It seems that this is a tough problem clearly $ \frac{1}{2}\ge\ h \ge\frac{1}{98}$

Don't you mean $ h = \frac{1}{98}$? It obviously can't be more, considering the extremal case $ P(x) = x(x - \frac {1}{98})(x - \frac {2}{98})...(x - 1)$

cosinator, in your example more than $ \frac {1} {98}$ can be achieved. One can look at larger lengths outside the interval between two consecutive roots!

Let $ p$ be a polynomial of degree $ n$ such that $ p(0)=P(1)=0$. Suppose there exists a $ r$ such that there exists no $ a,b ,0\le\ a\le\ b \le\ 1$ with $ p(a)=p(b)$, $ b-a=r$ I will prove that the distance between two consecutive roots of $ p(x)$ in $ [0,1]$ is less than or equal to r. suppose there exists two consecutive roots such that $ a,b$, $ a-b>r$ consider the polynomial $ g(x)=p(x)-p(x+r)$ $ g(b).g(a-r)>0$ gives $ p(b+r)p(a-r)<0$ so there is a root between $ b+r$ and $ a-r$. contradicting the fact that $ a$ and $ b$ are consecutive roots. now since a polynomial of degree $ n$ can have only atmost $ n$ roots in $ [0,1]$,and since there are two roots in $ [0,1]$ $ (n-1)r\ge 1$ gives $ r\ge\frac{1}{n-1}$ but clearly $ h>\frac{1}{n-1}$ for $ n\ge 4$ (Is it? ) Also for degree 4 ,$ h=\frac{1}{2}$

Can anyone give a solution or a hint for this hard problem?

