If $ |f(x)-f(y)| \leq 7|x-y|^{201} $ Then,

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that for any two real numbers $x$ and $y$ $$ |f(x)-f(y)| \leq 7|x-y|^{201} $$ Then,

(A) $f(101)=f(202)+8$

(B) $f(101)=f(201)+1$

(C) $f(101)=f(200)+2$

(D) None of the above.

My approach:-

$$|(f(x)-f(y)|\leq 7|x-y|^{201}$$ $$ \begin{array}{l}\frac{|f(x)-f(y)|}{|x-y|}|\leq 7| x-\left.y\right|^{200} \\ \lim_{x->y}\left|\frac{f(x)-f(y)}{x-y}\right| \leqslant 7|x-y|^{200} \\ f^{\prime}(x) = 0 \\ f(x)=C, \quad \text { Accordingly } f(y)=C\end{array} $$

So, option D is correct

Answer 1

Just for entertainment,

\begin{eqnarray} |f(y)-f(x)| &=& \sum_{k=0}^{n-1} |f( x+{k+1 \over n} (y-x)) ) - f( x+{k \over n} (y-x)) | \\ &\le& 7 \sum_{k=0}^{n-1} ({|y-x| \over n} )^{201} \\ &=& 7 {|y-x|^{201} \over n^{200}} \end{eqnarray}

Letting $n \to \infty$ we see that $f(y)=f(x)$.