About convolution, how to finish the proof Let $f\in L^1(T)$.
Show that the function $\phi: L^1(T)\rightarrow L^1(T)$ defined by
$\phi(g)=f*g$ is a linear map and that its norm equals $\|f\|_{L^1(T)}$.
Here $T$ is the complex unit circle.
Well, take two functions $g_1,g_2\in L^1(T)$ and $a,b\in C$, we have:
$\phi(ag_1+bg_2)=f*(ag_1+bg_2)=f*(ag_1)+bf*(bg_2)=af*g_1+bf*g_2=a\phi(g_1)+b\phi(g_2)$, by the linearity of the convolution.
$\|\phi(g)\|=\|f*g\|_1\leq={}$(a property of Convolution)${}=\|f\|_1 \|g\|_1$ So,
$\|\phi\|\leq \|f\|_1$.
What about the other direction?
Any help please
 A: Alright, here's a general sketch and answer of your problem. It got quite long, so I apologize for that, but I figured it's better to be more thorough than less thorough!
I'm going to keep your notation throughout and write
$$
T = \lbrace z \in \mathbb{C} \; : \; \lvert z \rvert = 1 \rbrace
$$
for the complex unit circle, $\lambda$ for the Haar measure on $T$ (which is also the Lebesgue measure on $\mathbb{C}$ restricted appropriately to $T$), and, for a fixed $f \in L^1(T,\lambda)$, $\phi:L^1(T,\lambda) \to L^1(T,\lambda)$ for the operator
$$
\phi(g) := f \ast g
$$
To adapt the hint from the link I posted, we need to consider what are called approximate identity operators in $L^1(T,\lambda)$. These are defined as follows.
Consider a collection of elements $(u_{\varepsilon})_{\varepsilon > 0}$ where each $u_{\varepsilon} \in L^1(T,\lambda)$ is given such that

*

*There exists a $c > 0$ for which
$$
\lVert u_{\varepsilon}\rVert = \int_{T} \lvert u_{\varepsilon}\rvert\,\mathrm{d}\lambda < c;
$$

*For all $\varepsilon > 0$,
$$
\int_{T} u_{\varepsilon} \,\mathrm{d}\lambda = 1;
$$

