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$begingroup$

Let $f(t)=sf min_xin [0,1](P(x)+tQ(x))$ where $tinmathbb R$, and $P,Q$ are two real polynomials of degree $leq n$. I have two questions :

(1) is it true that $mathbb R$ can be partitioned into finitely many intervals on which $f$ is monotonic ?

(2) If the answer to (1) is YES, can we control the size of the partition (i.e. the number of intervals) ? An upper bound of $Constanttimes n$ seems reasonable.

My thoughts : the answer to (1) is YES when $n=2$. Indeed, if we put $F_t(x)=P(x)+tQ(x)$, we have a first partition $mathbb R=I_- cup I_+$ such that $F_t$ has a nonpositive degree two coefficient (in $x$) when $tin I_-$ and $F_t$ has a positive degree two coefficient when $tin I_+$.

In turn, we have a partition $I_-=I_-,0 cup I_-,1$ such that the minimum of $F_t$ on $[0,1]$ is attained at $e$ when $tin I_-,e$ (for $e=0$ or $1$), and a partition $I_+=I_+,m lt 0 cup I_+,0 leq m leq 1 cup I_+,m gt 1$ where $m$ denotes the unique value in $mathbb R$ at which $F_t$ reaches its minimum when $tin I_+$.

So when $n=2$, we end up with a partition into $2+3=5$ intervals.

real-analysis polynomials

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asked Mar 21 at 8:19

Ewan DelanoyEwan Delanoy

42k443104

$endgroup$

add a comment | 

0

$begingroup$

Let $f(t)=sf min_xin [0,1](P(x)+tQ(x))$ where $tinmathbb R$, and $P,Q$ are two real polynomials of degree $leq n$. I have two questions :

(1) is it true that $mathbb R$ can be partitioned into finitely many intervals on which $f$ is monotonic ?

(2) If the answer to (1) is YES, can we control the size of the partition (i.e. the number of intervals) ? An upper bound of $Constanttimes n$ seems reasonable.

My thoughts : the answer to (1) is YES when $n=2$. Indeed, if we put $F_t(x)=P(x)+tQ(x)$, we have a first partition $mathbb R=I_- cup I_+$ such that $F_t$ has a nonpositive degree two coefficient (in $x$) when $tin I_-$ and $F_t$ has a positive degree two coefficient when $tin I_+$.

In turn, we have a partition $I_-=I_-,0 cup I_-,1$ such that the minimum of $F_t$ on $[0,1]$ is attained at $e$ when $tin I_-,e$ (for $e=0$ or $1$), and a partition $I_+=I_+,m lt 0 cup I_+,0 leq m leq 1 cup I_+,m gt 1$ where $m$ denotes the unique value in $mathbb R$ at which $F_t$ reaches its minimum when $tin I_+$.

So when $n=2$, we end up with a partition into $2+3=5$ intervals.

real-analysis polynomials

share|cite|improve this question

asked Mar 21 at 8:19

Ewan DelanoyEwan Delanoy

42k443104

$endgroup$

add a comment | 

$begingroup$

Let $f(t)=sf min_xin [0,1](P(x)+tQ(x))$ where $tinmathbb R$, and $P,Q$ are two real polynomials of degree $leq n$. I have two questions :

(1) is it true that $mathbb R$ can be partitioned into finitely many intervals on which $f$ is monotonic ?

(2) If the answer to (1) is YES, can we control the size of the partition (i.e. the number of intervals) ? An upper bound of $Constanttimes n$ seems reasonable.

My thoughts : the answer to (1) is YES when $n=2$. Indeed, if we put $F_t(x)=P(x)+tQ(x)$, we have a first partition $mathbb R=I_- cup I_+$ such that $F_t$ has a nonpositive degree two coefficient (in $x$) when $tin I_-$ and $F_t$ has a positive degree two coefficient when $tin I_+$.

In turn, we have a partition $I_-=I_-,0 cup I_-,1$ such that the minimum of $F_t$ on $[0,1]$ is attained at $e$ when $tin I_-,e$ (for $e=0$ or $1$), and a partition $I_+=I_+,m lt 0 cup I_+,0 leq m leq 1 cup I_+,m gt 1$ where $m$ denotes the unique value in $mathbb R$ at which $F_t$ reaches its minimum when $tin I_+$.

So when $n=2$, we end up with a partition into $2+3=5$ intervals.

real-analysis polynomials

share|cite|improve this question

asked Mar 21 at 8:19

Ewan DelanoyEwan Delanoy

42k443104

$endgroup$

share|cite|improve this question

asked Mar 21 at 8:19

Ewan DelanoyEwan Delanoy

42k443104

share|cite|improve this question

asked Mar 21 at 8:19

Ewan DelanoyEwan Delanoy

42k443104

asked Mar 21 at 8:19

Ewan DelanoyEwan Delanoy

42k443104

asked Mar 21 at 8:19

Ewan DelanoyEwan Delanoy

42k443104