1
$\begingroup$

NBHM PhD Screening Test 2005 Analysis

What is the least value of $K>0$ such that $|\sin^2x-\sin^2y|≤K|x-y|$ $\forall$ $x$,$y \in \mathbb R$

How can I solve this problem?

  • 0
    @BabakSorouh: I don't think so, but there is a simple way of calculating the $K$ for differentiable functions.2012-12-31

1 Answers 1

7

If $f$ is differentiable at $x$, then $\lim_{h\to 0} |\frac{f(x+h)-f(x)}{h}| = |f'(x)|$. Hence if $\epsilon >0$, there exists a $\delta>0$ such that if $|h| < \delta$, then $|\frac{f(x+h)-f(x)}{h}| \geq |f'(x)|-\epsilon$, or equivalently, $|f(x+h)-f(x)| \geq |h|(|f'(x)|-\epsilon)$.

It follows that if $f$ is differentiable, then if $K$ is such that $f$ satisfies the bound $|f(x)-f(y)| \leq K |x-y|$, then we must have $K \geq\sup_t |f'(t)|$. By the mean value theorem, we have $|f(x)-f(y)| \leq \sup_t |f'(t)| |x-y|$, hence the least value of $K$ that satisfies the bound is $\sup_t |f'(t)|$.

In the example above, $f(x) = \sin^2(x)$, hence $f'(x) = 2 \cos x \sin x = \sin (2x)$, and $\sup_t |f'(t)| = 1$. Hence $K=1$ is the least value of $K$ such that the bound holds.

  • 1
    @YadatiKiran: I'm not exactly sure what I was thinking when I wrote the above, if $|f(x)-f(y)| \le K|x-y|$ for all $x,y$ then you must have $K \ge \sup_t |f'(t)|$ just by taking limits. The mean value theorem shows that $\sup_t |f'(t)|$ is the smallest such value for $K$.2018-12-21