TITLE: Counterexamples for Hölder's inequality when $p$ and $q$ are not conjugate.
QUESTION [7 upvotes]: Hölder's inequality shows that, when $$ \frac{1}{p} + \frac{1}{q} = 1,$$ and $f\in L^p$ and $g\in L^q$, then $$\Vert f\,g\Vert_1 \le \Vert f \Vert_p \Vert g \Vert_q.$$
Is there an example of this inequality failing for $p=q=1$?  I have been unable to find one in the usual places (analysis texts, and various combinations of search terms online).

REPLY [2 votes]: For $f_{\alpha,c}(x) = x^{-\alpha}$ on $[c,\infty)$ and $0$ otherwise (and $\alpha > 1$) you have $$
  \|f_{\alpha,c}\|_1 = \int_{\mathbb{R}} f(x) \,dx = \frac{x^{-\alpha+1}}{1-\alpha}\bigg|_{x=c}^{c=\infty} = \frac{c^{1-\alpha}}{\alpha - 1}
$$
and $$
 \|f_{\alpha,c}\cdot f_{\alpha,c}\|_1 = \|f_{2\alpha,c}\|_1 = \frac{c^{1-2\alpha}}{2\alpha - 1} \text{,}
$$
which for $c = \sqrt[7]{\frac{1}{2}} < 1$, and $\alpha = 4$ yields $$
  \left(\|f_{\alpha,c}\|_1\right)^2 = \left(\frac{c^{1-\alpha}}{\alpha-1}\right)^2 = \frac{c\cdot c^{1 - 2\alpha}}{\alpha^2 - (2\alpha - 1)} = c\frac{2}{9} < \frac{2}{7} = \frac{c^{1 - 2\alpha}}{2\alpha - 1} = \|f_{\alpha,c}\cdot f_{\alpha,c}\| \text{.}
$$
The same works for every pair $c < 1$, $\alpha^2 > 4\alpha - 2$.