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It might sound silly, but I am always curious whether Hölder's inequality $$\sum_{k=1}^n |x_k\,y_k| \le \biggl( \sum_{k=1}^n |x_k|^p \biggr)^{\!1/p\;} \biggl( \sum_{k=1}^n |y_k|^q \biggr)^{\!1/q} \text{ for all }(x_1,\ldots,x_n),(y_1,\ldots,y_n)\in\mathbb{R}^n\text{ or }\mathbb{C}^n.$$ can be derived from the Cauchy-Schwarz inequality.

Here $\frac{1}{p}+\frac{1}{q}=1$, $p>1$.

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How about the other way round? – Fabian Jul 2 '11 at 19:53
@Fabian: ... Cauchy-Schwartz is a special case of Hölder's Inequality ... $p=q=2$ – Mark Bennet Jul 2 '11 at 20:03
@Sunni: I deleted my answer "No, as far as I know (...)" because apparently it was wrong. – Américo Tavares Jul 2 '11 at 22:28
That is fine. You are very welcome to post your answers. – Sunni Jul 2 '11 at 22:43
up vote 14 down vote accepted

Yes it can, assuming nothing more substantial than the fact that midpoint convexity implies convexity. Here are some indications of the proof in the wider context of the integration of functions.

Consider positive $p$ and $q$ such that $1/p+1/q=1$ and positive functions $f$ and $g$ sufficiently integrable with respect to a given measure for all the quantities used below to be finite. Introduce the function $F$ defined on $[0,1]$ by $$ F(t)=\int f^{pt}g^{q(1-t)}. $$ One sees that $$ F(0)=\int g^q=\|g\|_q^q,\quad F(1)=\int f^p=\|f\|_p^p,\quad F(1/p)=\int fg=\|fg\|_1. $$ Furthermore, for every $t$ and $s$ in $[0,1]$, $$ F({\textstyle{\frac12}}(t+s))=\int h_th_s,\qquad h_t=f^{pt/2}g^{q(1-t)/2},\ h_s=f^{ps/2}g^{q(1-s)/2}, $$ hence Cauchy-Schwarz inequality yields $$ F({\textstyle{\frac12}}(t+s))^2\le\int h_t^2\cdot\int h_s^2=F(t)F(s). $$ Thus, the function $(\log F)$ is midpoint convex hence convex. In particular, $1/p=(1/p)1+(1/q)0$ with $1/p+1/q=1$ hence $$ F(1/p)\le F(1)^{1/p}F(0)^{1/q}, $$ which is Hölder's inequality $\|fg\|_1\le\|f\|_p\|g\|_p$.

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Wow, that is amazing. It is rarely seen in literatures mention Holder's inequality is implied by CS. Is this consideration known before? – Sunni Jul 2 '11 at 22:32
That midpoint convex implies convex does require some regularity on a function such as measurability. There's very little chance it won't hold here of course but this is probably worth mentioning. – Zarrax Jul 2 '11 at 23:27
@Zarrax: Right. In the present case $F$ is also convex on $[0,1]$ hence continuous on $(0,1)$. This is enough to prove that $\log F$ is convex (and continuous) on $(0,1)$ hence, rather than measurability issues which are not a problem here, the important missing step is that convexity of $\log F$ should be used on $1/p=(1/p)t+(1/q)(1-t)$, then the limits of $F(t)$ and $F(1-t)$ when $t\to0$ should be compared to $F(0)$ and $F(1)$. I deliberately omitted these technicalities in my post, which is the reason why the first paragraph announces some indications of the proof, rather than a proof. – Did Jul 2 '11 at 23:43
@Sunni, I think this can interest you: YUAN-CHUAN LI AND SEN-YEN SHAW, A Proof of Hölder's Inequality using the C-S Inequality, Journal of Inequalities in Pure and Applied Mathematics. – leo Jul 3 '11 at 4:13
In the book: "THE CAUCHY–SCHWARZ MASTER CLASS" of J. MICHAEL STEELE, in the case of finite sums, as you post, is the exercise 9.3. – leo Jul 3 '11 at 4:23

I suggest the following consideration. We will prove the above inequality for rational $p,q\in(1,\infty)$ with $\frac1p+\frac1q= 1$, and the irrational cases follow by continuity.

If $p$ and $q$ are rational, let $p=\frac ab$ and $q=\frac ac$ with $b+c=a$ and $2^m\ge a$. Now by induction $$ \sum |x^{(1)}\dots x^{(2^m)}| \le \left( \sum |x^{(1)}\cdots x^{(2^{m-1})}|^2 \right)^{\frac12}\cdot\left( \sum |x^{(2^{m-1}+1)}\cdots x^{(2^{m})}|^2 \right)^{\frac12}$$ $$\le \left( \sum |x^{(1)}|^{2^m} \right)^{\frac1{2^m}}\cdots \left( \sum |x^{(2^m)}|^{2^m} \right)^{\frac1{2^m}}$$ where $x^{(i)}$ are sequences of length $n$, whose indices are omitted. By plugging in $$x^{(1)}= \dots= x^{(b)}= x^{\frac a{b2^m}},\quad x^{(b+1)}= \dots= x^{(b+c)}= y^{\frac a{c2^m}},\quad x^{(b+c+1)}= \dots = x^{(2^m)}=(xy)^{\frac1{2^m}}$$ we get $$\sum |xy|= \sum |x^{\frac a{2^m}}y^{\frac a{2^m}}(xy)^{\frac {2^m-a}{2^m}}| \le \left( \sum |x|^{p} \right)^{\frac{b}{2^m}}\cdot \left( \sum |y|^{q} \right)^{\frac c{2^m}}\cdot \left( \sum |xy| \right)^{\frac {2^m-a}{2^m}}$$ implying the inequality we aimed for by dividing the third term of the RHS, and putting the inequality to the $\frac{2^m}{a}$-th power.

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Your $p, q$ should not be in $(0,1)$. – Sunni Jul 2 '11 at 20:41
yea they arent, i dont know why i put it there ... oops – Peter Patzt Jul 2 '11 at 20:54

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