Answer to Question #310448 in Real Analysis for Nikhil Singh

Question #310448

Check whether the function, ,f defined below, is uniformly continuous or not:



f(x)=x^(1/2), x∈[1,2]

1
Expert's answer
2022-03-15T13:11:09-0400

ANSWER: The function f(x)=x12f(x)=x^ \frac{1}{2} is uniformly continuous on [1,2]

EXPLANATION 1.

Let ϵ>0\epsilon>0 . We need to show that exists δ>0\delta>0 such that

x,y[1,2]x,y \in[1,2] and xy<δ\left | x-y \right |<\delta imply f(x)f(y)<ϵ|f(x)-f(y)|<\epsilon .

We have

f(x)f(y)=xy=(xy)(x+y)x+y=xyx+yf(x)-f(y) =\sqrt{x}-\sqrt{y}=\frac{\left ( \sqrt{x} -\sqrt{y}\right )\cdot\left ( \sqrt{x}+\sqrt{y} \right )}{\sqrt{x}+\sqrt{y}}=\frac{x-y}{\sqrt{x}+\sqrt{y}} .

Because x1,y1\sqrt{x}\geq1, \sqrt{y}\geq1 , then xyx+yxy2\frac{\left | x-y \right |}{\sqrt{x}+\sqrt{y}}\leq\frac{\left | x-y \right |}{2} . If δ=2ϵ\delta=2\epsilon and xyδ|x-y| \leq\delta , then f(x)f(y)xy2δ2=2ϵ2=ϵ\left | f(x)-f(y) \right |\leq\frac{\left | x-y \right |}{2}\leq\frac{\delta}{2}=\frac{2\epsilon}{2}=\epsilon . So, we have shown, by the definition, that ff is uniformly continuous on [1,2].

EXPLANATION 2.

The function ff is continuous on a closed interval [1,2] , so by Cantor's theorem the function ff is uniformly continuous on [1,2].


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