1

If $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1 $ and $ f\in L_p $ $g\in L_q $ and $h\in L_r $ so how can I prove $$ ||fgh ||_1\le||f||_p\ ||g||_q\ ||h||_r $$

  • What have you tried so far? Do you know the statement of the normal Holder's inequality? – BigbearZzz May 27 '16 at 02:31
  • Yeah , I have already proved $||fg||_1\le ||f||_p\ ||g||_q.$ But for this prove we have used Young's inequality. Here we don't have any Young's inequality. That is stuck how to start. – user342666 May 27 '16 at 02:35
  • There is a Generalized Holder inequality, proofs of which can be found in abundance, and it will show you how to tackle this problem. – SquirtleSquad May 27 '16 at 02:41

1 Answers1

1

As one might expect, this works by applying Hölder's inequality twice:

Let $s=p$, $t=\frac{qr}{q+r}$. Then $\frac 1 s+\frac 1 t=1$, and Hölder's inequality yields $$ \|fgh\|_1\leq \|f\|_s\|gh\|_t=\|f\|_p\| |gh|^t \|_1^{\frac 1 t}. $$ Next let $\sigma=\frac q t=\frac{q+r}{r}$ and $\tau=\frac r t=\frac{q+r}{q}$. Then $\frac 1\sigma+\frac 1\tau=1$, and a second application of Hölder's inequality gives $$ \| |gh|^t \|_1^{\frac 1 t}\leq \| |g|^{t}\|_\sigma^{\frac 1 t}\| |h|^{t}\|_\tau^{\frac 1 t}=\| |g|^{t\sigma}\|_1^{\frac 1{t\sigma}}\| |h|^{t\tau}\|_1^{\frac 1{t\tau}}=\|g\|_q \|h\|_r. $$ Combine these two results to get the desired inequality.

To prove the generalized inequality for $n$ factors, just proceed by induction.

Thomas Ahle
  • 4,612
MaoWao
  • 15,139
  • 1
  • 18
  • 36