I have to prove that for any $g \in L^1(\mathbb{R})$ and for any function $u(t)$ bounded and continuous, the following inequality holds:
$$sup_{t \in \mathbb{R}}|(g \ast u)(t)| \leqslant ||g||_1\cdot||u||_\infty$$
I guess that the starting point is the Holder's inequality, which states that:
$$\bigl| \int_{-\infty}^{+\infty}g(t)\overline{u(t)}dt \bigr| \leqslant ||g||_1\cdot||u||_\infty$$
Then, it is sufficient to prove that:
$$sup_{t \in \mathbb{R}}|(g \ast u)(t)| \leqslant \bigl| \int_{-\infty}^{+\infty}g(t)\overline{u(t)}dt \bigr| $$
However I don't know how to proceed.
The version of Holder's inequality you wrote above is not the full power of the inequality. We really have the inequalities
$$ | \int_{- \infty}^{\infty} g(t) \overline{u(t)} dt | \leq \int_{- \infty}^{\infty} |g(t) \overline{u(t)} | dt = \int_{- \infty}^{\infty} |g(t) {u(t)} | dt \leq ||g||_1 ||u||_{\infty} $$
(where the second inequality is really Holder's inequality). Now, for any $x \in \mathbb{R}$, let $v_x(t) = u(x - t)$. Then observe that $||v_x||_{\infty} = ||\overline{v_x}||_{\infty} = ||u||_{\infty}$. It follows that
$$ | (g \ast u)(x)| = | \int_{- \infty}^{\infty} g(t) v_x(t) dt | \leq \int_{- \infty}^{\infty} |g(t) v_x(t) | dt = \int_{- \infty}^{\infty} |g(t) \overline{v_x(t)} | dt \leq ||g||_1 ||u||_{\infty} $$
Then taking the supremum preserves the inequalities.