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First, we prove that f+g has finite p-norm if f and g both do, which follows by
Indeed, here we use the fact that is convex over R+ (for p > 1) and so, by the definition of convexity,
This means that
Now, we can legitimately talk about . If it is zero, then Minkowski's inequality holds. We now assume that is not zero. Using the triangle inequality and then Hölder's inequality, we find that
We obtain Minkowski's inequality by multiplying both sides by
Minkowski's integral inequality
Suppose that and are two ?-finite measure spaces and is measurable. Then Minkowski's integral inequality is (Stein 1970, §A.1), (Hardy, Littlewood & Pólya 1988, Theorem 202) harv error: no target: CITEREFHardyLittlewoodPólya1988 (help):
with obvious modifications in the case . If , and both sides are finite, then equality holds only if a.e. for some non-negative measurable functions ? and ?.
If ?1 is the counting measure on a two-point set then Minkowski's integral inequality gives the usual Minkowski inequality as a special case: for putting for , the integral inequality gives
This notation has been generalized to
for , with . Using this notation, manipulation of the exponents reveals that, if , then .
When the reverse inequality holds:
We further need the restriction that both and are non-negative, as we can see from the example and : .
The reverse inequality follows from the same argument as the standard Minkowski, but uses that Holder's inequality is also reversed in this range.
See also the chapter on Minkowski's Inequality in.
Arthur Lohwater (1982). "Introduction to Inequalities". Missing or empty |url= (help)
^Bullen, Peter S. Handbook of means and their inequalities. Vol. 560. Springer Science & Business Media, 2013.
^Mulholland, H.P. (1949). "On Generalizations of Minkowski's Inequality in the Form of a Triangle Inequality". Proceedings of the London Mathematical Society. s2-51 (1): 294-307. doi:10.1112/plms/s2-51.4.294.