T_{k}f=\begin{cases}f(x)-k&\text{if }f(x)\geq k\\\ 0&\text{if }f(x)\leq
k\end{cases}$

and since $f-\\{f\\}_{k}$ is non-negative, when we compute
$\|f-\\{f\\}_{k}\|_{BLO}$ we can ignore all intervals having intersection with
$\\{x:f(x)<k\\}=(e^{-e^{k}},e^{-1})$.
In other words,

$\|f-T_{k}f|_{BLO(0,1/e)}=\|f-k\|_{BLO(0,1/e^{e^{k}})}=\|f\|_{BLO(0,1/e^{e^{k}})}.$

Let us set $J_{k}=(0,1/e^{e^{k}})$. Our goal is to show that
$\|f\|_{BLO(J_{k})}\to 0$ as $k\to\infty$. In [19], Garcia showed that
$e^{\|f\|_{BLO}}\leq A_{1}(e^{f})$.
In this case, since $e^{f(x)}=-\log(x)$:

${\|f\|_{BLO(J_{k})}}\leq\log[A_{1}(-\log(x))]_{J_{k}}$

where the subscript $J_{k}$ denotes the fact that we are considering the
$A_{1}$ constant of $-log(x)$ over $J_{k}$.
With the same strategy as before, we get:

$[A_{1}(-\log x)]_{J_{k}}=\sup\limits_{t\geq
e^{k}}\frac{e^{t}}{t}\int_{0}^{e^{-t}}[-log(x)]dx=\frac{e^{e^{k}}}{e^{k}}\int_{0}^{e^{-e^{k}}}[-log(x)]dx=\frac{1+e^{k}}{e^{k}}$

so that:

$0\leq\lim\limits_{k\to\infty}\|f-T_{k}f|_{BLO}\leq\lim\limits_{k\to\infty}\log[A_{1}(-\log(x))]_{J_{k}}=\lim\limits_{k\to\infty}\log(1+\frac{1}{e^{k}})=0.$

proving the strictness of the first inclusion in (39).

###  5.3 Functions in $\overline{L^{\infty}}^{BLO}$ which are not approximable
by truncation

There we show that there exists a function which is approximable by bounded
functions but not by truncation, i.e. a function
$f\in\overline{L^{\infty}}^{BLO}$ which does not belong to $T$.
To do so, we will first show that any function of the type

$\eta(x)=\sum\limits_{n=0}^{\infty}n\chi_{[a_{n+1},a_{n})}\quad\text{with
}a_{n}\text{ strictly decreasing to }0$ (40)

does not belong to $T$ and then we carefully choose $\\{a_{n}\\}$ in such a
way that $\eta(x)$ is in $\overline{L^{\infty}}^{BLO}$.
To prove the aforementioned claim, let us notice that, for a function $f(x)$
of the type in (40), the following inequality hold:

$\|f-\\{f\\}_{k}\|_{BLO}=\|f\|_{BLO(0,a_{k})}\geq 1,\quad\forall
k\in\mathbb{N}.$ (41)

This is because the $BLO$ norm of a jump function is at least equal to the
maximum size of its jumps: one can prove that it is by choosing intervals of
the tipe $[a_{n+1},a_{n}+\varepsilon]$ with sufficiently small
$\varepsilon>0$.
Our example of a function in $\overline{L^{\infty}}^{BLO}$ which is not
approximable by truncation will then be of the type described in (40) with the
choice $a_{n}=1/e^{e^{n}}$ (considered on the interval $J$).
Taking $f$ defined as

$f(x)=\sum\limits_{n=0}^{\infty}n\chi_{[1/e^{e^{n+1}},1/e^{e^{n}})}$

let us compute $A_{1}(e^{rf})$. Let us notice that $f$ is decreasing:

$A_{1}(e^{rf})=\sup\limits_{b\leq
1/e}\frac{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{0}^{b}e^{rf(x)}dx}{e^{rf(b)}}\leq\sup\limits_{n\in\mathbb{N}}\frac{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{0}^{a_{n+1}}e^{rf(x)}dx}{e^{rn}}$

where the second inequality comes from considering $a_{n+1}\leq b<a_{n}$ and
overextimating the integral average over $[0,b]$ with the one over
$[0,a_{n+1}]$ since $f$ is decreasing.
Explicitely computing the integral via the definition we have:

