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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.

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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 $f(x) = x/2$.

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

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  • $\begingroup$ can you check mine now? $\endgroup$ – Subhasis Biswas Jan 16 '19 at 20:45
  • $\begingroup$ 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. $\endgroup$ – Mike Jan 16 '19 at 20:50
  • $\begingroup$ The domain is compact!? So we cannot leave out the boundary point. $\endgroup$ – Subhasis Biswas Jan 16 '19 at 20:53
  • $\begingroup$ Does the result hold good in $(0,1)$? Can you help me out? $\endgroup$ – Subhasis Biswas Jan 16 '19 at 20:56
  • $\begingroup$ 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. $\endgroup$ – Mike Jan 16 '19 at 20:59

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