I've, once again, already worked out this problem, and am just looking for somebody to validate my proof, or point out any errors. Heres the problem:

If f is continuous on [0, 1] and f(0) = f(1), then prove there is some natural number number n such that:

f(x) = f(x + \frac{1}{n})

for some x \in [0, 1]


Now, heres my attempt at proving it:

Let

g(x) = f(x) - f(x + \frac{1}{n})

We know that g is continuous on [0, 1]. We also know that there exists a number z \in [0, 1] such that:

f(z) \geq f(x)

for all x \in [0, 1]

Suppose we let a = z - \frac{1}{n}; then we know that we can choose an n large enough to ensure that (since f(z) \geq f(x) for all x):


f(z - \frac{1}{n}) < f(z + \frac{1}{n} - \frac{1}{n})

Which means that:

f(a) < f(a + \frac{1}{n})

Which implies that:

g(a) < 0

Now, we can choose a b \in [0, 1] simmilarly. Suppose b = z, then we can ensure (by choosing a large enough or small enough n) ensure that:

f(z + \frac{1}{n}) < f(z)

Equivilantly...

f(b + \frac{1}{n}) < f(b)

Which implies that:

0 < g(b)

Now, by that one theorem (I don't remeber its name), since:

g(a) < 0 < g(b)

and g is continuous on [0, 1], and a, b \in [0, 1], then we a garunteed there exists some x \in [0, 1] such that:

g(x) = 0

This implies that we are also garunteed an x \in [0, 1] such that:

f(x) = f(x + \frac{1}{n})

This completes the proof,.... I hope

I'm not certain if this proof is correct... I'm alittle unsure about weather or not I can garuntee some of the inequalities involving f(z). I'm not even sure if this proof "goes in the right direction" persay. Any guidence, validation, error-pointing-out, and general advice would be much appreciated. Thanks in advance