Negation of monotonicity of a continuous functionSo I am given a continuous function mapping a connected domain to the reals, i.e f : ( a , b ) → R .I want to show that if f is not strictly monotone and f is continuous, we have x , y , z ∈ ( a , b ) with x &lt; y &lt; z such that: f ( x ) ≤ f ( y ) and f ( y ) ≥ f ( z )or f ( x ) ≥ f ( y ) and f ( y ) ≤ f ( z )As for what I've tried, using the negation of strictly monotone, we have that f must be increasing and decreasing in two intervals [ x 1 , x 2 ] and [ x 3 , x 4 ] i.e.: f ( x 1 ) ≤ f ( x 2 ) and f ( x 3 ) ≥ f ( x 4 ) or f ( x 1 ) ≥ f ( x 2 ) and f ( x 3 ) ≤ f ( x 4 )The problem comes when I look to combine the two into a single inequality. In an effort to combine these, the only way I see of doing it (directly) would be casewise. If anyone could provide a hint or (less preferred) a complete direct proof I would appreciate it.

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Negation of monotonicity of a continuous functionSo I am given a continuous function mapping a connected domain to the reals, i.e   f  :  (  a  ,  b  )  →      R  .I want to show that if f is not strictly monotone and f is continuous, we have   x  ,  y  ,  z  ∈  (  a  ,  b  ) with   x  &amp;lt;  y  &amp;lt;  z such that:  f  (  x  )  ≤  f  (  y  )   and   f  (  y  )  ≥  f  (  z  )or   f  (  x  )  ≥  f  (  y  )   and   f  (  y  )  ≤  f  (  z  )As for what I&#039;ve tried, using the negation of strictly monotone, we have that f must be increasing and decreasing in two intervals   [      x    1    ,      x    2    ] and   [      x    3    ,      x    4    ] i.e.:  f  (      x    1    )  ≤  f  (      x    2    )   and   f  (      x    3    )  ≥  f  (      x    4    ) or   f  (      x    1    )  ≥  f  (      x    2    )   and   f  (      x    3    )  ≤  f  (      x    4    )The problem comes when I look to combine the two into a single inequality. In an effort to combine these, the only way I see of doing it (directly) would be casewise. If anyone could provide a hint or (less preferred) a complete direct proof I would appreciate it.

Izabella Henson · Accepted Answer

Explanation:Use proof by contradiction. If your statement is false, then for every   x  &amp;lt;  y  &amp;lt;  z, we have   f  (  x  )  &amp;gt;  f  (  y  )         or       f  (  y  )  &amp;lt;  f  (  z  ) and   f  (  x  )  &amp;lt;  f  (  y  )         or       f  (  y  )  &amp;gt;  f  (  z  )  .

So I am given a continuous function mapping a connected domain to the reals, i.e f:(a, b) rightarrow R. I want to show that if f is not strictly monotone and f is continuous, we have x, y, z in (a,b) with x<y<z such that: f(x) leq f(y) and f(y) geq f(z) or f(x) geq f(y) and f(y) leq f(z).

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