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If I have understood the def. of well-defined correct. I must show that

i) $F$ is a relation from $L^1$ to $C_b(\mathbb{R})$

ii) Every element in $L^1$ is related to some element in $C_b(\mathbb{R})$

iii) if $f_1$ and $f_2$ are equal in $L^1$ then is must hold that $Ff_1 = Ff_2$ in $C_b(\mathbb{R})$

I would say that by def.

$$ Ff(\phi) = \int_{- \infty}^{\infty} f(x) e^{-i\phi x} dx $$

it holds that $F$ is a relation between $L^1$ and $C_b(\mathbb{R})$, ie i), and aswell every element in $L^1$ is related to some element in $C_b(\mathbb{R})$. Since $|Ff| < \infty$ because $f$ is in $L^1$ thus ii).

Now if $f_1 = f_2 \in L^1$ then we want to compare

$$ Ff_1(\phi) = \int_{- \infty}^{\infty} f_1(x) e^{-i\phi x} dx \ \ \ \ (1) $$ and $$ Ff_2(\phi) = \int_{- \infty}^{\infty} f_2(x) e^{-i\phi x} dx \ \ \ \ (2). $$ Since $f_1 = f_2$ and obviously $e^{-i \phi x } = e^{-i \phi x } $ iii) must hold.

Am I on the right track?

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    How can you see that $Ff(\phi)$ is continuous and the decay at infinity property? From what you have written it is only clear that $Ff$ is bounded. Also number iii) is not quite the same, $f_1=f_2$ in $L^1$ if $\int| f_1-f_2 |= 0$ (remember elements of $L^1$ are equivalence classes of functions). – 2017-01-06

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