Triangle inequality in $L_p$ Space

Loose proof

Using the \(convexity^1\) and absolutely scalability of norm function to prove it.

\(\|\alpha x\| + \|(1-\alpha) y\|\ge \|\alpha x+(1-\alpha )y\|\).

\(\forall v \in V,\alpha \in R: \|\alpha v\| = |\alpha| \|v\|\).

Let \(\alpha=1/2\), we get \(\|\frac{1}{2} x\| + \|\frac{1}{2}y\| \ge \|\frac{1}{2}(x + y)\|\) \(\Rightarrow \frac{1}{2} (\|x\| + \|y\|)\ge \frac{1}{2}\|(x + y)\|\)

thus \(\|x\|_p + \|y\|_p \ge \|x + y\|_p\)

Besides, \(\|x\|_p - \|y\|_p \le \|x - y\|_p \Leftarrow \|x\|_p = \|y+ (x - y)\|_p \le \|y\|_p + \|x-y\|_p\)

Rigorous proof

\(\begin{array}{l} \|f+g\|_{p}^{p}&=\int|f+g|^{p} \mathrm{~d} \mu \\ &=\int|f+g| \cdot|f+g|^{p-1} \mathrm{~d} \mu \\ &\leq \int(|f|+|g|)|f+g|^{p-1} \mathrm{~d} \mu \\ &=\int|f||f+g|^{p-1} \mathrm{~d} \mu+\int|g||f+g|^{p-1} \mathrm{~d} \mu \\ &\leq\left(\left(\int|f|^{p} \mathrm{~d} \mu\right)^{\frac{1}{p}}+\left(\int|g|^{p} \mathrm{~d} \mu\right)^{\frac{1}{p}}\right)\left(\int|f+g|^{(p-1)\left(\frac{p}{p-1}\right)} \mathrm{d} \mu\right)^{1-\frac{1}{p}} \\ &=\left(\|f\|_{p}+\|g\|_{p}\right) \frac{\|f+g\|_{p}^{p}}{\|f+g\|_{p}} \\ \end{array}\)
We obtain Minkowski's inequality by multiplying both sides by \(\frac{\|f+g\|_{p}^{p}}{\|f+g\|_{p}}\).

Hölder's inequality

Let \((S, \sigma, \mu)\) be a measure space and let \(p, q \in [1, \infty]\) with \(\frac{1}{p} + \frac{1}{q} = 1\). Then for all measurable real- or complex-valued functions \(f\) and \(g\) on \(S\), \(\|fg\|_1 \le \|f\|_p\|g\|_q\)

Reference

https://en.wikipedia.org/wiki/Minkowski_inequality
https://math.stackexchange.com/questions/2280341/why-is-every-p-norm-convex (This blog explained the convexity of norm detailedly.)