## 2013/11/14

### Young's, Hölder's and Minkowski's Inequalities

In this post, we prove Young's, Holder's and Minkowski's inequalities with full details. We prove Hölder's inequality using Young's inequality. Then we prove Minkowski's inequality by using Hölder.

Theorem 1. (Young's Inequality) For every $x,y \geq 0$ and $p > 0$, \begin{align} xy \leq \frac{x^p}{p} + \frac{y^q}{q} \end{align} where $p^{-1} + q^{-1} = 1$.

Proof. Put $t = 1/p$ and $1-t = 1/q$. Then by Jensen's inequality (since $\log$ is concave) \begin{align*} \log(t x^p + (1-t) y^q) &\geq t \log (x^p) + (1-t) \log(y^q) \\ &= \log (x^{tp}) + \log(y^{(1-t)q}) \end{align*} Now put $t =1/p$ and $1-t = 1/q$, \begin{align*} \log(x^p/p + y^q/q) &\geq \log (xy) \end{align*} Then exponentiate. $\blacksquare$

Now, fix a $\sigma$-finite measure space, $(\Omega,\cF,\mu)$.

Theorem 2. (Hölder's Inequality) If $p > 1$ and $p^{-1} + q^{-1} = 1$, then we have, \begin{align} \| f g \|_1 \leq \| f\|_p \| g \|_q \end{align} where $f \in L^p(\mu)$ and $g \in L^q(\mu)$.

Proof. The proof follows from the Young's inequality. Recall that we have \begin{align*} xy \leq \frac{x^p}{p} + \frac{y^q}{q} \end{align*} Now replace $x$ by $|f|/\|f\|_p$ and $|g|/\|g\|_q$. This corresponds, \begin{align*} \frac{|f| |g|}{\|f\|_p \|g\|_q} \leq \frac{|f|^p}{p \|f\|_p^p} + \frac{|g|^q}{q \|g\|_q^q} \end{align*} Now integrate both sides wrt $d\mu$, \begin{align*} \int \frac{|f| |g|}{\|f\|_p \|g\|_q} d\mu \leq \int \frac{|f|^p}{p \|f\|_p^p} d\mu + \int \frac{|g|^q}{q \|g\|_q^q} d\mu \end{align*} Note that, we are integrating wrt $d\mu(x)$ precisely. In the above inequality, $\|f\|_p$ and $\|g\|_q$ are numbers; not functions of $x$ whereas $|f|$ should be seen as $|f(x)|$ so as $g$. Therefore, integrals only operate on $|f|$ and $|g|$. Then, we precisely have, \begin{align*} \frac{1}{\|f\|_p \|g\|_q} \int |fg| d\mu &\leq \frac{1}{p \|f\|_p^p} \int |f|^p d\mu + \frac{1}{q \|g\|_q^q} \int |g|^q d\mu \\ &= \frac{\|f\|_p^p}{p \|f\|_p^p} + \frac{\|g\|_q^q}{q \|g\|_q^q} \\ &= \frac{1}{p} + \frac{1}{q} \end{align*} Noting $\int |fg| d\mu = \| fg \|_1$, and $1/p + 1/q = 1$, we have, \begin{align*} \| fg \|_1 \leq \|f\|_p \|g\|_q \end{align*} which concludes the proof. $\blacksquare$

Theorem 3. (Minkowski's Inequality) If $p \geq 1$, then \begin{align} \|f + g\|_p \leq \|f\|_p + \|g \|_p \end{align} where $f,g \in L^p(\mu)$.

Proof. We use the fact, $|x + y| \leq |x| + |y|$ as follows. \begin{align} \int |f + g|^p d\mu &= \int | f + g| |f+g|^{p-1} d\mu \nonumber \\ &\leq \int |f| |f+g|^{p-1} d\mu + \int |g| |f+g|^{p-1} d\mu \label{holderehazirlik} \end{align} Now notice $\int |f| |f+g|^{p-1} d\mu = \| f (f+g)^{p-1} \|_1$ (same true for $g$). This reminds us Hölder's inequality. Then we further bound the \eqref{holderehazirlik} as follows. \begin{align} \int |f| |f+g|^{p-1} d\mu + \int |g| |f+g|^{p-1} d\mu &= \| f |f+g|^{p-1} \|_1 + \| g |f + g|^{p-1} \|_1 \\ &\leq \| f\|_\alpha \| |f+g|^{p-1}\|_\beta + \|g\|_\alpha \| |f+g|^{p-1}\|_\beta \end{align} with $\alpha^{-1} + \beta^{-1} = 1$. Now let $\alpha = p$ and $\beta = q$. Therefore, we have, \begin{align*} \int |f + g|^p d\mu \leq \left\lbrace \|f\|_p + \| g\|_p\right\rbrace \left( \int |f+g|^{p} d\mu \right)^{(p-1)/p} \end{align*} Notice $1/q = (p-1)/p$ and $(p-1)q = p$. Also put $\int |f + g|^p d\mu = \|f + g\|_p^p$. Then, \begin{align*} \|f + g\|_p^p &\leq \left\lbrace \|f\|_p + \| g\|_p\right\rbrace \left( \left( \int |f+g|^{p} d\mu \right)^{1/p} \right)^{p-1} \\ & = \left\lbrace \|f\|_p + \| g\|_p\right\rbrace \| f + g\|_p^{p-1} \end{align*} With rearranging things, result follows. $\blacksquare$