# Exercise 3.4.14 Introduction to Real Analysis by Jiri Lebl

Suppose for $$f: [0,1] \to R$$, we have $$|f(x) - f(y) | \le K |x-y|$$, and $$f(0)= f(1) = 0$$. Prove that $$|f(x)| \le K/2$$. Further show by example that $$K/2$$ is the best possible, that is, there exists such a continuous function for which $$|f(x)| = K/2$$ for some $$x \in [0 ,1]$$.

I am trying to find $$x$$ and $$y$$ that satisfy $$|f(x)| \le K/2$$, but I constantly fails. Can you give some help? I also appreciate if you give some hint for the second part of the question (Further show ~).

Note that $$|f|$$ is continuous, so some $$M\in[0,1]$$ is such that $$|f(x)|\leq|f(M)|$$.
Assume that $$0\leq M\leq 1/2$$, then $$|f(M)|=|f(M)-f(0)|\leq K|M-0|=KM\leq(1/2)K$$, so $$|f(x)|\leq(1/2)K$$.
Assume that $$1/2\leq M\leq 1$$, then $$|f(M)|=|f(M)-f(1)|\leq K|M-1|=K(1-M)\leq K(1-1/2)=(1/2)K$$, so $$|f(x)|\leq(1/2)K$$ too.
Hint: for any $$x \in [0,1]$$ either $$x$$ is closer to $$0$$ or $$x$$ is closer to $$1$$. If $$y \in \{0,1\}$$ is the closer endpoint then $$|x - y| \le 1/2$$.
Hint2: $$|f(x) - f(0)| = |f(x)| \le K|x|$$. To make this an equality, we need $$f(x) = \pm Kx$$ for any suitable $$x$$. Don't forget about the condition $$f(1) = 0$$—if you can do the first hint you may have an idea what I mean by "suitable" $$x$$.
• Thanks for the answer. I am still struggling for the second part. How can I use $f(1) = 0$ to get the equality? – shk910 Dec 21 '19 at 3:48
• @shk910 Replace $0$ by $1$ in the hint to get $|f(x) - f(1)| = |f(x)| \le K|x - 1|$. – Trevor Gunn Dec 21 '19 at 4:04