Question-59
Let \(f:\mathbb{R}\rightarrow \mathbb{R}\) be such that \(|f( x) -f( y) |\leqslant ( x-y)^{2}\) for all \(x,y\in \mathbb{R}\). Then \(f( 1) -f( 0) =\)_____ (Answer in integer)
\(0\)
Let us rewrite the inequality as:
\[ \begin{aligned} -( x-y)^{2} & \leqslant f( x) -f( y) \leqslant ( x-y)^{2} \end{aligned} \]
Now, let \(x=y+h\), with \(h\neq 0\), then:
\[ \begin{aligned} -h^{2} & \leqslant f( y+h) -f( y) \leqslant h^{2}\\ & \\ -h & \leqslant \frac{f( y+h) -f( y)}{h} \leqslant h \end{aligned} \]
Taking limit \(h\rightarrow 0\) for all three expressions and using the sandwich theorem, we see that for all \(y\in \mathbb{R}\):
\[ \lim\limits _{h\rightarrow 0} \ \frac{f( y+h) -f( h)}{h} =0 \]
The limit is nothing but the derivative of \(f\) at \(y\), namely, \(f^{\prime }( y)\). Therefore, \(f^{\prime }( y) =0\) for all \(y\in \mathbb{R}\). This shows that \(f( y) =c\). The answer follows.