144
7 Device-Independent Quantum Cryptography
Now we apply Jensen’s inequality:
f (x + y) ≥
f (2x) + f (2y)
2
for f (x) = −x log x,
(7.118)
to the last term in (7.116):
h(λ
α
00 + λ
α
01 ) = −(λ
α
00 + λ
α
01 ) log(λ
α
00 + λ
α
01 ) + f (λ
α
10 + λ
α
11 )
≥ −(λ
α
00 + λ
α
01 ) log(λ
α
00 + λ
α
01 ) − λ
α
10 log(2λ
α
10 ) − λ
α
11 log(2λ
α
11 )
= −(λ
α
00 + λ
α
01 ) log(λ
α
00 + λ
α
01 ) − (1 − λ
α
00 − λ
α
01 )
− λ
α
10 log(λ
α
10 ) − λ
α
11 log(λ
α
11 ).
(7.119)
By combining (7.117) and (7.119) in (7.116) we get:
D ≥ −(1 − λ
α
01 ) log(1 − λ
α
01 ) − (λ
α
00 + λ
α
01 ) log(λ
α
00 + λ
α
01 ) − (1 − λ
α
00 − λ
α
01 )
+ λ
α
00 log λ
α
00
= −(λ
α
00 + λ
α
01 ) log(λ
α
00 + λ
α
01 ) − (1 − λ
α
01 ) log[2(1 − λ
α
01 )] + λ
α
00 log(2λ
α
00 )
=: g(λ
α
01 , λ
α
00 ).
(7.120)
In the last expression we defined the function g(x, y):
g(x, y) = −(x + y) log(x + y) − (1 − x) log[2(1 − x)] + y log(2y), (7.121)
and we will analyse it in the ranges of interest for the variables x = λ
α
01 and y = λ
α
00 ,
i.e.: 1/2 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x.
In these ranges the function in (7.121) is concave in x since its second derivative
is always negative:
∂
2 g(x, y)
∂ x 2
= −
1
ln(2)
1
1 − x
+
1
x + y
< 0.
(7.122)
Consider the points at the boundary x + y = 1, for which we get g(1 − y, y) = 0.
Thanks to the concavity of g(x, y), it holds that:
g
p
1
2
+ (1 − p)(1 − y), y
≥ pg
1
2
, y
+ (1 − p)g(1 − y, y), 0 ≤ p ≤ 1
or equivalently that:
g(x, y) ≥
1 − x − y
1
2
− y
g
1
2
, y
.
(7.123)