Definition of Fourier tranform I`ve reading the Deitmar's book  A first course in Harmonic Analysis.
In chapter three, section 5.
Before, I write some context.
Let $\rm L^1_{bc}(\mathbb R)$ be the set of all bounded continuous functions $f:\mathbb R\to\mathbb C$ satisfying
$$\| f\|_1:=\int_{-\infty}^\infty \lvert f(x) \rvert dx<\infty.$$
Similarly, $\rm L^2_{bc}(\mathbb R)$, but in this case
$$\| f\|_2:=\int_{-\infty}^\infty \lvert f(x) \rvert^2 dx<\infty.$$
Deitmar show that, $\rm L^1_{bc}(\mathbb R)\subseteq \rm L^2_{bc}(\mathbb R)$ and  $\rm L^2_{bc}(\mathbb R)$ is a pre-Hilbert space with the usual inner product.
Deitmar say: "By the lemma we see that $L^1_{bc}(\mathbb R)$ is a pre-Hilbert space. We write $L^2(\mathbb R)$ for its completation."
Plancherel's theorem say that
For every  $f\in \rm L^1_{bc}(\mathbb R)$ we have that  $\tilde f\in \rm L^2_{bc}(\mathbb R)$ and $$\| f\|_2=\|\tilde f\|_2.$$ In particular, the Fourier transform $f\mapsto \tilde f$ extends to a unitary map $L^2(\mathbb R)\to L^2(\mathbb R).$
$$For \ f\in L^1_{bc}(\mathbb R),\ \tilde f(y)=\int_{-\infty}^\infty f(x)e^{-2\pi xy}dx$$

1.) There exists a mistake for the definition of $L^2(\mathbb R)$ as completion of the space $L^1_{bc}(\mathbb R)$?
It should not be $L^2(\mathbb R)$ the completation of $L^2_{bc}(\mathbb R)$.


2.) In other book's, Plancherel's theorem is established as:
Let $f\in L^1(\mathbb R)\cap L^2(\mathbb R)$. Then $\tilde f\in L^2(\mathbb R)$. Furthermore $\| \tilde f\|^2_2=\| f\|_2^2$
Are the same?

Thanks you all
 A: It is true that every element of $L^1_{bc}(\mathbb{R})$ is also an element of $L^2_{bc}(\mathbb{R})$, and that's because every $f\in L^1_{bc}$ is assumed to be uniformly bounded by some constant $M_f$; therefore,
$$
              \int |f|^2dx  \le M_f \int |f|dx  < \infty,\;\;\; f \in L_{bc}^1(\mathbb{R}).
$$
So, in that sense, $L^1_{bc}(\mathbb{R}) \subseteq L^2_{bc}(\mathbb{R})$, but I do not like that notation because $L_{bc}^1, L_{bc}^2$ are not just sets to be compared: they are spaces with norms, one is an inner product space, and the other is a normed space. I think it would be better to leave the spaces out of the statement and to try to be precise about the set inclusion.
It is wrong to state that $L^1_{bc}(\mathbb{R})$ is a pre-Hilbert space because $L_{bc}^1$ carries a specific norm with it, and it is not a norm generated by an inner product. That's stretching the sloppiness past any reasonable point, at least so far as I am concerned.
It is true that if $f\in L^1(\mathbb{R})$ is bounded a.e. by a constant $M_f$, then $f$ is square integrable because $\int |f|^2dx \le \int |f|dx M_f < \infty$.
