
CHAPTER 1

P2, formula (1.1.5): the second ":=" should be only "=".

P5, L7 $(-1)^(k+1)$ should replace $(-1)^k$

P5, E.2b] The definition of Lexicographic order is screwed up.

\noindent
{\it Hint:}  For a symmetric polynomial $f$ in $n$ variables,
$$f(x_1, x_2, \ldots, x_n) = \sum^{m}_{\alpha_1 =0} \sum^{m}_{\alpha_2 = 0} \cdots
\sum\limits^{m}_{\alpha_n =0} c_{\alpha_1, \alpha_2, \ldots, \alpha_n}
x^{\alpha_1}_1 x^{\alpha_2}_2 \cdots  x^{\alpha_n}_n\,,$$
let
$$\sigma (f) : = (\nu_1,  \nu_2, \ldots, \nu_n)\,,
\qquad \nu_1 \geq  \nu_2 \geq \cdots \geq \nu_n \geq 0$$
if $(\nu_1,  \nu_2, \ldots, \nu_n)$ is the largest element in the
lexico-graphic ordering of the $n$-tuples formed by the exponents of
each term with a non-zero coefficient. By lexico-graphic ordering of
$n$-tuples of nonnegative integers we mean
$$(\nu_1,  \nu_2, \ldots, \nu_n) < (\tilde{\nu}_1,  \tilde{\nu}_2, \ldots, \tilde{\nu}_n)$$
if there is an $m \in {\Bbb N}$ such that
$\nu_j = \tilde{\nu}_j$ for each $j = 1, 2, \ldots, m-1$
and $\nu_m < \tilde{\nu}_m.$
This gives a well ordering of symmetric polynomials
in $n$ variables, that is, every set of symmetric polynomials in $n$
variables has  a minimal element.  Now use induction on $\sigma(f)$. \qed

P6, L10 Pisot should replace Salem. (For the name of the numbers).

P8: The basic Lagrange interpolation polynomials are
    denoted by $l_k$ here. Later, in Chapter 6 (although in the
    abstract Chebyshev space setting) these are denoted by
    $\ell_k$. It may be better to be consistent about it.


P9, L-3, L-5   change $z_i$ to $x_i$ (three times)

P9, L-3  change $p(z_i)$ to $p^(j)(x_i)$

P18, L4  there should be a power of two  on the last expression.

P18, c] for $1<m<n$

P18  E.7 b], p should be $p$.

P23, L14 replace $f^{\prime}_k$ by $f^{\prime}_{k-1}$

P23, L-1 replace $\beta_i\gamma_i$ by $-\beta_i\gamma_i$

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CHAPTER 2

P31, L18 there is an extra bracket in front of $\cos$

P37, E.10 h]  delete the $(-1)^k$ factor or leave it in and
replace  $(x^2-1)^k$ with $(1-x^2)^k$

P46, E.2 This exercise makes sense, but it is really for REAL C[a,b].

P47, E.4, b] $<f_i,f>$ should be $<f,f_i>$ in two places (at least the first one
needs changing).

P83  E.1, b] ... then p has no zeros in the cone
     {z\in C:|arg(z)|< \pi/n} (Ie replace 2n by n)

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CHAPTER 3

P93, L4 : $i=1,2,\ldots, n$ should be $i=0,1,\ldots, n$.

P94, line -12: "suppose a alternation" should be "suppose that an
alternation".

P94, line -11: "for $g-p$" should be "for $g-p$ on $A$" and in
line -9 "where x_i \in A$ and" should be deleted.

P94, line -10: delete the comma after $x_1$.

P94, line -4: $< \ldots <$ should be $< \cdots <$.

P94, line -2: "with" should be replaced by
"with $\max Y_{i-1} < \min Y_i, i=1,2, \ldots, m$ and".


