Suppose that $f: \mathbb{R}^n \to \mathbb{R}$ is continuous at $P$. Can anyone help me prove that there is an open ball $B$ in $\mathbb{R}^n$ with center $P$ such that $f$ is bounded on $B$.
Continuous Function
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0Sure. Read my "more-or-less" as "trivially consequent upon". – 2012-04-28
2 Answers
Proof:
Suppose that $f: \mathbb{R}^n \to \mathbb{R}$ is continuous at $P$. Thus, $ \lim_{x\to P}f(x) = f(P)$.
Let $\epsilon = 1 > 0$. Then, there is $\delta > 0$ such that when $x \in \mathbb{R}^n$ and $0 < \| x-P \| < \delta$ then, $|f(x) - f(P)| < 1$.
Then, $B(P, \delta) = $ {$x \in \mathbb{R}^n : \|x - P\| < \delta$ }.
So, if $x \in B(P, \delta)$, then $|f(x) - f(P)| < 1$.
Notice, $|f(x)| - |f(P)| \leq |f(x) - f(P)| < 1$. Hence, $|f(x)| < 1 + |f(P)|$.
Since, $f(P)$ is a constant in $\mathbb{R}$. Let $C = 1 + |f(P)|$. Then, $C \in \mathbb{R}$.
Therefore, we see for all $x \in B(P, \delta)$, we have $|f(x)| < C$.
Thus, there is an open ball $B$ with center $P$ such that $f$ is bounded on $B$.
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0That was nicely done. – 2012-04-28
Let's suppose that $f(p)=0$ If it's not, you just take the function $g(x)=f(x)-f(p)$. If f is continuous in $0$, then for every $\epsilon >0$ there is a $\delta >0$ such that, if $|x-0|<\delta$ then $|f(x)-f(0)|<\epsilon$.
That is, $|x|<\delta\rightarrow |f(x)|<\epsilon$
Consider the ball $B$ of center $0$ and the previously obtained $\delta$ radius. Therefore, you have an open ball with center $P=0$ such that $f(x)$ is limited (by $\epsilon$) for every $x$ in it.
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0Instead of $g(x) = f(x) - f(p)$, you should choose $g(x) = f(x-p)$. Then $g$ is continuous at $0$. You may then assume $g(0) = 0$, and then fix \epsilon > 0. There exists \delta > 0 so that |x| < \delta $\Rightarrow$ |g(x)| < \epsilon. So there is an open ball centered at $0$ such that |g(x)|< \epsilon (i.e. $g$ is bounded) on that ball. Unravel what we did, and you get continuity of $f$ at $p$ implies $f$ must be bounded on an open ball about $p$. – 2012-04-28