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For every pair of continuous functions $f, g$ $: \left[0, 1\right] \rightarrow R $, such that $max(f(x) : x \in \left[0, 1\right]) $$= $ $max( g(x) : x \in \left[0, 1\right] )$ , the correct statement(s) is(are)

(1)$(f(c))^2 + 3f(c) = (g(c))^2 + 3g(c)$ for some $ c \in [0, 1] $

(2)$(f(c))^2 + f(c) = (g(c))^2 + 3g(c)$ for some $ c \in [0, 1] $

(3)$(f(c))^2 + 3f(c) = (g(c))^2 + g(c)$ for some $ c \in [0, 1] $

(4)$(f(c))^2 = (g(c))^2 $ for some $ c \in [0, 1] $

My work

In this one possibility is that they have same maxima, assume it to be $c$ and this will satisfy option $1$ and $4$

Is there any other possibility than this ?

Aakash Kumar
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  • What tells you their respective maxima are attained at the same point $c$ (which appears to be what you assume in your last paragraph)? – Clement C. Jul 17 '16 at 15:49
  • It is the possibility. – Aakash Kumar Jul 17 '16 at 15:50
  • What do you mean? You are asked to state which statements are always true given the assumptions on $f,g$ not those that "could maybe be true for some particular choice of $f,g$." – Clement C. Jul 17 '16 at 15:51

2 Answers2

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tl;dr: 1. and 4. are true (but not for the reason you gave), 2. and 3. are false (there are counterexamples). Namely, you are asked to give the statements that are true for all $f,g$ satisfying the assumptions, not those that are true for some $f,g$ satisfying the assumptions.

  • Your last paragraph assumes that $f$ and $g$ attain their common maximum on the same point $c\in[0,1]$. This needs not be the case (and there are very simple counterexamples). However, we can show that there exists some $c\in[0,1]$ for which $f(c)=g(c)$ (but this may not be the maximum value).

    In more detail: let $h = f - g\colon[0,1]\to\mathbb{R}$. It is a continuous function; let $x_f$ (resp. $x_g$) be the point at which $f$ (resp. $g$) achieves its maximum. If $x_f=x_g$, we are done; otherwise, since $h(x_f) \geq 0$ and $h(x_g) \leq 0$, the intermediate value theorem (IVT) ensures that there exists $c\in[x_f,x_g]$ such that $h(c)=0$.

    Now, we have that $c\in[0,1]$ for which $f(c)=g(c)$, which indeed proves that 1. and 4. are true.

  • Let us focus on 2. and 3. now. Since the statements and the assumptions on $f,g$ are symmetric in $f,g$, it is enough to prove or disprove only one the the two statements. Let us disprove 2. by choosing simple functions $f,g$ possible: $f(x)=g(x)=x+1$. It is not hard to see that there is no $x\in[0,1]$ such that $(x+1)^2+(x+1)=(x+1)^2+3(x+1)$ (since the only solution to this equation is $-1$).

Clement C.
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As @Clement C. pointed in the comments, you're reasoning is wrong. You must show which of these statements are true for every such $f, g$

Let $f, g$ be such functions, $M = max \{f(x) : x \in [0, 1]\}$ and $a : f(a) = M, b : g(b) = M$. Then, if $a = b$ we're done. Otherwise, we can assume $a < b$ without loss of generality and consider $h = f - g$, having

$$h(a) = M - g(a) \geq M - M = 0$$ $$h(b) = f(b) - M \leq M - M = 0$$

Then, by Bolzano there must exist $c \in [a, b] : h(c) = 0 \iff f(c) = g(c)$, proving that both 1 and 4 are true.

For 2 and 3, consider $f(x) = g(x) = x + 1$. If such a $c$ existed we'd have $(x+1)^2+3(x+1)=(x+1)^2+(x+1) \iff 3x + 3 = x + 1 \iff x = -1$, which is impossible in $[0, 1]$. The other case is analogous.

cronos2
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