P95, line  -8: "Suppose there is" should be "Suppose
$p \in \text{span}\{f_0, \ldots , f_n\}$,  suppose there is"

P95, line -7: $p^*$ should be
$p^* \in \text{span}\{f_0, \ldots, f_n\}$.

P95, line -5: $p^*-p$ should be
$p^*-p \in \span\{f_0, \ldots, f_n\}$.


P100, line -3: $\{f_0, \ldots, f_m\}$ should be replaced by
$(f_0, \ldots, \f_m)$.

P104, line -2: $2m$ should be replaced by $2m+1$.

P110, L-1 n>0 should be n\ge 0.

P112, L-3 The span should be to f'_n (not n-1).

P119, Hint for c], Use part b] (not part a]).

P122, E.5: $T_{n,\lambda}$ and $T_{n,\gamma}$ should be defined
more carefully since the "sinh" system is not a Chebyshev system on
$[0,1]$, but only on $[\delta,1]$ with $\delta > 0$. So a limiting
argument is needed to talk about these function at all.


P123, L-5 Kro\'o and Szabados [94b]  (The b is missing).

P124, E.6 L6 By E.5 a], b] and c]  (That is add part c])

P124, E.6 L8 the right most formula should be  (-1)^m*2^(-n)*binomial(2*n,2*m)

P124, E.6 L9 replace d_{m,n} by |d_{m,n}|

P138, E.9 b] L_n does not need a (-1)^n in the definition.

P144, ThM 3.5.4 (L-2) A_{k,n} should always have a minus sign.

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CHAPTER 4

P164, line -5: The normalization of $S _n$ is not correct.
                $1/(2\pi)$ should be $1/\pi$

P172, the summation should start from $i=1$ rather than $i=0$, or else,
       we should say that $\lambda_0=0$ and $a_0=0$.

Lines 6 - 8 on page 172 should be reformulated.
The $C[0,1]$ case is NOT a special case of Theorem 4.2.4
(Full Muntz Theorem in $L_p[0,1]$), where only $1 \leq p < \infty$
is allowed.


P175, E.4 a] Assume sequence tends to infinity.

P178, E.3.b] (L-1)  This should just read "Deduce that"
             (L-2) in the product \lambda_i \neq \lambda_m
             (L-3) there are extraa brackets around (\lambda_m)
             (L-4) the span should include all $x^{\lambda_i}$, $i \in {\Bbb N}, i \neq m$.

P178, E.3 e] (L-7) change a_{i,n} to a_{i} in definition of f. Also we have to say somewhere that 
$a_{i,n} := 0$ for $i > k_n$.



P190, E.12 c] The zeros should be distinct.

P198, line 5: $3$ should be $12$.
P199, line 4: $3$ should be $12$.
P199, line -3: $3$ should be $12$.
P199, line -2: the first $3$ should be $12$, while the second $3$ should be $3/4$.

Changes to E.19 of Section 4.2.

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page 202, line -1: Add the following.

Choosing a subsequence if necessary, without loss
of generality we may assume that one of the following two
cases occurs.

$\noindent {\bf Case 1:}$ We have $\lambda_i > 0$ for each $i=0,1,2, \ldots$
with $\sum_{i=0}^\infty {\frac {1}{\lambda_i}} \= \infty \,.$

$\noindent {bf Case 2:}$ We have $-1 < \lambda_i < 0$ for each $i=0,1,2, \ldots$
with $\sum_{i=0}^\infty {(\lambda_i + 1)} \= \infty \,.$

In Case 1 we know from Theorem 4.2.1 (Full M\"untz Theorem in
$C[0,1]$) that ${\rm span} \{1, x^{\lambda_1}, x^{\lambda_2} \ldots \}$
is dense in $C[0,1]$. Show that this implies that
${\rm span} \{x^{\lambda_1}, x^{\lambda_2} \ldots \}$ is dense in
$L_1[0,1]$.

In Case 2 we argue as follows.

