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I would like to see a proof of when equality holds in Minkowski's inequality.

Minkowski's inequality. If $1\le p<\infty$ and $f,g\in L^p$, then $$\|f+g\|_p \le \|f\|_p + \|g\|_p.$$

The proof is quite different for when $p=1$ and when $1<p<\infty$. Could someone provide a reference? Thanks!

Martin Sleziak
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Sarah
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2 Answers2

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For $p = 1$, the proof uses the triangle inequality, $\lvert f(x) + g(x)\rvert \leqslant \lvert f(x)\rvert + \lvert g(x)\rvert$, and the monotonicity of the integral. You have equality $\lVert f+g\rVert_1 = \lVert f\rVert_1 + \lVert g\rVert_1$ if and only if you have equality $\lvert f(x) + g(x)\rvert = \lvert f(x)\rvert + \lvert g(x)\rvert$ almost everywhere. That means that almost everywhere at least one of the two functions attains the value $0$, or both values "point in the same direction", that is, have the same argument. You can simply formalise that as $f(x)\cdot\overline{g(x)} \geqslant 0$ almost everywhere.

For $1 < p < \infty$, in addition to the triangle inequality, the proof of Minkowski's inequality also uses Hölder's inequality,

$$\begin{align} \int \lvert f+g\rvert^p\,d\mu &\leqslant \int \lvert f\rvert\cdot\lvert f+g\rvert^{p-1}\,d\mu + \int \lvert g\rvert\cdot\lvert f+g\rvert^{p-1}\,d\mu\tag{1}\\ &\leqslant \lVert f\rVert_p \lVert f+g\rVert_p^{p-1} + \lVert g\rVert_p \lVert f+g\rVert_p^{p-1}.\tag{2} \end{align}$$

You then have equality $\lVert f+g\rVert_p = \lVert f\rVert_p + \lVert g\rVert_p$ if and only equality holds in $(1)$, and for both terms in $(2)$. Equality in $(1)$ is nearly the same as for the $p=1$ case, that gives a restriction $f(x)\cdot\overline{g(x)} \geqslant 0$ almost everywhere, except maybe on the set where $f(x)+g(x) = 0$ (but equality in Hölder's inequality forces $f(x) = 0$ and $g(x) = 0$ almost everywhere on that set, so at the end we really must have $f(x)\cdot \overline{g(x)} \geqslant 0$ almost everywhere if $\lVert f+g\rVert_p = \lVert f\rVert_p + \lVert g\rVert_p$). For equality of the terms in $(2)$, if $f+g = 0$ almost everywhere, we must have $f=g=0$ almost everywhere, and if $\lVert f+g\rVert_p > 0$, we have equality if and only there are constants $\alpha,\beta \geqslant 0$ with $\lvert f\rvert^p = \alpha \lvert f+g\rvert^p$, and $\lvert g\rvert^p = \beta\lvert f+g\rvert^p$ almost everywhere. We can of course take $p$-th roots, and together with the condition imposed by the triangle inequality, we obtain that

$$\lVert f+g\rVert_p = \lVert f\rVert_p + \lVert g\rVert_p$$

holds if and only if there are non-negative real constants $\alpha,\beta$, not both zero, such that $\alpha f(x) = \beta g(x)$ almost everywhere.

