# $|P(y)|\leq C \int_{-1}^1|P(x)|^2dx$

So this is an old pre lim question that I was stuck on. So suppose that $P(x)$ is a real-valued polynomial of degree $n$. Show that there exists a constant $C$ depending on $n$ such that $|P(y)|\leq C\int_{-1}^1|P(x)|^2dx$ for all $y\in (-1,1)$.

What I am trying to do is apply Cauchy Schwarz. So far I have the following

\begin{align} |P(y)|&\leq C_1\int_{-1}^1 |P(y)|dx\\ &\leq C_2\int_{-1}^1 |P(x)|dx\\ &\leq C_2(\int_{-1}^11dx)(\int_{-1}^1|P(x)|^2dx)^{1/2}\\ &\leq C_3(\int_{-1}^1|P(x)|^2dx)^{1/2} \end{align}

Basically its the second line I'm having trouble getting to. The change from integrating y to x. This is where I must use the fact our polynomials are of degree $n$, but how?

• It's a finite-dimensional real vector space, and there all norms are equivalent.... – Lord Shark the Unknown Aug 21 '17 at 4:56
• Are you referring to the vector space of polynomials of degree at most $n$? – thegamer Aug 21 '17 at 4:58
• If so then ${x^n,\cdots ,x,1}$ form a basis, so we just take the norm that is the supremum of the coefficients of $P$ correct? – thegamer Aug 21 '17 at 4:59
• Oh I get what you mean now. – thegamer Aug 21 '17 at 5:00
• However, there is a slight discrepancy. The inequality I gave above has $\int_{-1}^1 |P(x)|^2dx$ without the square root, so its not really a norm right? – thegamer Aug 21 '17 at 5:01

Legendre polynomials provide a base of $L^2(-1,1)$ with respect to the standard inner product. We also have $\left|P_n(x)\right|\leq 1$ for any $x\in[-1,1]$ (see these notes, from page 35), hence by assuming
$$P(x) = \sum_{k=0}^{n} c_k\,P_k(x) \tag{1}$$ for any $y\in(-1,1)$ we have:
$$\left|P(y)\right|\leq \sum_{k=0}^{n} |c_k|=\sum_{k=0}^{n}\sqrt{\frac{2k+1}{2}}\cdot\sqrt{\frac{2}{2k+1}}\left|c_k\right|\tag{2}$$ and by the Cauchy-Schwarz inequality: $$\left|P(y)\right|\leq \sqrt{\sum_{k=0}^{n}\frac{2k+1}{2}\sum_{k=0}^{n}\frac{2}{2k+1}c_k^2}=\sqrt{\frac{(n+1)^2}{2}\int_{-1}^{1}P(x)^2\,dx}\tag{3}$$ leading to: $$\left|P(y)\right|\leq \frac{n+1}{\sqrt{2}}\sqrt{\int_{-1}^{1}P(x)^2\,dx}\tag{4}$$ where $n$ is the degree of $P$. No inequality of the form $\left|P(y)\right|\leq C_n\int_{-1}^{1}P(x)^2\,dx$ may hold since it is not homogeneous: by replacing $P(x)$ with $\lambda P(x)$ the LHS gets multiplied by $|\lambda|$, but the RHS gets multiplied by $\lambda^2$.