I have a doubt about the proof of theorem 7.1.2 that appears in the book Fourier Analysis on Groups by Rudin, I will write the first part of the proof below
($G$ is locally compact abelian group)
Theorem 7.1.2 Every closed translation-invariant subspace of $L^1(G)$ is an ideal; conversely, every closed ideal in $L^1(G)$ is translation invariant.
Proof: For $f, g \in L^1(G)$ and $\phi \in L^\infty (G)$ we have
$$\int_G (f*g) (-x) \phi(x) dx = \int_G g(-y) (f*g) (y) dy, \qquad (1)$$
since each of these expressions is $(f*g*\phi)(0)$.
Suppose $I$ is closed an translation-invariant, $\phi$ annihilates $I$, and $f \in I$. Then $f*\phi = 0$, the right side of $(1)$ is $0$, hence $\phi$ annihilates $f*g$, for every $g \in L^1(G)$. Since this is true for every $\phi$ which annihilates $I$, the Hahn-Banach theorem implies that $f*g \in I$, and so $I$ is an ideal...
The doubt I have is about the Hahn-Banach theorem, why is it concluded that $f*g \in I$ when applying it?
This theorem is applied in a paper on which I am doing a project for my university. I have read the first part of the book and in my functional analysis class I saw some versions of the Hahn-Banach theorem, but I cannot find the connection with this step of the proof.
Thank you in advance for your help.