Let $E$ be the vector space of continuous functions on $[0,1]$. Show that the $L^1$-norm is not equivalent to the $L^2$-norm.
My thought was that, given a sequence of functions $f_n\in E$ which converges to the function $\frac{1}{\sqrt{x}}$, we can see that $$||f_n||_1=\int_0^1|f_n|dx \to \int_{0}^1 \frac{1}{\sqrt{x}}dx=2\sqrt{0}+2\sqrt{1}=2 $$ However, $$||f_n||_2=\left(\int_{0}^1 (f_n)^2 dx \right)^{1/2}\to \left(\int_0^1 \frac{1}{x}dx\right)^{1/2}$$ Since this sequence converges with respect to one norm but not the other we can conclude that they are not equivalent.
Does this argument make any sense? It feels like it doesn't make sense to talk about the norm of a function that isn't in the space $E$ since $\frac{1}{\sqrt{x}}\notin E$. But the hint for the problem says to consider truncating said function near 0.
Just consider the sequence $\{f_n\}_{n \ge 1} \subset C[0,1]$ defined by $f_n(x)=n x^n$
Then $||f_n||_1 = \int_{0}^1 nx^ndx=\frac{n}{n+1}$ . Hence $\lim_{n \to \infty} ||f_n||_1=1$
But $||f_n||_2=\int_{0}^1 |nx^n|^2dx=\frac{n^2}{2n+1} \to \infty \text{ as } n \to \infty$