Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For a general measure space, we define : $\|f\|_p= \left(\int\vert f\vert^p du\right)^{1/p}$. Let $0 < a < b < c < \infty$ and prove the following: $$ \|f\|_b \leqslant \max\{\|f\|_a, \|f\|_c\}. $$ Any help is appreciated because I dont understand the solution underneath

share|improve this question
    
The idea here is that $||f||_p \leq ||f||_q$ if $1 \leq p \leq q \leq \infty$ . But the inequality reverses in $0 <p,q \leq 1$. Hope this helps. –  Gautam Shenoy Nov 19 '12 at 7:39
    
That is to say $||f||_1$ is a point of minima. –  Gautam Shenoy Nov 19 '12 at 7:40
    
I think here we have to use Holder inequality in some way , so can you give more details please ? So do I have to take cases where a>1 and when a<1 with b,c > 1 ? –  user48931 Nov 19 '12 at 7:44
add comment

1 Answer

We have that

$$ \frac{1}{c}<\frac{1}{b}<\frac{1}{a}. $$ Therefore, there exists $\alpha\in (0,1)$ such that $$ \frac{1}{b}=\frac{\alpha}{c}+\frac{1-\alpha}{a}. $$ We claim that $$ \Vert f\Vert_b\leqslant \Vert f\Vert_c^\alpha\Vert f\Vert_a^{1-\alpha} (\ast) $$ from where it follows that $$ \Vert f\Vert_b\leqslant\max\{\Vert f\Vert_a, \Vert f\Vert_c\} $$ because $$ \Vert f\Vert_b\leqslant \Vert f\Vert_c^\alpha\Vert f\Vert_a^{1-\alpha}\leqslant \max\{\Vert f\Vert_a,\Vert f\Vert_c\}^{\alpha}\max\{\Vert f\Vert_a,\Vert f\Vert_c\}^{1-\alpha}=\max\{\Vert f\Vert_a,\Vert f\Vert_c\}. $$ Now, (*) can be proved easily using Holder's Inequality as follows $$ \int |f|^b=\int |f|^{\alpha b}|f|^{(1-\alpha)b}\leqslant \left(\int|f|^{\alpha b\frac{c}{\alpha b}}\right)^{\frac{\alpha b}{c}}\left(\int|f|^{(1-\alpha)b\frac{a}{(1-\alpha) b}}\right)^{\frac{(1-\alpha)b}{a}}=\left(\int|f|^{c}\right)^{\frac{\alpha b}{c}}\left(\int|f|^{a}\right)^{\frac{(1-\alpha)b}{a}}=\Vert f\Vert_c^{\alpha b}\Vert f\Vert_a^{(1-\alpha) b}, $$ where we have used the fact that $$ \frac{\alpha b}{c}+\frac{(1-\alpha)b}{a}=1. $$ Simiplifying we have the result.

share|improve this answer
    
Holder inequality states that for any 1<p,q<∞ , then ||f g ||1<= ||f||p ||g||q , so what is p and q here ? and the function g ? –  user48931 Nov 19 '12 at 8:00
    
How does the result follow from proving * ? I can't see it –  user48931 Nov 19 '12 at 8:28
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.