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Holder's inequality states that $\langle x, y\rangle \le \langle u, v\rangle \le \| x\|_p \|y \|_q$ where $u_i=|x_i|,v_i=|y_i|$ and $p+q=pq,p\ge 1$. We observe that equality occurs if and only if

$$\left(\frac{|x_i|}{\|x\|_p}\right)^{1/q}=\left(\frac{|y_i|}{\|y\|_q}\right)^{1/p}$$

and $\text{sign}(x_i)=\text{sign}(y_i)$.

Question: I see that signs equal make the first inequality an equality. But I couldn't get how the first statement makes the second relation an equality.

It seems that the $\langle u,v\rangle=\sum |x_i||y_i|:=\|x\|_p\|y\|_p$ implies that there would be equality while the condition is talking about $$\frac{|x_i|^p}{\|x\|_p^p}=\frac{|y_i|^q}{\|y\|_q^q}$$

I fail to see how these events relate?

$$\begin{split}\sum|x_i||y_i|&=(\sum |x_i|^p)^{1/p}(\sum|y_i|^q)^{1/q}\\ \frac{\sum|x_i||y_i|}{(\sum |x_i|^p)^{1/p}(\sum|y_i|^q)^{1/q}}=1\end{split}$$

I think this is really close

I found the answer to the first part in this post's diagram, after some contemplation.

$$\begin{split}\sum|x_i||y_i|&=(\sum |x_i|^p)^{1/p}(\sum|y_i|^q)^{1/q}\\ \Leftrightarrow\\ \sum\frac{|x_i|}{(\sum|x_i|^p)^{1/p}}\frac{|y_i|}{(\sum|y_i|^q)^{1/q}}&=1\\ \Leftrightarrow\\ \sum\frac{|x_i|}{(\sum|x_i|^p)^{1/p}}\frac{|y_i|}{(\sum|y_i|^q)^{1/q}}&=\frac 1p\sum{\frac{|x_i|^p}{\sum{|x_i|^p}}}+\frac 1q\sum\frac{|y_i|^q}{\sum|y_i|^q}\\ \Leftrightarrow\\ \frac{|x_i||y_i|}{(\sum{|x_i|^p})^{1/p}(\sum|y_i|^q)^{1/q}}&=\frac 1p\frac{|x_i|^p}{\sum |x_i|^p}+\frac 1q\frac{|y_i|^q}{\sum|y_i|^q},i=1,\dots,n\\ \Leftrightarrow\\ \end{split}$$

and I thought i had the last step but can't figure it out now. My question is: how did they get the last step?

Vons
  • 11,004

1 Answers1

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It comes from Young’s Inequality:

$$ab\le\frac{a^p}{p}+\frac{b^q}{q}$$

when $1/p+1/q=1$ and $a,b\ge0$ with equality only if $a^p=b^q\rightarrow a=b^{q/p}$.

I will now regurgitate the proof. Equality clearly holds if $a=b=0$ so let's consider $a,b>0$. Because the natural log is concave, $\ln(ta^p+(1-t)b^q)\ge t\ln(a^p)+(1-t)\ln(b^q)$. Put $t=1/p$ making $1-t=1/q$. We've got $\ln(ta^p+(1-t)b^q)\ge\ln(a)+\ln(b)=\ln(ab)$. Exponentiating both sides, we get $$a^p/p+b^q/q\ge ab$$

Now in the context of this problem, let

$$a=\frac{|x_i|}{\|x\|_p}=\frac{|x_i|}{(\sum |x_i|^p)^{1/p}}, b=\frac{|y_i|}{\|y\|_q}=\frac{|y_i|}{(\sum |y_i|^q)^{1/q}}$$

The result directly follows.

Vons
  • 11,004