In mathematical analysis, the universal chord theorem states that if a function f is continuous on [a,b] and satisfies                     f        (        a        )        =        f        (        b        )                , then for every natural number                     n                , there exists some                     x        ∈        [        a        ,        b        ]                 such that                     f        (        x        )        =        f        (        x        +                              1            n                          )                .
The theorem was published by Paul Lévy in 1934 as a generalization of Rolle's Theorem.
Let                     H        (        f        )        =        {        h        ∈        [        0        ,        1        ]        :        f        (        x        )        =        f        (        x        +        h        )                   for some                 x        }                 denote the chord set of the function f. If f is a continuous function and                     h        ∈        H        (        f        )                , then                                           h            n                          ∈        H        (        f        )                 for all natural numbers n. 
The case when n = 2 can be considered an application of the Borsuk–Ulam theorem to the real line. It says that if                     f        (        x        )                 is continuous on some interval                     I        =        [        a        ,        b        ]                 with the condition that                     f        (        a        )        =        f        (        b        )                , then there exists some                     x        ∈        [        a        ,        b        ]                 such that                     f        (        x        )        =        f        (        x        +        1                  /                2        )                .
In less generality, if                     f        :        [        0        ,        1        ]        →                  R                         is continuous and                     f        (        0        )        =        f        (        1        )                , then there exists                     x        ∈                  [          0          ,                                    1              2                                ]                         that satisfies                     f        (        x        )        =        f        (        x        +        1                  /                2        )                .