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On page 203 the text between lines 9 and -1 should be replaced by

$$g(z) := (z+1)^2 f(z)$$
is a bounded analytic function on the open unit disk that satisfies
$$g(\lambda_i) = 0 \qquad {\text with} \qquad
$\sum_{i=0}^\infty}{(1-|\lambda_i|)} = \infty$$
Hence Blaschke's Theorem (E.12 e]) yields that $g=0$ on the open unit disk.
Therefore $f=0$ on the open unit disk. Now observe that $f$ is an analytic
function on the half plane $\{z \in {\Bbb C}: {\rm Re}(z) > -1\}$,
hence $f(z) = 0$ whenever ${\rm Re}(z) > -1$, so

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Changes to E.20 of Section 4.2.

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page 204, line 5: Add the following.

${\it Outline.}$ Choosing a subsequence if necessary, without loss
of generality we may assume that one of the following two
cases occurs.

$\noindent {\bf Case 1:}$ We have $\lambda_i > 0$ for each $i=0,1,2, \ldots$
with $\sum_{i=0}^\infty {\frac {1}{\lambda_i}} \= \infty\,.$

$\noindent {\bf Case 2:}$ We have $-1 < \lambda_i < 0$ for each $i=0,1,2, \ldots$
with $\sum_{i=0}^\infty {\left( \lambda_i + \frac 1p} \right)} \= \infty\,.$

In Case 1 we know from Theorem 4.2.1 (Full M\"untz Theorem in
$C[0,1]$) that ${\rm span} \{1, x^{\lambda_1}, x^{\lambda_2} \ldots \}$
is dense in $C[0,1]$. Show that this implies that
${\rm span} \{x^{\lambda_1}, x^{\lambda_2} \ldots \}$ is dense in
$L_p[0,1]$.

In Case 2 we argue as follows.

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On page 204 the text between lines -8 and -1 should be replaced by

Use H\"older's inequality to show that
$$g(z) := (z+1)^2 f \left( z - 1 + \textstyle{\frac 1p} \right)$$
is a bounded analytic function on the open unit disk
that satisfies
$$g(\lambda_i) = 0 \qquad {\text with} \qquad 
\sum_{i=0}^\infty}{(1-|\lambda_i|)} = \infty$$
Hence Blaschke's Theorem (E.12 e]) yields that $g=0$ on the open unit disk.
Therefore $f=0$ on the open disk with diameter
$[-\frac 1p, 2-\frac 1p]$
Now observe that $f$ is an analytic
function on the half plane
$\left\{z \in {\Bbb C}: {\rm Re}(z) > -\frac 1p \right\}\,,$
hence $f(z) = 0$ whenever ${\rm Re}(z) > -\frac 1p$, so

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P205, E.20 c] change \lambda_t to \lambda_1 t

P221, E.1 b] (L-9) change  by E.3 c] to E.4 c].

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CHAPTER 5

The proof of Theorem 5.1.2 is flawed.  The result is correct and a correct
proof is in ``Remez-type inequalities on the size of generalized polynomials''
in J. London Math. Soc., 45 (1992), 255-264.

P232, L11: a parenthesis is missing in sign(...)).

P233, L4 \theta should be \tau

p.236, The conclusion of E.4 is incorrect. The exercise can be saved 
by the additional assumption "Suppose also that $f$ is monotone on each 
interval  $[x_1,x_2], [x_2,x_3], \ldots, [x_{2n},x_{2n+1}]$ where 
$x_{2n+1} := x_1 + 2\pi$. 

P236, line +6: delete one ) from $f(\theta))$.
P236, line +6: add one ) in $\text{sign}(g^{\prime} (\theta)$.

P237, L-2 change follows to follow.

P242, L6 the absolute value sign should be inside the summation.

$E_n(\Lambda)$ is defined differently on pages 276 and 289.
(Since these are local notations, I think it is not a problem.)
page 276, line -7:  := should be =.

