Obtain the triangle inequality from the Minkowski inequality

Question

I want to prove:

$$\|f-h\|_p \leq \|f-g\|_p + \|g-h\|_p$$

Minkowski inequality: $$\|f+g\|_p \leq \|f\|_p + \|g\|_p$$ Is $\|f-g\|_p \leq \|f\|_p + \|g\|_p$? It looks like we have:

$$\left(\int_a^b |f-g|^p\right)^{\frac1p}\leq \left(\int_a^b |f|^p\right)^{\frac1p}+\left(\int_a^b |-g|^p\right)^{\frac1p}$$ and since this is in the absolute value sigh, we obtain the result. Is this acceptable? Does the $f-g$ have $f+(-g)$, and hence I can just take the $-g$ like that?

Answer 1

The function names in $\|f+g\|_p \leq \|f\|_p + \|g\|_p$ are placeholders. Functions don't have to be named $f$ and $g$ in order for the inequality to work; they don't have to be named at all — you can plug in some algebraic combinations of functions in place of $f$ and $g$. For example, $$\|24f^2+3f\cos(g)\|_p \leq \|24f^2\|_p + \|3f\cos(g)\|_p$$

For another, simpler, example $$\|f+(-g)\|_p \leq \|f\|_p + \|-g\|_p$$ and you correctly observed that $\|-g\|_p=\|g\|_p$.

That said: you want $f-g$ and $g-h$ on the right side, don't you? How about putting these expressions in "$f$" and "$g$" placeholders?