TITLE: Relation between convolution and $L^p$ norms
QUESTION [0 upvotes]: 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.

REPLY [0 votes]: 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.
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