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.
 A: Both ways will always be possible for any two norms you care to choose. This is since the space of polynomials of degree less than or equal to one is finite dimensional, hence all norms are equivalent. 
In this particular case, you've shown $\lVert ax + b \rVert_{L^2} \leq \lVert ax + b \rVert$. The other direction is not very pretty, but I think the following works. Pick $\frac{1}{2} < \lambda < \frac{1}{\sqrt{3}}$. Then we have that $\mu = \frac{1}{3} - \lambda^2 > 0$ and $\nu = 1 - \frac{1}{4\lambda^2} > 0$. We can then check that
\begin{align*}
\lVert ax + b \rVert_{L^2}
&= \left(\lambda a + \frac{1}{2 \lambda}b\right)^2 + \mu a^2 +  \nu b^2
\end{align*}
Now pick $c$ such that $c > 2$, $c \mu > 2$ and $c \nu > 3$. Then we have
\begin{align*}
\lVert ax + b \rVert_{L^2}
&= \frac{1}{c} \left[ c\left(\lambda a + \frac{1}{2 \lambda}b\right)^2 + c\mu a^2 +  c\nu b^2 \right] \\
&\geq \frac{1}{c} \left[ 2\left(\lambda a + \frac{1}{2 \lambda}b\right)^2 + 2 a^2 +  3 b^2  \right] \\
&= \frac{1}{c} \left[ 2\lambda^2 a^2 + 2ab + \frac{1}{2\lambda^2} b^2 + 2a^2 + 3b^2 \right] \\
&\geq \frac{1}{c} \left[2a^2 + 3b^2  + 2ab\right] \\
&= \frac{1}{c} \lVert ax + b \rVert^2
\end{align*}
Like I said, not very pretty. Hence why having a general theorem saying all norms on a finite dimensional space are equivalent is a nice thing.
A: \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 $2$