*For any open neighborhood $U \subseteq T$ of the element $1 \in T$,
$$
\lim_{\varepsilon \to 0}\int_{T \setminus U} \lvert u_{\varepsilon}\rvert\,\mathrm{d}\lambda = 0.
$$
In the third point, the reason we only have to ask for the condition at the element $1 \in T$ is because $T$ is a group; we can write
$$
T = \lbrace e^{2\pi i\theta} \; | \; \theta \in \mathbb{R} \rbrace
$$
and define multiplication as $z \cdot w = e^{2\pi i\theta}e^{2\pi i \gamma} = e^{2\pi i (\theta + \gamma)}$. Because of this and the fact that we are using the Lebesgue measure on $T$, we can translate the open condition that we ask to hold at $1 \in T$ to hold at any $z \in T$ by simply taking the opens $z\cdot U$ and doing the same thing. It is also worth remarking that since the group $T$ is a first-countable compact Abelian group, it admits an approximate identity operator $(u_{\varepsilon})$; we will fix this throughout the remainder of this answer with the assumption (without loss of generality) that the constant $c$ we take has $0 < c \leq 1$.
We now want to show that there is an $L^1(T,\lambda)$-convergence
$$
\phi(u_{\varepsilon}) \to f
$$
so that we can give the bound $\lVert \phi\rVert \geq \lVert f\rVert$. For this we'll need to calculate
$$
\lVert \phi(u_{\varepsilon} - f\rVert = \int \lvert \phi(u_{\varepsilon}) - f\rvert\,\mathrm{d}\lambda,
$$
and this will itself require some manipulation. We first expand the modulus inside the integrand to get th estimate, for any $x \in T$ and any $\varepsilon > 0$,
$$
\lvert \phi(u_{\varepsilon})(x) - f(x) \rvert = \lvert (f \ast u_{\varepsilon})(x) - f(x) \rvert = \left\lvert \int_{T}f(xt^{-1})u_{\varepsilon}(t)\,\mathrm{d}\lambda(t) - f(x) \right\rvert \\
= \left\lvert \int_{T}f(xt^{-1})u_{\varepsilon}(t)\,\mathrm{d}\lambda(t) - \left(\int_{T}u_{\varepsilon}(t)\right)f(x) \right\rvert \\
= \left\lvert \int_{T}f(xt^{-1})u_{\varepsilon}(t)\,\mathrm{d}\lambda(t) - \int_{T}f(x)u_{\varepsilon}(t)\,\mathrm{d}\lambda(t) \right\rvert \\
= \left\lvert \int_{T}(f(xt^{-1}) - f(x))u_{\varepsilon}(t)\,\mathrm{d}\lambda(t) \right\rvert \\
\leq \int_{T} \left\lvert f(xt^{-1}) - f(x)\right\rvert \cdot \lvert u_{\varepsilon}(t)\rvert\,\mathrm{d}\lambda(t).
$$
Applying this estimate at the beginning of the second line we find that
$$
\lVert \phi(u_{\varepsilon}) - f\rVert = \int_{T} \lvert (f \ast u_{\varepsilon})(x) - f(x)\rvert\,\mathrm{d}\lambda(x) = \int_T\left\lvert\int_T f(xt^{-1})u_{\varepsilon}(t)\,\mathrm{d}\lambda(t) - \int_Tu_{\varepsilon}(t)f(x)\,\mathrm{d}\lambda(t) \right\rvert \,\mathrm{d}\lambda(x) \\
\leq \int_T \int_T \left\lvert f(xt^{-1})-f(x) \right\rvert\cdot \lvert u_{\varepsilon}(t)\rvert\,\mathrm{d}\lambda(t)\mathrm{d}\lambda(x).
$$
I now claim that
$$
\lim_{\varepsilon \to 0} \int_T \int_T \left\lvert f(xt^{-1})-f(x) \right\rvert\cdot \lvert u_{\varepsilon}(t)\rvert\,\mathrm{d}\lambda(t)\mathrm{d}\lambda(x) \overset{!}{=} 0.
$$
Assuming this for the moment, we find that it implies
$$
\lim_{\varepsilon \to 0} \lVert \phi(u_{\varepsilon}) - f\rVert = 0
$$
so $\phi(u_{\varepsilon}) \to f$ in $L^1(T,\lambda)$. Thus
$$
\lim_{\varepsilon \to 0} \lVert \phi(u_{\varepsilon}) \rVert = \lVert f \rVert
$$
so we get that
$$
\lVert \phi \rVert = \sup_{\substack{g \in L^1(T,\lambda) \\ \lVert g \rVert \leq 1}} \lVert \phi(g) \rVert \geq \sup_{\varepsilon \geq 0} \lVert \phi(u_{\varepsilon})\rVert \geq \lim_{\varepsilon \to 0} \lVert \phi(u_{\varepsilon})\rVert = \lVert f\rVert.
$$
I'm not going to prove the claim completely at the moment (I may come back and edit it in as a lemma), but I'll give you a quick rundown of how to do it. For any $r > 0$ you need to prove that there exists an open $U$ of $1$ for which
$$
\int_{T}\lvert f(xt^{-1}) - f(x)\rvert\,\mathrm{d}\lambda(x) \leq r
$$
for any $y \in U$. Once you have this lemma and you find $r > 0$ you can choose an appropriate $U$ and write $T = U \sqcup (X \setminus U)$ and break up your integral
$$
\int_T \int_T \left\lvert f(xt^{-1})-f(x) \right\rvert\cdot \lvert u_{\varepsilon}(t)\rvert\,\mathrm{d}\lambda(t)\mathrm{d}\lambda(x) = \int_T \int_T \left\lvert f(xt^{-1})-f(x) \right\rvert\cdot \lvert u_{\varepsilon}(t)\rvert\,\mathrm{d}\lambda(x)\mathrm{d}\lambda(t) \\
= \int_{U}\int_T \left\lvert f(xt^{-1}) - f(x)\right\rvert\cdot\lvert u_{\varepsilon}(t)\rvert\,\mathrm{d}\lambda(x)\mathrm{d}\lambda(t) + \int_{T \setminus U} \int_T\left\lvert f(xt^{-1}) - f(x)\right\rvert\,\lvert u_{\varepsilon}(t)\rvert\,\mathrm{d}\lambda(x)\mathrm{d}\lambda(t).
$$
with the change in order of integration justified by Fubini-Tonelli. From here you should use the lemma to bind the first summand above by $rc$; for the second summand use the fact that $(u_{\varepsilon})$ is an approximate identity operator to show that upper bound $2\lVert f\rVert \int_{T \setminus U}\lvert u_{\epsilon}(t)\rvert\,\mathrm{d}\lambda(t)$ can be made as small as you like.
