Suppose $f: [a,b]\longrightarrow \mathbb{R}$ is a continous function. If $\forall [c,d]\subset[a,b]$ there exists an open interval $U\subset[a,b]$ s.t. $f$ is constant on $U$, then can we infer that $f$ is constant on $[a,b]$? Motivation: Some days ago I asked a question: $f$ real continuous map in $[a,b]$. $∀ x_0∈[a,b], ∃ (c, d) : x∈ (c, d)⊂ [a,b]$ s.t. $x$ is minimum for $f$ in $(c,d)⇒f$ is constant map, and I wonder whether the statement is still true after we weaken its assumption. My try on this problem: I guess the statement is true. To prove it, we assume the contrary. If $a\leq\alpha<\beta\leq b$ and $f(\alpha)\neq f(\beta)$, then we divide $[\alpha, \beta]$ into $[\alpha,\frac{\alpha+\beta}{2}]$ and $[\frac{\alpha+\beta}{2}, \beta]$. $f$ is not constant on at least one of the two intervals. We then divide the interval on which $f$ is not constant into two equal parts... So we learn that $\exists\eta\in[a,b]$ s.t. $f$ is not constant on any closed interval containing $\eta$, but I do not know how this shall cause contradiction. Furthermore, if the foregoing statement is unfortunately false, what if we strengthen back the assumption: If $f\in C([a,b])$ and $\forall x\in(a,b)$ there exists an open interval $U$ containing $x$ s.t. $f$ is constant on $U$, can we infer that $f$ is constant on$[a,b]$? Applying the same strategy we learn that if we assume the contrary, then (wlog it's $a$) $\forall \varepsilon>0$, $f$ is not constant on $[a, a+\varepsilon]$, but I still can't find contradiction.
Asked
Active
Viewed 65 times
