For every polynomial $P$ of degree not greater than 2012 $\underset{3\leq t \leq 4}{\max}|P(t)|\leq C \underset{0\leq t \leq 1}{\max}|P(t)|$.

Question

Can you help me with the following?

Prove that there exists constant $C$ such that for every polynomial $P$ of degree not greater than 2012 $\underset{3\leq t \leq 4}{\max}|P(t)|\leq C \underset{0\leq t \leq 1}{\max}|P(t)|$.

I know that set of polynomials on $[a,b]$ is subset of $C[a,b]$ and that $C[a,b]$ is isometric isomorph to $C[0,1]$ for every $a,b$. So, $$ \underset{3\leq t \leq 4}{\max}|P(t)| = ||P||_{C[3,4]} = ||P||_{C[0,1]} = \underset{0\leq t \leq 1}{\max}|P(t)|,$$ so $C=1$.

I am very suspicious of this solution, can you tell me am I wrong and what should I do instead? Thank you!

Answer 1

The space of polynomials of degree less or equal $2012$ is finite dimensional (dimension is equal $2013).$ All norms on this space are equivalent. In particular the norms $$\|P\|_1=\max_{0\le x\le 1}|P(x)|,\qquad \|P\|_2=\max_{3\le x\le 4}|P(x)| $$ are equivalent. Therefore there is a constant $C$ satisfying the requirement. Moreover there is a constant $D$ such that $$\underset{3\leq t \leq 4}{\max}|P(t)|\geq D \underset{0\leq t \leq 1}{\max}|P(t)|$$

Remark For any $a<b$ we can interpret the polynomials as the subspace of $C[a,b]$ with norm $\|f\|_\infty =\displaystyle \max_{a\le x\le b}|f(x)|.$

Answer 2

$\textbf{Edit:}$ As Ryszard Szwarc pointed out there was a mistake so now the second try.

I think and explicit constant is easy to access threw watching shifted polynoms $$\underset{t \in [3,4]}{\max}|P(t)| = \underset{t \in [3,4]}{\max}|\sum_{k=0}^{2012}a_k\cdot t^k|=\underset{t \in [0,1]}{\max}|\sum_{k=0}^{2012}a_k \cdot(t+3)^k|\\=\underset{t \in [0,1]}{\max}|\sum_{k=0}^{2012}a_k \cdot \sum_{i=0}^k \binom{k}{i}t^{k-i}3^i|$$ $$ \leq \underset{t \in [0,1]}{\max}|\sum_{k=0}^{2012}a_k\cdot t^k|+\underset{t \in [0,1]}{\max}| \sum_{k=0}^{2012}a_k \cdot\sum_{i=1}^k \binom{k}{i}t^{k-i}3^i| \\ \leq \underset{t \in [0,1]}{\max}|P(t)|+ \sum_{k=0}^{2012}|a_k| \cdot\sum_{i=1}^k \binom{k}{i}3^i\\ $$ Choose a $C$ (only possible if $P(t)\neq0$) so that $$(C-1)\cdot\underset{t \in [0,1]}{\max}|P(t)|\geq\sum_{k=0}^{2012}|a_k| \cdot\sum_{i=1}^k \binom{k}{i}3^i$$ to get your upper bound.