$A_{1}(e^{rf})=\sup\limits_{n\in\mathbb{N}}\frac{1}{a_{n+1}}\sum\limits_{i=n+1}^{\infty}e^{r(i-n)}[a_{i}-a_{i+1}]$

and then recalling the choice of $a_{n}$ we get

$A_{1}(e^{rf})=\sup\limits_{n\in\mathbb{N}}e^{e^{n+1}}\sum\limits_{j=1}^{\infty}e^{rj}\left[\frac{1}{e^{e^{n+j}}}-\frac{1}{e^{e^{n+j+1}}}\right]=\sum\limits_{j=1}^{\infty}\frac{1}{e^{[e^{n}(e^{j}-e)-rj]}}.$

Noting that the series on the right hand side is dominated by a convergent one
that does not depend on $n$, i.e:

$\sum\limits_{j=1}^{\infty}\frac{1}{e^{[e^{n}(e^{j}-e)-rj]}}\leq\sum\limits_{j=1}^{\infty}\frac{1}{e^{[e^{j}-e-rj]}}=K(r)<+\infty,\quad\forall
r,n\in\mathbb{N}$

shows that $f(x)$ lies in $\overline{L^{\infty}}^{BLO}$, but, like every other
function of this type, is not approximable by truncation.
It is worth noticing that one can adapt the proof that $\log(-\log(x))$ is in
$T$ to every positive decreasing $VLO$ function, so that all positive
decreasing $VLO$ functions are in $T$.

###  5.4 Approximability by convolution

In [2] we also found an equivalent expression for $dist_{BLO}(f,VLO)$ showing
in particular that $VLO$ is a closed subset in $BLO$.
The procedure we used to construct a function in $VLO$ which was as close as
possible to a given $BLO$ function was not exclusively that of convolution.
A natural question then arises: does the sequence of convolutions of a $VLO$
function $f$ with a standard sequence of mollifiers $\varphi_{\varepsilon}$
converge (in $BLO$ norm) to the original function $f$ as $\varepsilon\to 0$?
We will prove that it does if we choose $\varphi$ with the following
properties:

  * 1.

$\varphi_{\varepsilon}>0$

  * 2.

$supp(\varphi_{\varepsilon})\subset(-\varepsilon,\varepsilon)$

  * 3.

$\int_{-\varepsilon}^{\varepsilon}\varphi_{\varepsilon}=1$

  * 4.

$\|\varphi_{\varepsilon}\|_{L^{\infty}}\leq\frac{C}{\varepsilon}$.

Since $VLO$ is closed in $BLO$ and the convolution of a $BLO$ function with a
continuous function with compact support lies in $VLO$, this will prove that a
function $f\in BLO$ can be approximated by a sequence of its convolutions with
a family of mollifiers if and only if $f\in VLO$.
Let us call $W_{a}(f)=\sup\limits_{|I|\leq
a}\left[\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}f-\inf_{I}f\right]$
and $f_{\varepsilon}(x)=(f\star\varphi_{\varepsilon})(x)$.
We have:

$f_{\varepsilon}(x)=\int_{-\varepsilon}^{\varepsilon}\varphi_{\varepsilon}(y)f(x-y)$

and so:

$f_{\varepsilon}(x)-f(x)=\int_{-\varepsilon}^{\varepsilon}\varphi_{\varepsilon}(y)[f(x-y)-f(x)]dy\leq\int_{-\varepsilon}^{\varepsilon}\varphi_{\varepsilon}(y)\left[f(x-y)-\inf_{[x-\varepsilon,x+\varepsilon]}f\right]dy$

Let us multiply and divide by $2\varepsilon$ to get:

$|f_{\varepsilon}(x)-f(x)|\leq
2C\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{-\varepsilon}^{\varepsilon}f(x-y)-\inf_{[x-\varepsilon,x+\varepsilon]}dy\leq
2CW_{2\varepsilon}(f).$

But the right hand side does not depend on $x$ anymore so that:

$0\leq\|f-f_{\varepsilon}\|_{BLO}\leq 2\|f-f_{\varepsilon}\|_{L^{\infty}}\leq
4CW_{2\varepsilon}(f)\to 0$

and the proof is concluded.

