Can we approximate a function $f\in L^1([0,1])$ by a function $g\in L^2([0,1])$ in $L^2$ norm?
I am solving a problem and the problem asked me to approximate a function $f$ to $g$ in $L^2$ norm as stated above. It seems not true since there is no condition like $L^1\subset L^2$ or $L^2\subset L^1$...... neither space cannot be dense in another space... I have no idea... I will be thanking for any hint or answer!
First, since $[0,1]$ is a finite measure space, we have that $L^2([0,1])$ is a proper subset of $L^1([0,1])$ (see e.g. here).
However, if $f \in L^1([0,1]) \setminus L^2([0,1])$ then for any $g \in L^2([0,1])$, $f-g \not \in L^2([0,1])$ and in particular, $\|f - g\|_{L^2}$ isn't even defined so it doesn't make sense to try to approximate $L^1$-functions in $L^2$-norm.
It is worth noting that you can approximate $f \in L^1([0,1])$ by functions from $L^2([0,1])$ in $L^1$-norm. For example, the simple functions are dense in $L^1([0,1])$ and are a subspace of $L^2([0,1])$.