I think our presentation of the proof of the Newman Inequality does not
really cover the trivial case when $\lambda_0 = 0$ and $\lambda_1 = 1$
(see (6.1.5) when the contour is changed for $|z| = 1$. One simple way
to avoid this to normalize so that the sum of the exponents $\lambda_j$
is $1 - \epsilon$ rather than $1$.

P243, in lines 7 and 8 $s^\widetilde$ should be
replaced by $s_n^\widetilde$ three times.

P244, in lines 9 and 11 $s^\widetilde$ should be
replaced by $s_n^\widetilde$ twice.

P244, in line 10 $s$ should be replaced by $s_n$
three times.

P281, L-4 the square bracket after the infinity is wrong.

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CHAPTER 6

P288, in line 3,
 $$\lambda_k := 1 + \frac{(k-1)u}{v}\,, \qquad k=1,2, \ldots \,.$$

P288,in line 5, $M_{nv-v}(\Lambda)$ should be
$M_{nv-v+1}(\Lambda)$.

On page 288 the text between 8 and 18 should be replaced by
the following.

Without loss of generality we may assume that
$1/2 < \delta < u/v < 1$. Observe that $p_n$ is of the form
$$p_n(x) = xr_{nv-v}(x^{u/v}) = (x^{u/v})^{v/u}r_{nv-v}(x^{u/v})$$
with an $r_{nv-v} \in {\Cal P}_{nv-v}$. Use A.4.8 (Markov
Inequality for ${\rm GAP}_N$) to deduce that there is an
absolute constant $c_1 > 0$ such that
$$\|p\|_{[0,1]} = \|p\|_{[y,1]}$$
with
$$y:= (c_1v^2n^2)^{-v/u}\,.$$
Hence, using the definition of $p_n$, we obtain
$$p_n\|_{[0,1]} \leq 2^v y^{1-u} = 2^v(c_1v^2n^2)^{(u-1)v/u}i\,.$$
Therefore
$$|\frac{p_n^{\prime}(0)|}{\left( \sum_{j=0}^{nv-v+1}{\lambda_j} \right)
\|p_n\|_{[0,1]}} \geq
\frac{(2n^2)^v}{\left( \sum_{j=0}^{nv-v {\left(1 + \frac{ju}{v} \right)
\right 2^v(c_1v^2n^2)^{(u-1)v/u}}
\geq \frac{(n^2)^{v/u}}{(1+nu)nv} c_1^{(1-u)v/u} \rightarrow \infty$$
as $n \rightarrow \infty$.


P290, L+7 The maximum should be taken over $f$ rather than $p$.

P293, L+8 $\|Q_{2n,\epsilon}\|_{[-1,1]} \leq e^{n\epsilon t}$
t should be dropped, of course.

P293, L+ 11 $2n-1-n\epsilon$ should be $\frac{\epsilon}{e^\epsilon - 1}\,(2n-1)$.

P301 in lines -11 and -13, E.3 g] must be E.4 g]

P304, In Lemma 6.2.3: "if" should be "for" three times.

The proof of Lemma 6.2.3 in our book should contain some  more
details. At least when I reread it recently, it gave me some difficulty to
reproduce what we mean. I suggest making the proof a bit longer as follows.

\demo{Proof}  It is sufficient to prove the lemma in the case that there
is an
index $m$ so that $1 \leq m \leq n$, $m \neq k$, and
$$\matrix \format \c & \quad \l\\
x_j = \widetilde{x}_j & \text{if } j = 0, 1, \ldots, n\,; \quad j \neq m\,;\\
x_m < \widetilde{x}_m & \text{if } m < k\,;\\
x_m \, > \widetilde{x}_m & \text{if } m > k\,.
\endmatrix $$
The general case of the lemma then follows from repeated applications of
the above special cases. We may also assume that $m < k$; the case $m > k$
is similar. Note that in the above special cases
$$\ell_k - \widetilde{\ell}_k \in M_n(\Lambda) $$
has a zero at each of the points
$$x_0, x_1, \ldots, x_{m-1}, x_{m+1}, x_{m+2}, \ldots, x_n\,,$$
hence it changes sign at each of these points, and does not have any other
zeros in $[0,\infty)\,.$

