Showing that $f(x)=x$ for at least one point in $[0,1]$

Problem: Let $$I=[0,1]$$ be the closed unit interval. Suppose $$f$$ is a continuous mapping of $$I$$ into $$I$$. Prove that $$f(x)=x$$ for at least one $$x \in I$$.

Attempt:

We have a known result:

Let $$g$$ be a continuous real valued function on a metric space $$X$$. Then $$Z(g)=\{p \in X: g(p)=0 \}$$, i.e. the zero set of $$g$$ is closed. [Proof is simple using "inverse image of a closed set is closed under continuous map"]

Now, we construct $$F:I \to I$$, such that $$F(x) =f(x)-x$$, which is definitely continuous, being a linear combination of two continuous functions.
[It is assumed that $$F(x)=f(x)-x>0$$, $$\forall x\in I$$. [ If $$x>f(x)$$, consider the function $$F_1(x)=x-f(x)$$ ]. Otherwise, if the function $$F(x)$$ changes sign somewhere within the interval, it must attain the value $$0$$, giving us $$f(x)=x$$ ].

But we know that $$Z(F)$$ is closed, hence it cannot be $$\phi$$. We are done.

Is this at all a valid proof?

Edit: Being doubtful, I write up another approach:

The function $$f$$ maps into $$I$$, i.e. itself. Hence, to be $$f(x)>x$$ for every value of $$x$$, its range set would have to exceed $$I$$, and the best case scenario would be the identity map, for which $$f(x)=x$$ for all $$x$$ . Otherwise, $$F(x)$$ must change sign.

No, it is not at all valid. The empty set is also closed.

Note that your "proof" would also apply to $$F(x) = f(x) - x/2$$, but there is not necessarily a solution to $$f(x) = x/2$$.

• Could you please check my "proof" now? – Subhasis Biswas Jan 16 '19 at 20:47
• Not only that, using my argument would be so disastrous, that we can "Prove" every continuous function attains zero, which is ridiculous. – Subhasis Biswas Jan 16 '19 at 20:51

No, the proof is not valid. It is not clear (nor is it necessarily true) that $$F$$ is a mapping from $$I$$ to $$I$$.

I would approach it this way: Letting $$F$$ be as you specified, it follows that $$F(0) \geq 0$$ while $$F(1) \le 0$$. Furthermore, $$F$$ is also continuous. Therefore by the intermediate value theorem there is an $$x \in [0,1] = I$$ s.t. $$F(x) = 0$$, or equvialently, $$f(x) = x$$.

• can you check mine now? – Subhasis Biswas Jan 16 '19 at 20:45
• Well, the writing really does need to be improved. And, the inequality $f(x) > x$ could hold for every point in $I \setminus \{1\}$. So I would advise you to try again. – Mike Jan 16 '19 at 20:50
• The domain is compact!? So we cannot leave out the boundary point. – Subhasis Biswas Jan 16 '19 at 20:53
• Does the result hold good in $(0,1)$? Can you help me out? – Subhasis Biswas Jan 16 '19 at 20:56
• To answer your question, the result does not necessarily hold in $(0,1)$. For example, consider the function $f(x)= 1/2 + x/2$. The idea for showing that it does hold in $[0,1]$ is this: The function $F$ is nonnegative at $x=0$ and it is nonpositive at $x=1$. so as $F$ is also continuous, somewhere in $[0,1]$ (even if it is at precisely one endpoint $x=0$ or the other endpoint $x=1$ or at both endpoints of $I$, it must take on the value 0. – Mike Jan 16 '19 at 20:59