Showing that $\|f\|_1$ is not equivalent to $\|f\|_2$

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 1

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.

Answer 2

HINT: To find a counterexample, think about $1/x$ on $(0,1]$.

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

Answer 3

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.)