It is also obvious that
$$\text{sign}(\ell_k(x)) = \text{ sign} (\widetilde{\ell}_k(x))\,, \qquad
x \in [0, x_0]\,, $$
$$\text{sign}(\ell_k(x)) = \text{ sign} (\widetilde{\ell}_k(x))\,, \qquad
x \in [x_m, \widetilde{x}_m]\,, $$
and
$$|\ell_k(x)| < |\widetilde{\ell}_k(x)|\,, \qquad x \in (x_{m-1},
\widetilde{x}_m)\,, $$
where $x_{-1} := 0$.
Collecting our observations, we deduce that
$$|\ell_k(0)| \leq |\widetilde{\ell}_k(0)|\,,$$
and the lemma is proved.
\qed \enddemo


P305, L+2: $1 \leq m \leq n$ is replaced by $0 \leq m \leq n$.

P305, L+3: "if" should be "for".

P305, L+13: comma is replaced by period.

P305, L+14: "which" is replaced by "This".

P305, L-11: "which finishes the proof" is replaced by
      "otherwise $\ell_k - \tilde{\ell}_k$ would have a zero in $(x_{m-1},x_{m+1})$,
      where $x_{-1} := 0$ and $x_{n+1} := \infty$, a contradiction."
      (We have to say where the assumption is used.)

P305, In Lemma 6.2.4: "if" should be "for" three times.

P306, lines +10,11,12: "if" should be "for".

P306, line 8: $\beta x$ should be $x/\beta$

P307, end of proof: we may think of adding the short justification of
      each line in the last sequence of inequalities, by copying the arguments
      from the JAMS paper.

P307, lines 3 and 4: the parentheses outside should be
replaced by absolute value signs.

It seems to that the technical Lemma 6.2.6 can be completely dropped
from the proof of Remez in the book.

Observe that in Lemmas 6.2.3, 6.2.4, and 6.2.5, the point $0$
can be replaced by any $y \geq 0$ as long as $y$ is to the left
of each of the points $x_j$. This would imply a few minor
changes. One has to be a bit careful with Lemmas 6.2.4 and 6.2.5. For
example the conclusion of Lemma 6.2.4 would be
$$\ell_k(y) \leq \widetilde{\ell}_k(0)\,.$$ (Check the proof.)
Similarly, in Lemma 6.2.5 we should estimate $|p(y)|$ by the
value of the Chebyshev polynomial STILL at $0$.

This makes the book half a page shorter. But more importantly, when
you actually teach this, it makes a huge difference to be able to
omit the ugly detail formulated in Lemma 6.2.6.

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CHAPTER 7

P338, L4 f should be \tilde f


P338, the last paragraph in the hint to E.8 should read :

"To estimate $|f^{\prime}(y)|$ for $y \in [-1,1]$, without loss of
generality we may assume that $y \geq 0$. Now use a linear shift...

P339, L-12 ASR^C_n is redundently defined.

P342, line 2: A comma is missing in front of n.

P343, line 6: A comma is missing in front of n.


P347, L-7 change less than or equal to $2\Pi$ to less than

APPENDIX 1

P361, L-2  O(N) should be O(n)

P364, f3]  x^(2*m) should be x^(2^m)

APPENDIX 3

P385, L-8 $L(P^*)$ should be $L(p^*)$.

P386, E.1  "co" should be plain in $co(A)$ in the second line of E.1.

P389, E.2 change Reformation to Reformulation in the title

APPENDIX 1

P361, L-2  O(N) should be O(n)

P364, f3]  x^(2*m) should be x^(2^m)

APPENDIX 3

P385, L-8 $L(P^*)$ should be $L(p^*)$.