##  6 Duality

As we said in the introduction, thanks to [15] we know that the predual of
$BMO$ is the Hardy space $H^{1}$, but it is still unknown who is $BMO^{*}$. To
identify the dual space of $BMO$ is a very hard and interesting question.
Obviously, since $BLO$ and $VLO$ are not vectorial spaces, we can not talk
about their dual in classical sense, and we define the following classes of
linear and bounded functionals and we call them "dual" the same:

######  Definition 6.1.

Let us call "dual" of $BLO$ the following class:

$BLO^{\ast}=\\{\phi:\phi\,\text{is a linear functional defined
on}\,BLO\,\text{such that}\,|\phi(f)|\leq K||f||_{BLO}\\}$444i.e. they respect
$\phi(au+bv)=a\phi(u)+b\phi(v)$ for every $a,b,u,v$ such that both sides of
the equality make sense

and

######  Definition 6.2.

Let us call "dual" of $VLO$ the following class:

$VLO^{\ast}=\\{\phi:\phi\,\text{is a linear functional defined
on}\,VLO\,\text{such that}\,|\phi(f)|\leq K||f||_{BLO}\\}$.

We are able to prove the following result:

######  Theorem 11.

$VLO^{\ast}$ is isometric to $VMO^{\ast}$.

###### Proof.

Let us observe that for $\psi\in BMO^{\ast}$ we have that $\psi_{|VLO}$ is
still linear on $VLO$ and

$\psi_{|VLO}(f)\leq||\psi||_{VMO}^{\ast}||f||_{BMO}\leq||\psi||_{VMO^{\ast}}2||f||_{BLO}.$

It means that $\psi_{|VLO}$ is continuous as a functional on $VLO$, with
operator norm at most doubled in the process.
Let us now observe that for every $\phi\in VLO^{\ast}$, we can define
$\hat{\phi}$ acting on $VMO$ by using a decomposition theorem from Korey [23].
Actually, M.B. Korey showed that for every $g\in VMO$, there exist
$g_{1},g_{2}\in VLO$ such that $g=g_{1}-g_{2}$ and $g_{1},g_{2}$ can be chosen
such that

$||g_{1}||_{BLO}+||g_{2}||_{BLO}\leq||g||_{BMO}.$ (42)

Inspired by this observation we define
$\hat{\phi}(g)=\phi(g_{1})-\phi(g_{2})$.
First, we have to prove that the definition is well posed, because the
decomposition is not unique.
So let us notice that:

$g_{1}-g_{2}=h_{1}-h_{2}\Rightarrow
g_{1}+h_{2}=g_{2}+h_{1}\Rightarrow\phi(g_{1}+h_{2})=\phi(g_{2}+h_{1})\Rightarrow\phi(g_{1})+\phi(h_{2})=\\\
=$$\phi(g_{2})+\phi(h_{1})\Rightarrow\phi(g_{1})-\phi(g_{2})=\phi(h_{1})-\phi(h_{2})$$$
(43)

so that $\hat{\phi}$ is well defined on $VMO$ (in the sense that it does not
depend on the decomposition).
Of course $\hat{\phi}$ is linear on $VMO$ and using (42) we also obtain that
it is bounded, with operator norm at most equal to the operator norm of
$\phi$.
In formulas:

$|\hat{\phi}(g)|\leq|\phi(g_{1})|+|\phi(g_{2})|\leq||\phi||_{VLO^{\ast}}(||g_{1}||_{BLO}+||g_{2}||_{BLO}\leq||\phi||_{VLO^{\ast}}||g||_{BMO}.$

We now notice that ${\hat{\psi}}_{|VLO}=\psi$ for all $\psi\in VMO^{\ast}$ and
$\hat{\phi}_{|VLO}=\phi$ for all $\phi$ in $VLO^{\ast}$, so that we
constructed a bijection between $VLO^{\ast}$ and $VMO^{\ast}$.
Since the action to put the hat on a function that belongs to $BLO$ and the
restriction are linear actions it is an isomorphism of vector spaces and by
the fact we were able to control the operator norm in the process it is an
isometry. ∎

######  Remark: 2.

$VMO^{\ast}$ is isomorphic to the Hardy space $H_{1}$ (see [28] p.180).

##  7 Acknowledgements

The author is member of the Gruppo Nazionale per l’Analisi Matematica, la
Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta
Matematica (INdAM).
The author would like to warmly thank Professor Carlo Sbordone for having
suggested her studying this problem.

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