8
$\begingroup$

In a paper that I am reading, an inequality is given and the justification is Hölder's inequality: $$\left(\sum_{i=1}^n a_{ii}\right)^m \le n^{m-1} \sum_{i=1}^n a_{ii}^m$$ where $a_{ii}\ge0$ for all $i$.

However, I do not know why the justification works. I've tried justifying it by writing $a_{ii}$ as $a_{ii}\cdot1$, then applying the inequality. I get somewhat close, since $n$ raised to a power comes out as a factor, but unfortunately I'm not arriving at the above inequality.

$\endgroup$
4
$\begingroup$

Assuming $m > 1$ and the $a_{ii}$ are nonnegative, use conjugate exponents $m$ and $m/(m-1)$ in Hölder's inequality to get

$$\sum_{i = 1}^n a_{ii} \le \left(\sum_{i = 1}^n 1^{m/(m-1)}\right)^{(m-1)/m} \left(\sum_{i = 1}^n a_{ii}^m\right)^{1/m} = n^{(m-1)/m}\left(\sum_{i = 1}^n a_{ii}^m\right)^{1/m}$$

The inequality is then obtained by raising to the $m$th power.

$\endgroup$
2
$\begingroup$

Holder's inequality in the discrete form is the following.

Let $a_1$, $a_2$,...,$a_n$, $b_1$, $b_2$,...,$b_n$, $\alpha$ and $\beta$ be positives. Prove that: $$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}$$

It's just homogenization of the Jensen inequality for $f(x)=x^k$, where $k>1$.

Now by Holder we obtain: $$n^{m-1}\sum\limits_{i=1}^na_{ii}^m=\left(\sum\limits_{i=1}^n1\right)^{m-1}\sum\limits_{i=1}^na_{ii}^m\geq\left(\sum\limits_{i=1}^n\left(1^{m-1}a_{ii}^m\right)^{\frac{1}{m}}\right)^m=\left(\sum\limits_{i=1}^na_{ii}\right)^m$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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