Answer to Question #153628 in Calculus for Arora

Question

Show that there does not exist a non-constant continuous function f : R2 → R such that f(x, y) = 5 for all (x, y) ∈ R2 with x^2+y^2<1

Answer 1

Let the function

$f:\R^2\rightarrow \R\ni\\ f(x,y) = \begin{cases} 5 &\text{if } x^2+y^2<1 \\ x^2+y^2+4 &\text{if } x^2+y^2\geq1 \end{cases}$

We can see the continuity of $f(x,y)$ by two processes.

(i)

for $x^2+y^2<1$ , $f(x,y)$ is a constant function.

for $x^2+y^2<1 , f(x,y)= x^2+y^2+4$ which is sum of three continuous functions,$\;x^2,y^2\;and \;4.$ So it is continuous.

for $x^2+y^2=1$ , $\lim_{\{(x,y)|x^2+y^2\rightarrow 1\}}f(x,y)=5$ from all sides.

Therefore we can conclude that, $f(x,y)\; is \; continuous \;\forall(x,y)\in \R^2$ .

(ii) [By $\epsilon-\delta$ definition]

Let a point be $(h,k)$

Given $\epsilon>0,$

If $\; h^2+k^2<1 ,$

Take $\;\delta = min\{\sqrt{ h^2+k^2},1-\sqrt{ h^2+k^2}\}$

Let $(x,y)\in\R^2\ni d((h,k),(x,y))<\delta$

$\Rightarrow f(x,y)=5\\ \Rightarrow |f(x,y)-f(h,k)|=|5-5|=0<\epsilon$

If $h^2+k^2\geq1,$

Take $\delta =\sqrt{ h^2+k^2}-\sqrt{ h^2+k^2-\epsilon}$

Let $(x,y)\in\R^2\ni d((h,k),(x,y))<\delta$

$\Rightarrow (x,y)\in \{(p,q)|(x-h)^2+(y-k)^2<\delta^2\}\\ \Rightarrow f(x,y)\in \left[f(p_1,q_1),f(p_2,q_2)\right]\\ where,\; p_1^2+q_1^2=(\sqrt{h^2+k^2}-\delta)^2=(\sqrt{h^2+k^2-\epsilon})^2=h^2+k^2-\epsilon\\ and, p_2^2+q_2^2=(\sqrt{h^2+k^2}+\delta)^2\leq h^2+k^2+\epsilon\\ \Rightarrow f(x,y)\in [h^2+k^2+4-\epsilon,h^2+k^2+4+\epsilon]\\ \Rightarrow f(x,y)\in [f(h,k)-\epsilon,f(h,k)+\epsilon]\\ \Rightarrow |f(x,y)-f(h,k)|<\epsilon$

So, $d((h,k),(x,y))<\delta \Rightarrow d(f(x,y),f(h,k))<\epsilon$

Hence $\pmb{f(x,y)}$ is continuous $\pmb{\forall(x,y)\in \R^2}$ .