$L^2$ is a subspace of $L^1$

Question

I know that $L^2\subset L^1$ and also $L^1, L^2$ are vector spaces so one thing is clear $L^2$ is a subspace of $L^1$. I was reading the definition of subspace of a normed linear space and it says that, Let $(X,||.||_X)$ be a normed linear space, then $Y\subset X$ is a subspace of a normed linear space if it is a vector space and the norm define in it is $||.||_{X|Y}$ that the same norm defined in $X$ but restricted to $Y$.
My question is we have $L^2$ is a subspace of $L^1$ and $L^1$ is a normed linear space with norm $||.||_1$ and $L^2$ is also a normed linear space with norm $||.||_2$ then how we will get here $||.||_2=||.||_{2|1}$,that is we should get the norm of $L^2$ by restricting the norm of $L^1$.Where I am making mistake ,please help.

Answer 1

First, as mentioned in comments, $L^2$ isn't subspace of $L^1$ in general. If there is an infinite sequence $X_i$ of sets that are pairwise disjoint and each has measure greater than $\varepsilon$, then $f(x) = \sum_i \frac{\mathbb I_i(x)}{i\cdot\sqrt{\mu(X_i)}}$ is in $L^2$, but not in $L^1$.

Second, there is a conflict of definitions.

You can consider algebraic definition of subspace, that cares only about elements and algebraic operations (sum of vectors and multiplication by scalar). In this sense, $L^2$ is subspace of $L^1$ for finite measure spaces.

Or you can consider definition of subspace for normed spaces, that also requires norm on subspace to be induced by norm on the space. In this sense, $L^2$ isn't subspace of $L^1$. Moreover, norm in $L^2$ even isn't equivalent to norm on $L^1$ (and $L^2$ isn't even closed wrt $L^1$ norm).