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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 ~).

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2 Answers 2

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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.

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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" $x$.

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    $\begingroup$ Thanks for the answer. I am still struggling for the second part. How can I use $f(1) = 0$ to get the equality? $\endgroup$
    – shk910
    Dec 21, 2019 at 3:48
  • $\begingroup$ @shk910 Replace $0$ by $1$ in the hint to get $|f(x) - f(1)| = |f(x)| \le K|x - 1|$. $\endgroup$ Dec 21, 2019 at 4:04

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