Two norms on $C_b([0,\infty])$ $C_b([0,\infty])$ is the space of all bounded, continuous functions.
Let $||f||_a=(\int_{0}^{\infty}e^{-ax}|f(x)|^2)^{\frac{1}{2}}$
First I want to prove that it is a norm on $C_b([0,\infty])$. The only thing I have problems with is the triangle inequality, I do not know how to simplify 
$||f+g||_a=(\int_{0}^{\infty}e^{-ax}|f(x)+g(x)|^2)^{\frac{1}{2}}$
The second thing I am interested in is how to show that there are constants $C_1,C_2$ such that $||f||_a\le C_1||f||_b$ and $||f||_b\le C_2||f||_a$ so for $a>b>0$ the norms $||.||_a$ and $||.||_b$ are not equivalent.
 A: For triangular inequality, write, using Cauchy-Bunyakovsky-Schwarz,
\begin{align}\lVert f+g\rVert_a^2&=\int_0^{+\infty}e^{-ax}(|f(x)|^2+|g(x)|^2+2f(x)g(x))dx\\
&\small\leqslant\int_0^{+\infty}e^{-ax}(|f(x)|^2+|g(x)|^2)dx+2\left(\int_0^{+\infty}e^{-ax}|f(x)|^2dx\right)\left(\int_0^{+\infty}e^{-ax}|g(x)|^2dx\right)\\
&=\left(\lVert f\rVert_a^2+\lVert g\rVert_a^2\right)^2.
\end{align}
Let $0<a<b$. As $e^{-bx}\leqslant e^{-ax}$, we have for all $f\in C_b([0,+\infty))$ that $\lVert f\rVert_b\leqslant \lVert f\rVert_a$.
For non equivalence, take $f_n$ a function which is $e^{b/2x}$ on $[0,n]$, $0$ for $x\geqslant n+1$ and linear in $[n,n+1]$.
A: For the first one: Use
$$e^{-\alpha \cdot x} \cdot |f(x)+g(x)|^2 = \left|e^{-\frac{\alpha}{2} \cdot x} \cdot f(x)+ e^{-\frac{\alpha}{2} \cdot x} \cdot g(x) \right|^2$$
and apply the triangel inequality in $L^2$. 
Concerning the second one: For $a>b>0$ you have
$$e^{-a \cdot x} \cdot |f(x)|^2 \leq e^{-b \cdot x} \cdot |f(x)|^2 $$
i.e. $\|f\|_a \leq \|f\|_b$. The inequality $\|f\|_b \leq C \cdot \|f\|_a$ does not hold in general. To see this let $c \in (b,a)$ and
$$f_n(x) := \min\{n,e^{c \cdot x}\} (\in C_b)$$
Then $\|f_n\|_a < \infty$,$\|f_n\|_a \to c<\infty$, but $\|f_n\|_b \to \infty$.
A: For the triangle inequality, note  $\|f\|_a=\|e^{-ax/2}f\|_2$, so you can apply Minkowski. 
To see that the norms are not equivalent (and so such constants cannot both exist), let
$$
f_n(x)=\begin{cases}e^{bx/2},&\text{ if } x\in[0,n] \\ \text{line},&\text{ if } x\in[n,n+e^{-bn/2}] \\ 0,&\text{ if }x>n+e^{-bn/2}\end{cases}
$$
Then
$$
\|f_n\|_a\leq\left(\int_0^ne^{(b-a)x}\,dx +1\right)^{1/2}\leq\left(\frac1{b-a}+1\right)^{1/2},
$$
$$
\|f_n\|_b\geq\left(\int_0^n 1\right)^{1/2}=\sqrt n
$$
