Let $ f $ be a continuous function such that $f([0,1] \supset [0,1]$.

then does this containment make sense that such a continuous function have at least one fixed point?

how can Prove whether this is true of false...!

Any suggestions will be deeply appreciable.


"Barebones" proof: since $f([0,1]) \supset [0,1]$, we have that there are $a, b\in [0,1]$ such that $f(a) \leq 0 \leq a$ and $f(b) \geq 1 \geq b$. Without loss of generality, we may assume $a < b$.

Now define $g(x) := f(x) - x$ on $[a,b]$; then $g$ is continuous, and we have $g(a)= f(a) - a \leq 0$, and $g(b) = f(b) - b\geq 0$. Therefore, by the intermediate value theorem, there is a $c \in (a,b)$ such that $g( c) = 0$, i.e. $f( c) = c$.


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