Given $f(x+iy):=\int_{-a}^{a}g(t)e^{2\pi i(x+iy)t}dt$ is it true that $\forall M>0, \sup_{y\in[-M,M]}|f(x+iy)|\rightarrow0, |x|\rightarrow\infty$? If $a>0$, and $g\in L^1(-a,a)$ define:
$$f:\mathbb{C}\rightarrow\mathbb{C}, z\mapsto \int_{-a}^{a}g(t)e^{2\pi izt}\operatorname{d}t.$$
I know from Riemann-Lebesgue lemma that:
$$\forall y\in\mathbb{R}, |f(x+iy)|\rightarrow0, |x|\rightarrow\infty.$$
Does the same relation hold uniformly in $y$ on compact sets? I.e.:

Is it true that $$\forall M>0, \sup_{y\in[-M,M]}|f(x+iy)|\rightarrow0, |x|\rightarrow\infty?$$
  If not, what about if we strengthen the condition $g\in L^1(-a,a)$ to $g\in L^p(-a,a)$ for $p>1$?

 A: This is true and here are some hints: if the result is false there exist sequence $\{y_n\}$,$\{x_n\}$ such that $|y_n| \leq M$, $|x_n| \to \infty$ and $|f(x_n+iy_n)|$ is bounded away from $0$. Choose a subsequence of $(y_n)$ converging to some $y$. Note that $\int_{-a}^{a} |g(t)e^{-2\pi y_n t}-g(t)e^{-2\pi y t}|\, dt \to 0$. Convergence in $L^{1}$ implies uniform convergence of Fourier transforms. Now use the fact that $f(x_n+iy) \to 0$ to get a  contradiction. 
A: I've found another answer based on a different approach.
Fix $M>0$. Fix $\delta>0$.
Observe that $f$ is an entire function, so it satisfies the mean value theorem.
Then, denoting with $\mu$ the Lebesgue measure on $\mathbb{C}$ we get:
$$\forall x\in \mathbb{R}, \forall y \in [-M,M], |f(x+iy)|=\left|\frac{1}{\pi\delta^2}\int_{D_{\delta}(x+iy)}f(w)\operatorname{d}\mu(w)\right|\le \frac{1}{\sqrt{\pi}\delta}\left(\int_{D_{\delta}(x+iy)}|f(w)|^2\operatorname{d}\mu(w)\right)^{1/2}\le \frac{1}{\sqrt{\pi}\delta}\left(\int_{[x-\delta,x+\delta]\times[M-\delta,M+\delta]}|f(w)|^2\operatorname{d}\mu(w)\right)^{1/2}.$$
So:
$$\sup_{y\in[-M,M]}|f(x+iy)|\le \frac{1}{\sqrt{\pi}\delta}\left(\int_{[x-\delta,x+\delta]\times[M-\delta,M+\delta]}|f(w)|^2\operatorname{d}\mu(w)\right)^{1/2}$$
and then if we manage to show somehow that $f\in L^2(\mathbb{R}\times[-M-\delta,M+\delta])$ we are done thanks to dominated convergence theorem.
Actually, this is the case if $g \in L^2(\mathbb{R})$. In fact, thanks to Plancherel theorem:
$$\forall y\in[-M-\delta,M+\delta], \int_\mathbb{R}|f(x+iy)|^2\operatorname{d}x = \int_\mathbb{R}\left|\int_\mathbb{R} \chi_{[-a,a]}(t)g(t)e^{-2\pi y t} e^{2\pi i x t }\operatorname{d}t\right|^2\operatorname{d}x = \int_\mathbb{R} \left|\chi_{[-a,a]}(t)g(t)e^{-2\pi y t} e^{2\pi i x t }\right|^2\operatorname{d}t \le e^{4 \pi(M+\delta)a}\|g\|_{L^2(-a,a)}^2,$$
and so:
$$\int_{\mathbb{R}\times[-M-\delta,M+\delta]}|f(w)|^2\operatorname{d}\mu (w) \le 2(M+\delta)e^{4 \pi(M+\delta)a}\|g\|_{L^2(-a,a)}^2<+\infty.$$
Now, for the general case, thanks to the density of $L^2(-a,a)$ in $L^1(-a,a)$, we can get a sequence $(g_n)_{n\in\mathbb{N}}\subset L^2(-a,a)$ that converges in $L^1((-a,a))$ to $g$. Noticing that:
$$\forall x \in \mathbb{R}, \forall y \in [-M,M], \left|\int_{-a}^a g_n(t) e^{2\pi i (x+iy)t}\operatorname{d}t - \int_{-a}^a g(t) e^{2\pi i (x+iy)t} \operatorname{d}t \right| \le \int_{-a}^a|g(t)-g_n(t)|e^{-2\pi y t}\operatorname{d}t\le e^{2\pi Ma} \|g-g_n\|_1,$$
we get that we can find a sequence of function $f_n:\mathbb{R}\times[-M,M]\rightarrow\mathbb{C}$ that converges uniformly on $\mathbb{R}\times[-M,M]$ to $f$ and such that:
$$\forall n\in\mathbb{N}, \sup_{y\in[-M,M]}|f_n(x+iy)|\rightarrow 0, |x|\rightarrow \infty.$$
So, if $\varepsilon>0$ and $n\in\mathbb{N}$ is such that $$\|f_n-f\|_{L^\infty(\mathbb{R}\times[-M,M])} < \varepsilon/2$$ and $R>0$ is such that $$\forall |x|>R, \sup_{y\in[-M,M]}|f_n(x+iy)| < \varepsilon/2$$
we get that:
$$\forall |x|>R, \sup_{y\in[-M,M]}|f(x+iy)| \le \sup_{y\in[-M,M]}|f(x+iy)-f_n(x+iy)|+\sup_{y\in[-M,M]}|f_n(x+iy)| \\ \le \|f_n-f\|_{L^\infty(\mathbb{R}\times[-M,M])}+\sup_{y\in[-M,M]}|f_n(x+iy)| < \varepsilon $$
and, by the arbitrariness of $\varepsilon>0$, we get the conclusion.