Here is a possible approach.... The general solution will be $\frac{1}{\lfloor \frac{n-1}{2}\rfloor\ +1 }$ where $n$ is the degree of polynomial step 1:proving that $h\geq \frac{1}{\lfloor \frac{n-1}{2}\rfloor \ +1}$ (*) suppose there exists an $r$ such that there exists no $a,b,0\leq a,b\leq 1$ with $p(a)=p(b)$ and $b-a=r$ (1) i will prove that the distance between two consecutive roots of $p(x)$ in $[0,1]$ is not greater than $r$ (2) consider the polynomial $g(x)=p(x)-p(x+r)$ since $a,b,b+r,a-r\in\ [0,1]$ , $g(b).g(a-r)>0$ [If it was less than or equal to zero,then $g(x)$ has a root in $[0,1-r]$,contradicting assumption (1)] so $p(b+r).p(a-r)<0$ which means $a$ and $b$ are not consecutive roots which is a contradiction.This proves (2) Now let $ \frac{1}{k+1}\leq r\leq \frac{1}{k}, k\in \mathbb{N}$ since $g(x)$ has no roots in $[0,1-r],$ assume WLOG $g(x)>0$ for all $x\in \ [0,1-r]$ For $ m\in \{1,2,\text{...}k-1\}$,we have $\sum _{i=1}^m g((i-1)r)>0\Longrightarrow p(\text{mr})<0$ Now $p(x)$ has a root in $(0,r)$, but $p(mr).p(mr+r)>0$,this means $p(x)$ has atleast two roots in $(mr,mr+r)$ which means $p(x)$ has atleast $2+1+2(k-1)$ zeros $\Longrightarrow $ $2k+1\leq\n$ if $n$ is odd and $2k+2\leq\n$, if $n$ is even this proves (*) Step 2: proving that the functional equation $Q(x+r)-Q(x)=P(x),Q(0)=0$ where $P$ is a given polynomial, has a polynomial solution for $Q$ Let degree of $P$ be $n$ From the functional equation we can get the allowed values for $Q(r), Q(2r),........$ Now there exists a unique polynomial of degree $n+1$ passing through the points ${(0,Q(0)),(r,Q(r)),..........((n+1)r,Q((n+1)r))}$ I argue that this polynomial (let us call it $R(x)$)satisfies the functional equation for the reason below Let $R(x+r)-R(x)=T(x)$, now $T(x)$ is a polynomial of degree $n$. so T(x)-P(x) is also a polynomial of degree atmost $n$ and so have atmost $n$ real zeros or zero everywhere. But we have $T(0)-P(0)=T(r)-P(r)=...........=T(nr)-P(nr)=0$,so as said above $T(x)-P(x)$ is zero everywhere .So $R(x+r)-R(x)=T(x)=P(x)$, $R(x)$ satisfies the functional equation step 3 Consider the case when $n$ is odd, let $n=2m-1$ let $q(x)$ be the polynomial solution to equation $q(0)=0$, $q(x+\frac{1}{m})-q(x)=x(x-\frac{m-1}{m})((\prod _{i=1}^{m-2} (x-\frac{i}{m})){}^2)$ the solution has the following properties The solution as mentioned has degree $n$. $q(\frac{i}{m})=0,$ for $i=0,1,2,\text{...}m$ $q'(\frac{i}{m}+\frac{1}{m})=q'(\frac{i}{m})\neq 0$,for $i=0,1,2,\text{...}m-2$. [if it was zero,then $q(x)$ has atleast $2m$ linear factors which is a contradiction] Also since $q'(x)$ has same sign at $\frac{1}{m}$ and $\frac{2}{m}$ we have a zero for $q(x)$ between $\frac{1}{m}$ and $\frac{2}{m}$ similarly for $\frac{2}{m}$ and $\frac{3}{m}$ and so on till $\frac{m-2}{m}$ and $\frac{m-1}{m}$ so we got all zeroes of $q(x)$. now since leading coefficent of $q(x)$ is positive,it is increasing at $0$.also $q'(x)$ is negetive at $\frac{1}{m}$.......$\frac{m-1}{m}$ .......(3) step 4 consider $q(x+\frac{1}{m}+\epsilon )-q(x)=x(x-\frac{m-1}{m})((\prod _{i=1}^{m-2} (x-\frac{i}{m})){}^2)+q(x+\frac{1}{m}+\epsilon )-q(x+\frac{1}{m})$, consider the set closed ${ S:[0,1-(\frac{1}{m}+\epsilon )]-\underset{i=1}{\overset{m-1}{\cup }}[\frac{i}{m}-\lambda ,\frac{i}{m}+\lambda ]}$ where $\lambda$ is a positive number less than half of min{$|a-b|:q(a)=0,q'(b)=0$} It is easy to note that if $\epsilon$ is less than half of min{$|a-b|:q(a)=0,q'(b)=0$},then for $i=0,1,.... m-1$ , $q(\frac{1+i}{m}+\epsilon )-q(\frac{1+i}{m})=\epsilon\ q'(c)$, for some $c\in [\q(\frac{1+i}{m}+\epsilon ),q(\frac{1+i}{m})]$ which is less than zero [because of (3)] which means $q(\frac{1+i}{m}+\epsilon )-q(\frac{i}{m})<0$ So,choose $\lambda$ such that $q(x+\frac{1}{m}+\epsilon )-q(x)$ is less than zero in $\underset{i=1}{\overset{m-1}{\cup }}[\frac{i}{m}-\lambda ,\frac{i}{m}+\lambda ]$ Note that $S$ is independent of $\epsilon$ if $\epsilon\leq\lambda$ Now $x(x-\frac{m-1}{m})((\prod _{i=1}^{m-2} (x-\frac{i}{m})){}^2)$ is negetive in $S$ and so has a negetive maximum $\alpha$ in that set Now choose an $\epsilon$ such that $ |q(x+\frac{1}{m}+\epsilon )-q(x+\frac{1}{m})|<-\alpha $ and $\epsilon\leq\lambda$ So in $S$ ,$q(x+\frac{1}{m}+\epsilon )-q(x)<0$ So $q(x+\frac{1}{m}+\epsilon )-q(x)<0$ in $[0,1-(\frac{1}{m}+\epsilon )]$ for all small enough $\epsilon$ Hence $h\leq \frac{1}{\lfloor \frac{n-1}{2}\rfloor \ +1}$ Similarly we can prove it for even $n$s

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Anybody has a shorter solution???