P386, E.1  "co" should be plain in $co(A)$ in the second line of E.1.

P389, E.2 change Reformation to Reformulation in the title

APPENDIX 5

P423, line 15: "see E.12" should be "see E.11".


p432, part c] of E.10 should be replaced by 

###############################################################################

\noindent {\bf c]} Suppose $p \in {\Cal P}_{n,0}$ has only real zeros, $p^{\prime}$
does not vanish in $[-1,1]$, $p(-1) = 0$, and $p(1)=1$. Show that
$$\|p^{\prime}\|_{[-1,1]} \leq \frac{n}{2} \, \left( \frac{n}{n-1} \right)^{n-1} \|p\|_{[-1,1]}\,.$$

\noindent{\it Hint:}
Let the zeros of $p$ be denoted by $x_1 = -1,x_2,\ldots,x_n$.
For $x \in [-1,1]$ we have
$$\split p(x) & = \frac{p(x)}{p(1)} = \frac{1+x}{2} \prod_{j=2}^n{\frac{x-x_j}{1-x_j}} = \frac{1+x}{2} \prod_{j=2}^n{\left( 1 - \frac{1-x}{1-x_j} \right)} \cr
& \leq \frac{1+x}{2} \, \left( \sum_{j=2}^n{\left( 1 - \frac{1-x}{1-x_j} \right)} \right)^{n-1} 
\leq \frac{1+x}{2} \, \left( \frac{1}{n-1} \left( n-1 -\sum_{j=2}^n{\frac{1-x}{1-x_j}} \right) \right)^{n-1} \cr
& \leq \frac{1+x}{2} \, \left( \frac{1}{n-1} \left( n-1 + \frac{1-x}{2} \right) \right)^{n-1} 
\leq \frac{1+x}{2} \, \left(1 +\frac{1}{n-1} \right)^{n-1} \cr
& \leq \frac{1+x}{2} \, \left( \frac{n}{n-1} \right)^{n-1}\,. \cr \endsplit $$
Here we used the inequality to compare the geometric and arithmetic means of nonnegative numbers,
and we also used the fact that
$$\frac 12 + \sum_{j=2}^n{\frac{1}{1-x_j}} = \frac{p^{\prime}(1)}{p(1)} \geq 0\,.$$
We conclude that
$$\split 0 \leq p^{\prime}(x) & = p(x) \, \sum_{j=1}^n{\frac{1}{x-x_j}} \leq 
\frac{1+x}{2} \, \left( \frac{n}{n-1} \right)^{n-1} \sum_{j=1}^n{\frac{1}{x-x_j}} \cr  
& \leq \frac{1+x}{2} \, \left( \frac{n}{n-1} \right)^{n-1} \frac{n}{x+1} \cr 
& \leq  \frac{n}{2} \, \left( \frac{n}{n-1} \right)^{n-1} \|p\|_{[-1,1]}\,. \cr \endsplit $$

################################################################################


P438, b] surely it should be n \|p\|_D in the conclusion

P438, line -10 what Lemma? The words of the lemma should be removed

P439, line 2: there is an extra || in the norm

p440, line 12: "\|p\|_{[-1,1]}" should be "\|p\|_{[0,1]}".  

p441, line 11: "|r(c)|" should be "|r(d)|".

p441, line -8: in the denominator of the second ratio "|q(c)|" should be "(c-a)|q(c)|". 

P446, E.21 Then each closed disc OF RADIUS ONE

P447, E.22 d] $z+1$ should be $z-1$ in  line -6

P456, In the reference to He and Li  the journal is the Rocky Mountain J.







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BIBLIOGRAPHY

Change A.M.S. to Amer. Math. Soc. in "The $L_p$ Version of Newman's Inequality for lacunary polynomials.''
also make the and "L" lowercase.

In Borwein, Chen and Dilcher the volume is 47 (not 11)




