Minkowski's Inequality

Apply the standard proof deriving Minkowski's inequality from Hölder's inequality to this scene's concrete functions $f(t) = t$ , $g(t) = 1-t$ ( $a = b = 1$ ) with $p = 2$ , proving it completely. Verify each step of the proof (the Hölder application, $(p-1)q = p$ , the division) numerically as well.

\|f+g\|_p^p = \int_0^1 (f+g)^p = \int_0^1 f(f+g)^{p-1} + \int_0^1 g(f+g)^{p-1}

\int_0^1 f(f+g)^{p-1} \le \|f\|_p\,\|(f+g)^{p-1}\|_q = \|f\|_p\,\|f+g\|_p^{p-1} \quad \big((p-1)q = p\big)

p = 2

,

a = b = 1

:

\|f\|_2 = \|g\|_2 = \frac{1}{\sqrt{3}} \approx 0.577350

, and

f+g \equiv 1

gives

\|f+g\|_2 = 1

. The Hölder step:

\int_0^1 f(f+g) = \frac12 \le \frac{1}{\sqrt{3}} \approx 0.577350

.

Show proof

With $p = 2$ the conjugate exponent is $q = \frac{p}{p-1} = 2$ , and $(p-1)q = 1 \cdot 2 = 2 = p$ . First split $\|f+g\|_2^2 = \int_0^1 (f+g)^2$ into two terms: $\int_0^1 (f+g)^2 = \int_0^1 f(f+g) + \int_0^1 g(f+g)$ .
Apply Hölder's inequality $\int |uv| \le \|u\|_p\|v\|_q$ to each term with $u = f$ , $v = f+g$ : $\int_0^1 f(f+g) \le \|f\|_2\,\|f+g\|_2$ . Numerically, $\int_0^1 f(f+g) = \int_0^1 t \cdot 1\,dt = \frac12 = 0.5$ while $\|f\|_2\,\|f+g\|_2 = \frac{1}{\sqrt{3}} \cdot 1 \approx 0.5773502692$ , so $0.5 \le 0.5773502692$ . The $g$ term is identical by symmetry ( $\int_0^1 g(f+g) = \int_0^1 (1-t)\,dt = \frac12$ ).
$\frac12 + \frac12 = 1 = \|f+g\|_2^2 \le \big(\|f\|_2 + \|g\|_2\big)\,\|f+g\|_2 = \frac{2}{\sqrt{3}} \approx 1.1547005384$
Adding the two terms gives $\|f+g\|_2^2 \le \big(\|f\|_2 + \|g\|_2\big)\,\|f+g\|_2^{2-1}$ . Since $\|f+g\|_2 = 1 \ne 0$ , dividing by $\|f+g\|_2$ yields $\|f+g\|_2 \le \|f\|_2 + \|g\|_2$ — i.e. $1 \le \frac{2}{\sqrt{3}} \approx 1.1547005384$ . The gap is $\frac{2}{\sqrt{3}} - 1 \approx 0.1547005384$ , matching the scene's number. (For general $p$ the same argument uses $(p-1)q = p$ to get $\|f+g\|_p^p \le \big(\|f\|_p + \|g\|_p\big)\,\|f+g\|_p^{p-1}$ , then divides by $\|f+g\|_p^{p-1}$ for Minkowski's inequality.)

Prove completely, starting from homogeneity $\|cf\|_p = c\|f\|_p$ , that Minkowski's inequality is an equality when $g = c\,f$ for a constant $c \ge 0$ . In this scene that is the case $a = b = 0$ ( $f = g \equiv 1$ , i.e. $c = 1$ ); confirm numerically that the gap is 0.

g = cf\ (c \ge 0) \;\Rightarrow\; \|f+g\|_p = (1+c)\|f\|_p = \|f\|_p + \|g\|_p

a = b = 0:\quad \|f\|_p = \|g\|_p = 1,\quad \|f+g\|_p = 2

p = 2

,

a = b = 0

:

f = g \equiv 1

, so

g = 1 \cdot f

— proportional.

\|f\|_2 = \|g\|_2 = 1

,

\|f+g\|_2 = 2

, gap

= 0

.

