Estimation of a function by a linear function

Question

I am stuck on this problem:

Problem: Let $f:[0,1]\rightarrow\mathbb{R}$. Suppose that there exists $\epsilon>0$ such that for all $x,y\in[0,1]$ we have $|f(x)-f(y)|\leq \epsilon$. Let $g$ be the linear function with $g(0)=f(0)$ and $g(1)=f(1)$, i.e. $$g(x)=(f(1)-f(0))x+f(0).$$ Prove that for all $x\in [0,1]$ we have $|f(x)-g(x)|\leq \epsilon$.

So far, I have been able to get a $2\epsilon$ upper bound by \begin{align*} |f(x)-g(x)|&\leq |f(x)-(f(1)-f(0))x-f(0)| \\ & \leq |f(x)-f(0)|+|f(1)-f(0)|\cdot|x| \\& \leq \epsilon+\epsilon\cdot 1 =2\epsilon. \end{align*} But I am not sure how to get the $\epsilon$ bound. I tried other forms of the triangle inequality, but often I get a $|g(x)-g(0)|$ term which is not generally small.

Answer 1

Let $m=\inf \{f(x): 0\leq x\leq 1\}$ and $M=\sup \{f(x): 0\leq x\leq 1\}$. Then $M-m \leq \epsilon$. Since $g(0)$ and $g(1) \in [m,M]$ and $g$ is linear it follows that $g(x) \in [m,M]$ for all $x$. Now the distance between any two points of $[m,M]$ is at most $M-m \leq \epsilon$. Hence $|f(x)-g(x)| \leq \epsilon$.

Answer 2

Hint : show that there exists an interval $[a,a+\varepsilon]$ such that for all $x \in [0,1]$, $f(x) \in [a,a+\varepsilon]$.

Then deduce that for all $x \in [0,1]$, $g(x) \in [a,a+\varepsilon]$. And finally, deduce that $|f(x)-g(x)| \leq \varepsilon$.

Answer 3

Because we know that $|f(x)-f(y)|\le\epsilon$, we can say that $f$ is bounded. Now, WLOG, assume that $f(0)\le f(1)$. So, we know that $g(x)\in[f(0),f(1)]$. So, $f(x)-g(x)\in[f(x)-f(1),f(x)-f(0)]$. Now, because $|f(x)-f(y)|\le\epsilon\forall x,y\in[0,1]$, $|f(x)-g(x)|\le\epsilon$. $\blacksquare$

Another way of thinking about this is that $A:=\{g(x)|x\in[0,1]\} = [f(0),f(1)]$ represents the extent of $g(x)$. Now, the extent of $f$ shown by $B:=\{f(x)|x\in[0,1]\}= [\min_{x\in[0,1]}f(x),\max_{x\in[0,1]}f(x)]$. Now, as $A$ is by definition a proper subset of $B$, we have proven the lemma.