Minkowski's Inequality

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 $u = f$, $v = f+g$: $\int_0^1 f(f+g) \le \|f\|_2\,\|f+g\|_2$.

The form without absolute values is valid here because in this scene $f(t) = t \ge 0$ and $f+g \equiv 1 \ge 0$, so $|f| = f$, $|f+g| = f+g$ and $\int f(f+g) = \int |f(f+g)|$.

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 ($\int_0^1 g(f+g) = \int_0^1 (1-t)\,dt = \frac12$).

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.)

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: equality here holds because $g = cf$ with $c \ge 0$ (same direction), not by a general $p = 1$ linearity claim: for sign-changing functions, e.g. $g = -f$, $\|f+g\|_1 = 0 < \|f\|_1 + \|g\|_1$. The deeper equality condition comes from the Hölder pairs actually used in deriving Minkowski, $(f, |f+g|^{p-1})$ and $(g, |f+g|^{p-1})$: equality requires $|f|^p$ proportional to $|f+g|^p$ and $|g|^p$ proportional to $|f+g|^p$, i.e. $|f|$ and $|g|$ both proportional to $|f+g|$, which includes this proportional case.)

Step 1 (reduction): it suffices to show that if $\sum_n \|f_n\|_p < \infty$ then $\sum_n f_n$ converges in $L^p$. The reason: a Cauchy sequence with a subsequence converging to $h$ converges to $h$ itself (shown by the $\varepsilon/2$ estimate at the end of Step 4), and Step 4 builds a subsequence of an arbitrary Cauchy sequence exactly as a sum of differences $d_k$ with $\sum_k \|d_k\|_p < \infty$. So Steps 2–3 prove the series claim, and Step 4 applies it.

Step 2: let $M = \sum_n \|f_n\|_p < \infty$, $S_N = \sum_{n \le N} |f_n|$, and $F_N = \sum_{n \le N} f_n$. Applying Minkowski's inequality (problem 1 of this scene, by finite induction) to $S_N$ gives $\|S_N\|_p \le \sum_{n \le N} \|f_n\|_p \le M$. The sequence $S_N$ increases pointwise to $g := \sum_n |f_n| \in [0, \infty]$, so by the monotone convergence theorem $\int g^p = \lim_N \int S_N^p = \lim_N \|S_N\|_p^p \le M^p$. Hence $g \in L^p$ and $g < \infty$ almost everywhere, so $\sum_n f_n(x)$ converges absolutely for almost every $x$. Call its sum $f(x)$ (and set $f(x) = 0$ on the exceptional null set). Then $|f| \le g$, so $f \in L^p$.

Step 3 (tail estimate): apply the same argument to the tail series $\sum_{n > N} |f_n|$: its pointwise sum $g_N := \sum_{n > N} |f_n|$ satisfies $\|g_N\|_p \le \sum_{n > N} \|f_n\|_p$, by applying Minkowski to finite tail sums and passing to the limit with the monotone convergence theorem as in Step 2. Pointwise almost everywhere, $|f - F_N| = \left|\sum_{n > N} f_n\right| \le g_N$. Therefore $\|f - F_N\|_p \le \|g_N\|_p \le \sum_{n > N} \|f_n\|_p \to 0$ as $N \to \infty$, since this is the tail of the convergent series $\sum_n \|f_n\|_p$. Hence $\sum_n f_n \to f$ in $L^p$.

Step 4 (subsequence): let $(h_k)$ be Cauchy in $L^p$. Choose indices $n_1 < n_2 < \cdots$ with $\|h_{n_{k+1}} - h_{n_k}\|_p < 2^{-k}$ for every $k$: this is possible inductively because $(h_k)$ is Cauchy. Put $d_k = h_{n_{k+1}} - h_{n_k}$. Then $\sum_k \|d_k\|_p < \sum_k 2^{-k} < \infty$, and the telescoping identity $h_{n_1} + \sum_{k=1}^{K} d_k = h_{n_{K+1}}$ holds. So Steps 1–3, applied to the series with first term $h_{n_1}$ followed by the $d_k$, give $h_{n_k} \to h$ in $L^p$ for some $h \in L^p$.

Step 4 (conclusion): finally, let $\varepsilon > 0$. Choose $N_0$ so that $k, l \ge N_0$ implies $\|h_k - h_l\|_p < \varepsilon/2$ (the Cauchy condition), and choose $j$ with $n_j \ge N_0$ and $\|h_{n_j} - h\|_p < \varepsilon/2$. Then for every $k \ge N_0$, $\|h_k - h\|_p \le \|h_k - h_{n_j}\|_p + \|h_{n_j} - h\|_p < \varepsilon/2 + \varepsilon/2 = \varepsilon$. Hence $h_k \to h$ in $L^p$. Therefore $L^p[a,b]$ is complete. ∎