Bourbaki

Proposition 1

Tag 08QN, permanent. Cite it as [Bourbaki, Tag 08QN].

Integration, chapter IV, § 3. LP spaces

— For every real number p⩾1p \geqslant 1 and every pair of positive functions f,gf, g (finite or not) defined on X,

(∣μ∣∗((f+g)p))1/p⩽(∣μ∣∗(fp))1/p+(∣μ∣∗(gp))1/p (|\mu|^*((f + g)^p))^{1/p} \leqslant (|\mu|^*(f^p))^{1/p} + (|\mu|^*(g^p))^{1/p}

(Minkowski’s inequality).

The inequality (1) is obvious when one of the terms of the second member is equal to +∞+\infty. In the contrary case, ff and gg are finite almost everywhere (§2§ 2, No. 3, Prop. 7). If f1f_1 and g1g_1 are finite positive functions equivalent to ff and gg, respectively, then f1p,g1pf_1^p, g_1^p and (f1+g1)p(f_1 + g_1)^p are equivalent to fp,gpf^p, g^p and (f+g)p(f + g)^p, respectively, and since equivalent positive functions have the same upper integral (§2§ 2, No. 3, Prop. 6), we are reduced to proving the inequality (1) in the case that ff and gg are everywhere finite; but in this case the inequality is a special case of the general Minkowski inequality proved in Ch. I, No. 2, Prop. 3.

We shall also have occasion to make use of the following elementary inequality: if p⩾1p \geqslant 1 then, for any numbers a⩾0, b⩾0a \geqslant 0,\ b \geqslant 0,

ap+bp⩽(a+b)p. a^p + b^p \leqslant (a + b)^p .

The inequality is obvious if a=b=0a = b = 0 or if one of the numbers a,ba, b is +∞+\infty; if a,ba, b are finite and a+b>0a + b > 0, it may be written

(aa+b)p+(ba+b)p⩽1, \left( \frac{a}{a + b} \right)^p + \left( \frac{b}{a + b} \right)^p \leqslant 1 ,

which follows from the fact that

(aa+b)p⩽aa+b,(ba+b)p⩽ba+bandaa+b+ba+b=1. \left( \frac{a}{a + b} \right)^p \leqslant \frac{a}{a + b} , \quad \left( \frac{b}{a + b} \right)^p \leqslant \frac{b}{a + b} \quad \text{and} \quad \frac{a}{a + b} + \frac{b}{a + b} = 1 .

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.