Question: I have the following norms:
$$
\|f\|\_1=\int\_{t\_0}^{t\_1}\|f(t)\|\_2dt
$$
$$
\|f\|\_2=\sqrt{\int\_{t\_0}^{t\_1}\|f(t)\|\_2^2dt}
$$

I need to show their non-equivalence, i.e. that there do not exist numbers $a,b\in\mathbb R$, $a\ge b>0$ such that $b\|f\|\_2\le\|f\|\_1\le a\|f\|\_2$.

I would like to do the proof by myself (don't show the full proof please). I am almost sure this has to be a proof by contradiction, so I somehow need to write a few inequalities that give something like this:

$$
\begin{array}{rll}
\|f\|\_1 & = & \int\_{t\_0}^{t\_1}\|f(t)\|\_2dt \\
 & \le & ??? \\
 & \vdots & \\
 & \le & \text{something}\cdot\|f\|\_2 \\
\end{array}
$$

and show that this "something" can take a value that would contradict the assumption that an $a$ exists. However, I do not know how to make the journey from $\|f\|\_1$ to $\|f\|\_2$ using inequalities, primarily due to the presence of the square root. Could you please help me get started? Thanks a lot!

Answer: 
Think of
$$
f(t)=\frac{1}{\sqrt{t-t\_0}}.
$$
You can show that $f$ belongs to $L^1$ but not $L^2$. 

If you for some reason would like a sequence of functions $f\_n\in L^1\cap L^2$ such that
$$
\lim\_{n\to+\infty}\frac{\|f\_n\|\_{L^2}}{\|f\_n\|\_{L^1}}=+\infty,
$$
then you can cut the example above off, and consider
$$
f\_n(t)=\chi\_{(t\_0+1/n,t\_1)}\frac{1}{\sqrt{t-t\_0}}.
$$
Here $\chi\_{(t\_0+1/n,t\_1)}$ denotes the characteristic function of the interval $(t\_0+1/n,t\_1)$. (If you want, you might consider only large $n$ so that $(t\_0+1/n,t\_1)$ is not empty, but that is a minor detail.)
HINT: To find a counterexample, think about $1/x$ on $(0,1]$. 

HINT TO A HINT: $1/x$ is $f^2$, not $f$.