# How localized can a polynomial be in the L1 norm?

Let $$0 be a parameter, $$\Omega = [-1,1]$$, and $$\Omega_s\subset \Omega$$ be a set of measure $$s$$. I would like to bound the following ratio from above:

$$\sup_{p\in\mathcal{P}_n} \frac{\int_{\Omega_s} |p(x)| dx}{\int_{\Omega \setminus \Omega_s} |p(x)| dx},$$

where $$\mathcal{P}_n$$ is the space of polynomials of degree $$\leq n$$.

I can partially answer this question with Markov's brother's inequality. It can be used to show that if $$0\leq s\leq 1/(6n^2)$$, then

$$\sup_{p\in\mathcal{P}_n} \frac{\int_{\Omega_s} |p(x)| dx}{\int_{\Omega \setminus \Omega_s} |p(x)| dx} \leq \frac{1}{2}.$$

However, the bound above seems very weak and only applies when $$s$$ is very small. Ideally, I would like to have an upper bound that is attained by some $$\Omega_s$$ and polynomial $$p$$. Thank you in advance.

EDIT: I was interested in thinking about the $$L_p$$ analogues of the Remez inequality (see Theorem 5.1.1 of [1]), which tells us how concentrated a polynomial can be in the $$L_\infty$$-norm. It shows that

$$\sup_{p\in\mathcal{P}_n} \frac{\|p(x)\|_{L_\infty(\Omega_s)}}{\|p(x)\|_{L_\infty(\Omega\setminus \Omega_s)}} \leq T_{n}\left(\frac{2+s}{2-s}\right),$$

where $$T_n$$ is the degree $$n$$ Chebyshev polynomial. The inequality is attained when $$\Omega_s = [1-s,1]$$ and one takes

$$p(x) = \pm T_n \left(\frac{2x+s}{2-s}\right).$$

PARTIAL ANSWER (added 27th December 2018): If one assumes that $$\Omega_s = [-1,1-s]$$ or $$\Omega_s=[-1+s,1]$$, then one can use the $$L_\infty$$-Remez inequality together with Nikolskii's inequality to obtain a bound. Intuitively, I imagine that $$\Omega_s = [-1,1-s]$$ and $$\Omega_s=[-1+s,1]$$ are worst-case intervals but I haven't managed to prove it.

By the Remez inequality, we have

$$\int_{\Omega_s} |p(x)| dx \leq s \sup_{x\in\Omega_s} |p(x)| \leq sT_{n}\left(\frac{2+s}{2-s}\right)\sup_{x\in\Omega\setminus\Omega_s} |p(x)|.$$

Moreover, if $$\Omega_s = [-1,1-s]$$ or $$\Omega_s=[-1+s,1]$$, then by Nikolskii's inequality (see Theorem 2.6 of [2]) we find

$$\sup_{x\in\Omega\setminus\Omega_s} |p(x)| \leq \frac{4n^2}{2-s}\int_{\Omega\setminus\Omega_s} |p(x)|dx.$$

We conclude that

$$\int_{1-s}^{1} |p(x)|dx \leq \frac{4sn^2}{2-s}T_{n}\left(\frac{2+s}{2-s}\right)\int_{-1}^{1-s} |p(x)|dx$$

for any polynomial of degree $$\leq n$$. This bound is not tight even in this restricted situation, though.

REFERENCES

[1] Peter Borwein and Tamás Erdélyi. Polynomials and polynomial inequalities. Vol. 161. Springer Science & Business Media, 2012.

[2] R. A. DeVore and G. G. Lorentz. Constructive approximation. Vol. 303. Springer Science & Business Media, 1993.

• The exact bound will depend on $\Omega_s$ and there is little hope to find it for arbitrary $\Omega_s$. – Alexandre Eremenko Dec 21 '18 at 16:08
• I am just asking for an upper bound on the ratio so I am happy with worst-case $\Omega_s$ (whatever that may be). – alext87 Dec 21 '18 at 19:22
• For $n=1$ it seems to be optimal when $(2-s-\sqrt{{2-s)^2+4})/2$ is a root of $p$. – user35593 Dec 22 '18 at 17:27
• Did you check Theorem A.4.10 p.401 of [1] ? – user111 Dec 25 '18 at 17:28
• That's certainly an interesting inequality. Though, I am not sure it can be used to answer the question above. Are you suggesting that there is a way to select the $\chi$ in Theorem A.4.10 so that the result applied to all polynomials of degree $\leq N$? – alext87 Dec 26 '18 at 23:27