Counterexamples for Hölder's inequality when $p$ and $q$ are not conjugate.

Question

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).

Answer 1

Consider the real line with Lebesgue measure and $f$, $g$ the characteristic functions of the interval $[0,x]$ for a fixed $x$. Since $\lVert f\rVert_p= x^{1/p}$ and similarly for the $q$ norm, the Hölder's inequality would read $x\leqslant x^{1/p+1/q}$.

If $p=q=1$, just pick some $x$ such that $x\gt x^2$.

This proves more in the general case: if $1/p+1/q\neq 1$, there is not universal constant $C$ such that for any $(f,g)\in \mathbb L^p\times\mathbb L^q$, $\lVert fg\rVert_1\leqslant C\lVert f\rVert_p\cdot \lVert g\rVert_q$.

Answer 2

Let $f(x)=x^2, g(x)=\frac{1}{1-x}$ on $[0,\frac{1}{2}]$

Here $\|f\|_1=\frac{1}{3}\frac{1}{8}=\frac{1}{24}$, $\|g\|_1=\ln 2$

Now $\|fg\|_1=\ln 2-\frac{5}{8}\gt\frac{\ln2}{24}$

Answer 3

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 $c < 1$, $\alpha^2 > 4\alpha - 2$.