<s>[INST] Show the equivalence of norms on polynomial space Let $\|ax+b\|^2:=2a^2+3b^2+2ab$ be a norm on the space of all polynomials of degree one, where $a,b\in\mathbb{R}$.
Hence, $\|ax+b\|_{L^2} ^2:=\int_0^1(ax+b)^2\,\mathrm{d}x$ be another norm ($L^2$-norm).
I want to show equivalence, in the sense that, for all polynomials of degree one
$\exists\, c>0\colon \|\cdot\|\le c\|\cdot \|_{L^2}$ holds and vice versa.
The first part was not too hard,
\begin{align*}
\|ax+b\|_{L^2}^2 &= \int_0^1 (ax+b)^2\,\mathrm{d}x = a^2/3+ab+b^2\le (a+b)^2-ab\le (a+b)^2+\frac{a^2+b^2}{2}\\
&\le \frac{3}{2}a^2+2ab+\frac{3}{2}b^2\le 2a^2+2ab+3b^2=\|ax+b\|^2
\end{align*}
Is there any way to do it better? I noticed, that I got $c=1$, is that in some way significant?
The other way was resistant against all my attempts.
$\|ax+b\|^2=2a^2+2ab+3b^2=\ ...$
Here, I thought that finding a real constant $\alpha$ such that $\alpha(a^3/3+ab+b^2)\ge 2a^2+2ab+3b^2$ would yield success, since then I can find a suitable antiderivative to connect the norms. I was not successful, nevertheless.
 [/INST] \begin{align*}
\|ax+b\|_{L^2}^2 &= \int_0^1 (ax+b)^2\,\mathrm{d}x = a^2/3+ab+b^2\le  a^2 + ab + b^2
\le
2(a^2+ab+b^2)  \le 2a^2+2ab+3b^2=\|ax+b\|^2
\end{align*}
Therfore c is atleast $2$
</s>