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?
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?
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.