Answer to Question #294907 in Real Analysis for Dhruv bartwal

Question #294907

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

1
Expert's answer
2022-02-09T17:32:28-0500

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

ε>0δ>0x,yA:xy<δf(x)f(y)<ε\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)=1x2.f(x)=\frac{1}{x^{2}}.

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

f(x)f(y)=1x21y2=y2x2x2y2=1x2y2y2x2=1x2y2y+xyx|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>0y+x>0 and yx=xy|y-x|=|x-y| we get

1x2y2y+xyx=x+yx2y2xy=(1xy2+1x2y)xy<xy\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 1xy2<12\frac{1}{x y^{2}}<\frac{1}{2} and 1x2y<12 for x,y[1,).\frac{1}{x^{2} y}<\frac{1}{2}\ \text{for}\ x, y \in[1, \infty).

Therefore if you set δ=ε\delta=\varepsilon we get

f(x)f(y)<xy<δ=ε|f(x)-f(y)|<|x-y|<\delta=\varepsilon


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment