1 Minkowski Inequality
In mathematical analysis, the Minkowski inequality establishes that the \(L^p\) spaces are normed vector spaces.
Let \(S\) be a measure space, let \(1\leq p\leq \infty\) and let \(f\) and \(g\) be elements of \(L^p(S)\). Then \(f+g\) is in \(L^p(S)\) and we have the Minkowski inequality \[\|f+g\|_{p}\leq \|f\|_{p}+\|g\|_{p}.\]
Proof: Firstly, we prove that \(f+g\) has finite \(p\)-norm if \(f\) and \(g\) both do, which follows by the fact that \[ |f+g|^p\leq 2^{p-1}(|f|^p+|g|^p). \] Here we use the fact that \(h(x)=|x|^p\) is convex over \(\mathbb{R}^+\) (for \(p>1\)) and so, by definition \[ \left| \frac{1}{2}f+ \frac{1}{2}g\right|^p\leq \left| \frac{1}{2}|f|+ \frac{1}{2}|g|\right|^p\leq \frac{1}{2}|f|^p+\frac{1}{2}|g|^p, \] which means that \[ |f+g|^p \leq \frac{1}{2}|2f|^p+ \frac{1}{2}|2g|^p=2^{p-1}|f|^p+2^{p-1}|g|^p. \] Next, considering \(\|f+g\|_{p}\). If it is zero, then Minkowski’s inequality holds. Assuming that \(\|f+g\|_p\) is not zero, using the triangle inequality and then Hölder's Inequality we have that \[ \begin{align} \|f+g\|_{p}^p & = \int |f+g|^p \, d\mu = \int |f+g|\cdot |f+g|^{p-1} \, d\mu\leq \int (|f|+|g|)|f+g|^{p-1} \, d\mu \\ & = \int |f\mid f+g|^{p-1} \, d\mu +\int |g\mid f+g|^{p-1} \, d\mu \\ & \leq \left( \left( \int |f|^p \, d\mu \right)^{1/p} + \left( \int |g|^p \, d\mu \right)^{1/p}\right) \left( \int |f+g|^{(p-1)( \frac{p}{p-1})} \, d\mu \right)^{1-1/p} \tag{Holder} \\ & = (\|f\|_{p}+\|g\|_{p})\frac{\|f+g\|_{p}^p}{\|f+g\|_{p}}. \end{align} \] Multiplying both sides by \(\frac{\|f+g\|_{p}}{\|f+g\|_{p}^p}\) we obtain the result.
For random variables \(X,Y\) with \(p \in (1,\infty)\) then the above result can be interpreted in terms of expectations as \[(\mathbb{E}|X+Y|^p)^{1/p} \leq (\mathbb{E}|X|^p)^{1/p} + (\mathbb{E}|Y|^p)^{1/p}.\]
Proof: First, we notice that \[ \mathbb{E}|X+Y|^p = \mathbb{E}[|X+Y| |X+Y|^{p-1}]\leq \mathbb{E}[|X| |X+Y|^{p-1}]+\mathbb{E}[|Y| |X+Y|^{p-1}]. \] Using Hölder's Inequality on the right hand terms for \(q\) selected to satisfy \(\frac{1}{p}+\frac{1}{q}=1\) gives \[ \begin{align} \mathbb{E}|X+Y|^p & \leq (\mathbb{E}|X|^p)^{1/p}(\mathbb{E}|X+Y|^{q(p-1)})^{1/q}+(\mathbb{E}|Y|^p)^{1/p}(\mathbb{E}|X+Y|^{q(p-1)})^{1/q} \\ \implies\frac{\mathbb{E}|X+Y|^p}{(\mathbb{E}|X+Y|^{q(p-1)})^{1/q}} &\leq (\mathbb{E}|X|^p)^{1/p}+(\mathbb{E}|Y|^p)^{1/p}, \end{align} \] which gives the result since \(q(p-1)=p\) and \(1-\frac{1}{q}=\frac{1}{p}\).