Show proof

First prove homogeneity. For $c \ge 0$ , $\|cf\|_p^p = \int_0^1 |cf|^p = \int_0^1 c^p |f|^p = c^p \|f\|_p^p$ , so taking $p$ -th roots gives $\|cf\|_p = c\|f\|_p$ .
If $g = cf$ with $c \ge 0$ , then $f + g = (1+c)f$ , so by homogeneity $\|f+g\|_p = \|(1+c)f\|_p = (1+c)\|f\|_p = \|f\|_p + c\|f\|_p = \|f\|_p + \|cf\|_p = \|f\|_p + \|g\|_p$ — equality holds.
For the scene: $a = b = 0$ gives $f(t) = t^0 \equiv 1$ and $g(t) = (1-t)^0 \equiv 1$ , so $g = 1 \cdot f$ . Hence $\|f\|_2 = \left(\int_0^1 1\,dt\right)^{1/2} = 1$ , $\|g\|_2 = 1$ , $\|f+g\|_2 = \|2\|_2 = 2$ , matching the right side $\|f\|_2 + \|g\|_2 = 2$ — the gap is exactly 0. (Note: at $p = 1$ equality always holds by linearity $\int (f+g) = \int f + \int g$ . The deeper equality condition — that the Hölder steps need $|f|^p$ proportional to $|g|^q$ — includes this proportional case.)

Show that Minkowski's inequality fails for $0 < p < 1$ . First prove that for $f, g \ge 0$ the inequality reverses to $\|f+g\|_p \ge \|f\|_p + \|g\|_p$ , then compute the concrete counterexample $p = \frac12$ , $f(t) = t$ , $g(t) = 1-t$ directly and verify numerically that $\|f+g\|_{1/2} = 1 > \frac89 = \|f\|_{1/2} + \|g\|_{1/2}$ .

0 < p < 1:\quad (f+g)^p \ge f^p + g^p \;\Rightarrow\; \|f+g\|_p \ge \|f\|_p + \|g\|_p

p = \tfrac12:\quad \|f\|_{1/2} = \|g\|_{1/2} = \tfrac49,\quad \|f+g\|_{1/2} = 1 > \tfrac89

This problem is about a separate computation at

p = \frac12

, outside the scene's slider range (

1 \le p \le 4

) — the preset is the scene default

p = 2

,

a = b = 1

.

Show proof

Let $0 < p < 1$ and $f, g \ge 0$ . Pointwise $(f+g)^p \ge f^p + g^p$ (the base $f+g \ge \max(f, g)$ with positive exponent), so integrating gives $\|f+g\|_p^p \ge \|f\|_p^p + \|g\|_p^p$ . With $r = \frac1p > 1$ we have $(A+B)^r \ge A^r + B^r$ for $A, B \ge 0$ (e.g. $(1+t)^r \ge 1 + t^r$ for $t \ge 0$ ), hence $\|f+g\|_p \ge \big(\|f\|_p^p + \|g\|_p^p\big)^{1/p} \ge \|f\|_p + \|g\|_p$ — the inequality reverses.
Concretely, take $p = \frac12$ , $f(t) = t$ , $g(t) = 1-t$ . Since $\int_0^1 \sqrt{t}\,dt = \frac23$ , we get $\|f\|_{1/2} = \left(\frac23\right)^2 = \frac49 \approx 0.444444$ , and by symmetry $\|g\|_{1/2} = \frac49$ . So the right side is $\frac49 + \frac49 = \frac89 \approx 0.888889$ . But $f + g \equiv 1$ , so $\|f+g\|_{1/2} = \left(\int_0^1 1\,dt\right)^2 = 1$ . Hence $1 > \frac89$ — Minkowski's inequality breaks, and the gap is $-\frac19 \approx -0.111111$ , negative (also confirmed by numerical integration).
The exact failure point of the standard proof is also clear: Hölder's inequality needs the conjugate exponent $q = \frac{p}{p-1} > 1$ , but for $p < 1$ we get $q < 0$ , so the Hölder step itself is unavailable. That is why Minkowski's inequality essentially requires $p \ge 1$ .

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