Daniel Fischer
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    What is the bar notation above $g(x)$? I'm a little confused by your exposition on the first paragraph about at least one of the two functions attaining the value 0 or "pointing in the same direction". Are you assuming they are complex-valued? – mathjacks Apr 25 '15 at 23:46
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    Yes, I'm assuming complex-valued functions, so $\overline{g(x)}$ denotes the complex conjugate. For (extended) real-valued functions, drop the conjugate and replace "point in the same direction" with "have the same sign" (where, just in case, $0$ has the same sign as any [extended] real number). – Daniel Fischer Apr 25 '15 at 23:49
  • I don't know if you will see this, but I think there is a typo in the (2) equation. The Hölder Inequality implies that (1) $\leq |f|_p||f+g|^{p-1}|_q+|g|_p||f+g|^{p-1}|_q$ – Affonso Jun 27 '17 at 02:50
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    @Affonso Yes, but since $(p-1)q = p$ (and $p/q = p-1$), we have $$\lVert h^{p-1}\rVert_q = \Biggl(\int h^{(p-1)q},d\mu\Biggr)^{1/q} = \Biggl(\int h^p,d\mu\Biggr)^{1/q} = \lVert h\rVert_p^{p/q} = \lVert h\rVert_p^{p-1}.$$ – Daniel Fischer Jun 27 '17 at 09:30
  • Why does the equaltiy $\int \lvert f\rvert\cdot\lvert f+g\rvert^{p-1},d\mu + \int \lvert g\rvert\cdot\lvert f+g\rvert^{p-1},d\mu = \lVert f\rVert_p \lVert f+g\rVert_p^{p-1} + \lVert g\rVert_p \lVert f+g\rVert_p^{p-1}$ imply "term by term" equality? – JKEG Oct 25 '19 at 23:56
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    @JKEG Each term on the left is $\leqslant$ the corresponding term on the right (Hölder's inequality). Since all terms are finite (by the assumption that $f,g \in L^p$) the sums can only be equal if the corresponding terms are equal. If either of the inequalities were strict, there'd be a strict inequality between the sums. – Daniel Fischer Oct 26 '19 at 00:01
  • For $p>1$ the equality in (1) case may need some additional argument to handle the case of $f(x)+g(x)=0$ on a set of positive measure? Because we can't conclude that $f(x)\overline{g(x)}\ge0$ in that case. – PatrickR Mar 31 '20 at 23:53
  • @PatrickR Indeed, on the set where $f(x) + g(x) = 0$ we can only conclude $f(x)\overline{g(x)} \geqslant 0$ almost everywhere with the equality in Hölder's inequality. Thanks for pointing that out. – Daniel Fischer Apr 24 '20 at 21:25
  • In the first paragraph, if the values attained by $x$ and $y$ point in the same direction almost everywhere, or if at least one of them is equal to zero, can we write that $y = \theta$ or $x = \theta y$ for some constant $\alpha$? Couldn't we have, say, $x(t) = t$ and $y(t) = t^2$ where $|x|_1 + |y|_1 = |x + y|_1$ but there is no constant for which $x = \alpha y$? – akm Sep 21 '20 at 18:19
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    @akm For $p = 1$ (and for $p = \infty$) we can have equality without one function being a nonnegative multiple of the other (almost everywhere). In the $p = 1$ case all we need is that $f \cdot \overline{g} \geqslant 0$ almost everywhere, there need not even exist a nonnegative function $q$ such that $g = q\cdot f$ or $f = q\cdot g$ (but we can write $\alpha f = \beta g$ [a.e.] with two nonnegative functions $\alpha,\beta$ if equality holds, e.g. $\alpha = \lvert g\rvert$ and $\beta = \lvert f\rvert$ work). – Daniel Fischer Sep 22 '20 at 11:18
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Let $q$ be a conjugate exponent of $p$, meaning $\frac{1}{q} + \frac{1}{p} = 1$. Now $\|f + g\|_p = \|f\|_p + \|g\|_p$ iff $\|f+ g\|_p^p = \|f + g\|_p^{p - 1}(\|f\|_p + \|g\|_p)$.

So since $$ \begin{split} \|f + g\|_p^p & = \int_X |f + g|^p d\mu = \int_X (|f + g|)|f + g|^{p - 1} d\mu \\ &\leq \int_X \big(|f| + |g|\big)|f + g|^{p - 1} d\mu = \int_X |f||f + g|^{p-1} d\mu + \int_X |g||f + g|^{p-1} d\mu \\ & = \|f(f + g)^{p - 1}\|_{L^1} + \|g(f + g)^{p - 1}\|_{L^1} \\ &\leq \|f\|_p\|(f + g)^{p - 1}\|_q + \|g\|_p \|(f + g)^{p - 1}\|_q \\ & = \big(\|f\|_p + \|g\|_p\big)\|(f + g)^{p - 1}\|_q \\ & = \big(\|f\|_p + \|g\|_p\big)\left(\int_X \big(|f(x) + g(x)|^{p - 1}\big)^q d\mu\right)^{1/q} \\ & = \big(\|f\|_p + \|g\|_p\big)\left(\int_X |f(x) + g(x)|^{pq - q} d\mu\right)^{1/q} \\ & = \big(\|f\|_p + \|g\|_p\big)\left(\int_X |f(x) + g(x)|^{p} d\mu\right)^{1/q} \\ & = \big(\|f\|_p + \|g\|_p\big)\left(\int_X |f(x) + g(x)|^{p} d\mu\right)^{1 - 1/p} \\ & = \big(\|f\|_p + \|g\|_p\big)\left(\int_X |f(x) + g(x)|^{p} d\mu\right)^{\frac{p - 1}{p}} = \|f + g\|_p^{p - 1}\big(\|f\|_p + \|g\|_p\big) \end{split} $$ where the $3^{rd}$ inequality follows from the triangular inequality and the $6^{th}$ inequality holds from Hölder's inequality.

Therefore we have that Minkowski's inequality gives us equality iff the triangular inequality is an equality almost everywhere and Hölder's inequality is an equality. Now the triangular inequality is an equality almost everywhere, iff $g(x)$ and $f(x)$ have the same sign almost everywhere, or if on every set of positive measure where the two differ by sign at least one of them is zero. We have equality for Hölder’s if $|f|^p$ is a constant multiple of $|g|^q$ almost everywhere.

Daniele Tampieri
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Samantha Wyler
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