<s>[INST] Show that: $\frac{1}{c}\sqrt{h} \Big\|\begin{pmatrix} f(0) \\f(h) \end{pmatrix} \Big\|_2 \le \Big\|f\Big\|_{L^2} \le c\sqrt{h}\ldots$ Let $\mathcal P:=\left\{f:\Omega \rightarrow \mathbb R \space \vert  \space f(x)=ax+b, a,b \in \mathbb R \right\}$ with $\Omega:=[0,h], h\in \mathbb R_+$. I want to show this inequality holds for some suitable constant $c\in \mathbb R$ and I want to determine this constant $c$. I have seen this before in equivalent norms but I am having some trouble with the arithmetic.
$$\frac{1}{c}\sqrt{h} \Big\|\begin{pmatrix} f(0) \\f(h) \end{pmatrix} \Big\|_2 \le \Big\|f\Big\|_{L^2\left( \Omega\right)} \le c\sqrt{h}\Big\|\begin{pmatrix} f(0) \\f(h) \end{pmatrix} \Big\|_2$$
My attempt:
$$\begin{aligned}&\frac{1}{c} \sqrt{h}\sqrt{\left(f(0)\right)^2+\left(f(h)\right)^2}\le\int_0^h (ax+b)^2 dx\le c\sqrt{h} \sqrt{\left(f(0)\right)^2+\left(f(h)\right)^2} \\ &\iff \frac{1}{c}\sqrt{h}\sqrt{b^2+(ah+b)^2} \le \int_0^h (a^2x^2+2abx+b^2) dx \le c\sqrt{h}\sqrt{b^2+(ah+b)^2} \\ &\iff \frac{1}{c}\sqrt{a^2h^3+2bh^2+2b^2h}\le  \int_0^h (a^2x^2+2abx+b^2) dx \le  c\sqrt{a^2h^3+2bh^2+2b^2h} \\ & \iff \frac{1}{c}\sqrt{a^2h^3+2bh^2+2b^2h}\le a^2\left[\frac{x^3}{3}\right]_0^h+ ab\left[x^2\right]_0^h+b^2 \left[x\right]_0^h \le  c\sqrt{a^2h^3+2bh^2+2b^2h} \\ & \iff \frac{1}{c}\sqrt{a^2h^3+2bh^2+2b^2h}\le \frac{a^2h^3}{3}+abh^2+b^2h \le c\sqrt{a^2h^3+2bh^2+2b^2h} \\ & \iff \ldots?\end{aligned}$$
I feel like I am very close but I am not sure how to proceed. Can $c$ be any expression involving $a,b,h$? Getting rid of the the square roots seems to be the problem. Is this even the right approach or should I have used some inequality (Cauchy-Schwarz, etc.) earlier to simplify the $L^2$-norm after squaring the entire inequality?
 [/INST] I made a mistake writing down the $L^2$-Norm. We have:
$$\vert \vert f \vert\vert_{L^2} = \left( \int_0^h \vert f(x) \vert^2dx \right)^{\frac{1}{2}} $$
Therefore the inequality becomes:
$$\frac{1}{c}\sqrt{a^2h^3+2bh^2+2b^2h}\le \sqrt{\frac{a^2h^3}{3}+abh^2+b^2h} \le c\sqrt{a^2h^3+2bh^2+2b^2h}$$
Let $c=\sqrt{4}=2$
$$\implies \sqrt{\frac{a^2h^3}{4}+\frac{bh^2}{2}+\frac{b^2h}{2}}\le \sqrt{\frac{a^2h^3}{3}+abh^2+b^2h} \le \sqrt{4a^2h^3+8bh^2+8b^2h}$$
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