$\hat{g}(n)=\hat{f}(n)$ if $n \ge 0$ and $\hat{g}(n)=0$ otherwise. Prove there is a $M_p$ depending only on $p$ such that $\|g\|_p \le M_p\|f\|_p$ Given $1<p<\infty$, $f \in L^p(T)$ and $T$ is the unit circle. Let $\hat{f}(n)$ be the Fourier coefficients of $f$. If there exists an $g \in L^p(T)$ such that $\hat{g}(n)=\hat{f}(n)$ for $n \ge 0$ and $\hat{g}(n)=0$ for $n<0$, prove that there is a $M_p$ depending only on $p$ such that $\|g\|_p \le M_p\|f\|_p$. My thought is that since
\begin{equation}
\frac{1}{e^{it}-z}=\frac{1}{e^{it}}\sum_{n=0}^\infty (\frac{z}{e^{it}})^n,
\end{equation} we can get
\begin{equation}
F_1(z)=\frac{1}{2\pi} \int_{-\pi}^\pi \frac{e^{it}+z}{e^{it}-z} f(e^{it}) dt=2\sum_{n=0}^\infty \hat{f}(n)z-\hat{f}(0).
\end{equation} Define
\begin{equation*}
F_2(z)=\frac{F_1(z)+\hat{f}(0)}{2}=\sum_{n=0}^\infty \hat{f}(n)z^n.
\end{equation*} Then we can apply M. Riesz's Theorem to show that the function $g(e^{it})=F_2(e^{it})$ satisfies the requirement that  $\hat{g}(n)=\hat{f}(n)$ for $n \ge 0$ and $\hat{g}(n)=0$ for $n<0$. However, I am stuck with showing that the constant $M_p$ depends only on $p$, not on $f$.
 A: For the sake of clarity, I reformulate your construction in terms of functional analysis.
Let $\mathrm{H}^p(\mathbb{T})$ denote the space of functions $g \in \mathrm{L}^p(\mathbb{T})$ satisfying $\hat{g}(n) = 0$ for all $n < 0$ and $\mathrm{H}^p(\mathbb{D})$ the usual Hardy space. There is a linear isometry between these spaces given by the map $\iota: f \mapsto f^{\star}$ where $f^{\star}$ denote the radial function of $f$. Consider the projection $\pi : \mathrm{L}^p(\mathbb{T}) \to \mathrm{H}^p(\mathbb{T})$ defined as $\iota \circ \theta$ where $\theta : \mathrm{L}^p(\mathbb{T}) \to \mathrm{H}^p(\mathbb{D}), f \mapsto \sum_{n=0}^{\infty} \hat{f}(n)z^n$. You want to prove that $\pi$ is a bounded operator.
But $\iota$ is a bounded operator, so we will prove that $\theta$ is bounded. Since $\theta$ is a linear map between two normed spaces, we only have to prove that $\theta$ is continuous. Furthermore, both $\mathrm{H}^p(\mathbb{T})$ and $\mathrm{L}^p(\mathbb{D})$ are Banach spaces. Thus, according to the closed graph theorem, it is sufficient to show that the graph of $\theta$ is closed.
Let $\{f_k\}$ be a sequence in $\mathrm{L}^p(\mathbb{T})$ tending towards $f$ and assume $F_k = \theta(f_k)$ tends to $G \in \mathrm{H}^p(\mathbb{D})$. We have to prove that $G = \theta(f)$. But $\theta(f)$ is an analytic function on the unit disk. So $\theta(f)$ depends only on its Taylor coefficients at the origin, which are $\hat{f}(n)$ by construction. Since the maps $\mathrm{L}^p(\mathbb{T}) \to \mathbb{C},\ h \mapsto \hat{h}$ are continuous, we have done.
EDIT: Alas, it seems difficult to avoid using the closed graph theorem. We can prove that $\hat{f_k}(n)$ converges to $\hat{f}(n)$ uniformly in $n$ as follows: since $p > 1$, by the Jensen convexity inequality we have:
\begin{align}|\hat{f}_k(n)-\hat{f}(n)|^p &= \Big|\int_0^{2\pi} (f_k(e^{it})-f(e^{it}))e^{-int} \frac{dt}{2\pi} \Big|^p \leqslant \int_0^{2\pi} |f_k(e^{it}-f(e^{it})|^p \frac{dt}{2\pi} \\ &\leqslant \|f_k-f\|_p^p.\end{align}
Hence
\begin{align}
\Big(\int_0^{2\pi} |F_k(re^{it})-\theta(f)(re^{it})|^p \frac{dt}{2\pi}\Big)^{\frac{1}{p}} &\leqslant \sum_{n=0}^{\infty} \Big(\int_0^{2\pi} |\hat{f_k}(n)-\hat{f}(n)|^p r^{pn} \frac{dt}{2\pi}\Big)^{\frac{1}{p}} \\
&\leqslant \sum_{n=0}^{\infty} \|f_k-f\|_p r^n
\end{align}
Unfortunately, we must prove the above integral converges uniformly in $r$ to zero when $k$ goes to the infinity.
