(i) d1(f, f) = R 1 0
|f(x) − f(x)| dx =
R 1
0
0 dx = 0. And, if f 6= g, then
F(x) = |f(x)−g(x)| is not identically zero. Hence ∃x0 so F(x0) = 2 >
0. And by continuity, ∃δ such that ∀x, x0−δ < x < x0+δ, F(x) > . So,
d1(f, g) = R 1
0
F(x) dx =
R x0−δ
0
F(x) dx +
R x0+δ
x0−δ
F(x) dx R 1
x0+δ
F(x) dx ≥
R x0+δ
x0−δ
, dx = 2δ > 0

0 = d1(f, g) = R 1
0
|f(x) − g(x)| dx, implies |f(x) − g(x)| = 0 for all x,
and thus f = g.
(ii) d1(f, g) is symmetric in definition, as |f − g| = |g − f|, and thus
equals d1(g, f).
(iii) d1(f, h)+d1(h, g) = R 1
0
|f(x)−h(x)|+|h(x)−g(x)| dx ≥
R 1
0
|f(x)−
g(x)| dx = d1(f, g).