Answer to Question #294907 in Real Analysis for Dhruv bartwal

Question

Show that the function f defined by f(x)= 1/x^2 is uniformly continuous on [a, ∞] , a> 0

Answer 1

For a function $f: A \rightarrow B$ to be uniformly continuous we need that

$\forall \varepsilon>0 \exists \delta>0 \forall x, y \in A:|x-y|<\delta \Rightarrow|f(x)-f(y)|<\varepsilon$

We have in your case $f(x)=\frac{1}{x^{2}}.$

We have $\forall x, y \in[1, \infty)$

$|f(x)-f(y)|=\left|\frac{1}{x^{2}}-\frac{1}{y^{2}}\right|=\left|\frac{y^{2}-x^{2}}{x^{2} y^{2}}\right|=\frac{1}{x^{2} y^{2}}\left|y^{2}-x^{2}\right|=\frac{1}{x^{2} y^{2}}|y+x||y-x|$

Now, since $y+x>0$ and $|y-x|=|x-y|$ we get

$\frac{1}{x^{2} y^{2}}|y+x \| y-x|=\frac{x+y}{x^{2} y^{2}}|x-y|=\left(\frac{1}{x y^{2}}+\frac{1}{x^{2} y}\right)|x-y|<|x-y|$

since $\frac{1}{x y^{2}}<\frac{1}{2}$ and $\frac{1}{x^{2} y}<\frac{1}{2}\ \text{for}\ x, y \in[1, \infty).$

Therefore if you set $\delta=\varepsilon$ we get

$|f(x)-f(y)|<|x-y|<\delta=\varepsilon$