I want to show explicitly that $L^2[0,1]$ subspace of $L^1[0,1]$ has no internal points

Question

I want to show explicitly that $L^2[0,1]$ subspace of $L^1[0,1]$ has no internal points.

If S subspace of X, I have the following result:

S = X if and only if the interior of S $\neq \emptyset$

Then to prove that the interior is empty, I just need to prove S is different from X.

Can I just pick $f(x) = \frac{1}{\sqrt{x}}$ and say that it belongs to $L^1$ but not to $L^2$ and be done?

Is there something that I am missing?

Answer 1

No, you are not missing anything. If $X,Y$ are normed spaces, and $X$ is a proper subspace of $Y$, then $X$ has empty interior relative to $Y$. So if you show that $L^2[0,1]$ is a proper subspace of $L^1[0,1]$, you are done. The function $f(x)=\frac{1}{\sqrt{x}}$ is indeed an element of $L^1[0,1]$ that does not belong to $L^2[0,1]$.

Answer 2

I think 'explicitly' here means you have to show that if $f \in L^{2}$ then $B(f,\epsilon)$ is not contained in $L^{1}$ for any $\epsilon >0$. General theory is not supposed to be used here. For this just consider $f+\frac 1 {n\sqrt x}$ and show that this function lies in the ball $B(f,\epsilon)$ for $n$ sufficiently large but it does not belong to $L^{2}$ for any $n$.