Show that if f is continuous and 0‰¤f(x)‰¤1 for all xˆˆ[0,1] then there is a number cˆˆ[0,1] such that f(c)=c.

Hello!
To find some `c` such that `f(c) = c` is actually the same as to solve for `c` the equation
`f(c) - c = 0,` or to find root(s) of `f(c) - c.`

There is a well-known
Bolzano theorem about a root of a continuous function: if a function g is continuous on `[ a, b
]` and has different signs on `a` and `b,` then it has at least one root somewhere in
between.

It would be good to use it here, but is it possible? We clearly have
`a = 0, b = 1` and `g(x) = f(x) - x.` Is `g` continuous? Yes, because it is a difference of
continuous functions (it is a simple exercise for you that `h(x) = x` is continuous).

What about different signs? `g(0) = f(0) - 0 = f(0),` and it is given that `0 lt= g(0)
= f(0) lt= 1.` Further, `g(1) = f(1) - 1,` and again `0 lt= f(1) lt= 1,` so `-1 lt= g(1) = f(1)
- 1 lt= 0.` This way, `g(0) gt= 0` and `g(1) lt= 0.`

Well, if `g(0) gt 0` and
`g(1) lt 0,` then we can apply Bolzano theorem. But what if `g(0) gt= 0` but not `gt 0,` of if
`g(1) lt= 0` but not `lt 0` ? This is the question for you to think on.

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