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: 
**Hint**: Try to construct a sequence of functions $f\_n$ in the vector space you work in such that $\frac{||f\_n||\_1}{||f\_n||\_2} < \frac{1}{n}$. This will show that a constant $b > 0$ such that $b ||f||\_2 \leq ||f||\_1$ for **all** $f$ does not exist.
HINT: To find a counterexample, think about $1/x$ on $(0,1]$. 

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