Holder's inequality

Question

Suppose that $f$ and $g$ are two non negative real valued functions defined on a measure space $(X,\mu)$.

Let $0<p<\infty$. Holder's inequality says that $\int fg d\mu\le \|f\|_p \|g\|_q$ where $\frac{1}{p}+\frac{1}{q}=1$ with equality iff $a f^p=b g^q$ for some constants $a$ and $b$.

So, in general, $\int fg d\mu=\|f\|_p \|g\|_q- x$ and that $x=0$ iff $a f^p=b g^q$ for some constants $a$ and $b$.

My question is, is there a way to find $x$? (apart from $x=\|f\|_p \|g\|_q-\int fg d\mu$)

If you don't want $x = |f|_p|g|_q - \langle f,g\rangle$, what sort of expressions are you after? — Willie Wong, Jun 05 '12 at 07:34
I am looking for an equality involving $|f|_p |g|_q$ and $\int fg d\mu$. — Kumara, Jun 05 '12 at 08:51
@Kumara Do you wish to improve Hölder's inequality somehow? — AD - Stop Putin -, Jun 05 '12 at 09:01
No, not really. While establishing an identity I used Holder's inequality. So it ends up with an inequality. But the identity which I am trying to establish is supposed to be an equality. That is why I was wondering whether there is any identity (equality) involving $|f|_p |g|_q$ and $\int fg d\mu$. — Kumara, Jun 05 '12 at 09:09
See "A stability version of Holder's inequality" by J. Aldaz, http://dx.doi.org/10.1016/j.jmaa.2008.01.104 — , Jun 05 '12 at 15:13
@LeonidKovalev: Thank you for sharing the information. This article might be helpful to me. — Kumara, Jun 06 '12 at 09:05

score 1 · Accepted Answer · answered Jul 27 '12 at 20:57

Though the thread linked by @leo has an excellent discussion of the equality case, I'd like to add an estimate for the quantity $x=\|f\|_p\|g\|_q-\langle f,g\rangle$ introduced above. The estimate is due to J.M. Aldaz, and it applies only in the reflexive range $1<p<\infty$. By exchanging $p$ and $q$ we may assume $1<p\le 2$. It is also convenient to normalize $\|f\|_p=1=\|g\|_q$. In this case, $$\frac{1}{q}\left\| |f|^{p/2}-|g|^{q/2} \right\|_2^2 \le 1-\int|fg|\le \frac{1}{p}\left\| |f|^{p/2}-|g|^{q/2} \right\|_2^2$$ The lower bound is of most interest (it gives the equality case), but it's also neat that the upper bound is the same when $p=q=2$. The proof is based on the corresponding refinement of Young's inequality: $$ \frac{1}{q}(u^{p/2}-v^{q/2})^2\le \frac{u^p}{p}+\frac{v^q}{q}-uv\le \frac{1}{p}(u^{p/2}-v^{q/2})^2 $$ which holds for all $u,v\ge 0$ and $1<p\le 2$, with $q$ being the conjugate exponent.

Holder's inequality

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