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Can I please receive feedback on my proof/solution below? Thank you.

$\def\n{{\bf n}} \def\R{{\mathbb R}} \def\N{{\mathbb N}} \def\x{{\bf x}} \def\y{{\bf y}} \def\a{{\bf a}} \def\f{{\bf f}} \def\0{{\bf 0}}$ Let $\f\colon D\to \R^m$, $D\subseteq\R^n$, and let $\a$ be a cluster point of $D$. Prove $\displaystyle{\lim_{\x\to\a}\f(\x)}$ exists if and only if for every $\epsilon>0$ there exists a $\delta>0$ such that for all $\x_1,\,\x_2\in D\cap (B(\a,\delta)\setminus\{\a\})$, $\|\f(\x_1) - \f(\x_2)\|<\epsilon$.

$\textbf{Solution:}$ Suppose $\displaystyle{\lim_{\x\to \a}\f(\x)}$ exists, and let $\epsilon >0$. Then, there exists $\delta > 0$ such that $||\f(\x)-\f(\a)|| <\frac{\epsilon}{2}$ for all $\x \in (B(\a,\delta)\setminus\{\a\}) \cap D$. Let $\x_1,\,\x_2\in (B(\a,\delta)\setminus\{\a\})\cap D$. Then, $||\f(\x_1)-\f(\a)|| <\frac{\epsilon}{2}$ and $||\f(\x_2)-\f(\a)|| <\frac{\epsilon}{2}$. Therefore, by triangle inequality, $$||\f(\x_1)-\f(\x_2)|| \le ||\f(\x_1)-\f(\a)|| + ||\f(\x_2)-\f(\a)|| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$

Conversely, we must show there do not exist two sequences $\x_n \to \a$ and $\y_n \to \a$, with $\f(\x_n)\to L_1$ and $\f(\y_n) \to L_2$ and $L_1 \neq L_2$, because if they do exist, then the limit can not exist.

Let $L_2>L_1$. Then, we can choose $\epsilon>0$ such that for any $x\in B(L_2;\epsilon)=(L_2-\epsilon,L_2+\epsilon)$ and for any $y\in B(L_1;\epsilon)=(L_1-\epsilon,L_1+\epsilon)$, we have $|x-y|>\epsilon$. This follows since $|\x-\y|\geq \x-\y>(L_2-\epsilon)-(L_1+\epsilon)=(L_2-L_1)-2\epsilon$, which is greater than $\epsilon$ if $\epsilon<\frac{L_2-L_1}{3}$.

Choose such an $\epsilon$. By hypothesis, there exists $\delta_1>0$ such that if $\x,\y \in B(\a;\delta_1)\backslash\{\a\}$, then $|\f(\x)-\f(\y)|<\epsilon$. Since $\x_n \to \a$ and $\f(\x_n)\to L_1$, there exists $N_{\x}$ such that for all $n\geq N_{\x}$, we have $\x_n \in B(\a;\delta)\backslash\{\a\}$ and $\f(\x_n)\in B(L_1;\epsilon)$. Similarly, there exists $N_{\y}$ such that for all $n\geq N_{\y}$, we have $\y_n \in B(\a;\delta)\backslash\{\a\}$ and $\f(\y_n) \in B(L_2;\epsilon)$. Take any $n_0 \geq \max\{N_x,N_{\y}\}$. Then $\x_{n_0},\y_{n_0}\in B(\a;\delta_1)\backslash\{\a\}$, but $$ |\f(\x_{n_0})-\f(\y_{n_0})|\geq \f(\x_{n_0})-\f(\y_{n_0})>(L_2-L_1)-2\epsilon>\epsilon, $$ a contradiction.

In the higher dimension case, the argument follows that $\epsilon>0$ can be chosen small so that $d(B(L_1;\epsilon),\, B(L_2;\epsilon))>\epsilon$.We wish to have $\epsilon$ sufficiently small such that $d(B(L_1, \epsilon), B(L_2,\epsilon)) > \epsilon$. So, let $F(\epsilon) = d(B(L_1, \epsilon), B(L_2,\epsilon)) - \epsilon$. Then $F(0) = ||L_2-L_1|| > 0$. Hence, for $\epsilon$ sufficiently small, $F(\epsilon) > 0$.

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    $\begingroup$ Wait what is the definition of $\lim_{x \rightarrow a} f(x)$? Because the definition of the limit is the epsilon-delta definition. $\endgroup$ Apr 12, 2020 at 23:33
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    $\begingroup$ @SpencerKraisler Here we are using the sequential characterization of limit to prove the problem (Cauchy Criterion). $\endgroup$ Apr 12, 2020 at 23:58

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Your proof for the first part is correct. For the second part, you basically showed that if $x_n \to a$, then $f(x_n)\to L$ for some $L$. The problem though is that, a priori, this $L$ that arose depends on the sequence you picked. You have to show there aren't two sequences $x_n \to a$ and $y_n \to a$, with $f(x_n)\to L_1$ and $f(y_n) \to L_2$ and $L_1 \neq L_2$.

I'll sketch the idea assuming we are in $\mathbb{R}$. WLOG, suppose $L_2>L_1$. Then I claim we can choose $\epsilon>0$ so that for any $x\in B(L_2;\epsilon)=(L_2-\epsilon,L_2+\epsilon)$ and for any $y\in B(L_1;\epsilon)=(L_1-\epsilon,L_1+\epsilon)$, we have $|x-y|>\epsilon$. This follows easily since $|x-y|\geq x-y>(L_2-\epsilon)-(L_1+\epsilon)=(L_2-L_1)-2\epsilon$, which is greater than $\epsilon$ if $\epsilon<\frac{L_2-L_1}{3}$.

Choose such an $\epsilon$. By hypothesis, there exists $\delta_1>0$ such that if $x,y \in B(a;\delta_1)\backslash\{a\}$, then $|f(x)-f(y)|<\epsilon$. Since $x_n \to a$ and $f(x_n)\to L_1$, there exists $N_x$ such that for all $n\geq N_x$, we have $x_n \in B(a;\delta)\backslash\{a\}$ and $f(x_n)\in B(L_1;\epsilon)$. Similarly, there exists $N_y$ such that for all $n\geq N_y$, we have $y_n \in B(a;\delta)\backslash\{a\}$ and $f(y_n) \in B(L_2;\epsilon)$. Take any $n_0 \geq \max\{N_x,N_y\}$. Then $x_{n_0},y_{n_0}\in B(a;\delta_1)\backslash\{a\}$, but $$ |f(x_{n_0})-f(y_{n_0})|\geq f(x_{n_0})-f(y_{n_0})>(L_2-L_1)-2\epsilon>\epsilon, $$ a contradiction.

In the higher dimension case, it's an easy argument that $\epsilon>0$ can be chosen small so that $d(B(L_1;\epsilon),\, B(L_2;\epsilon))>\epsilon$.

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  • $\begingroup$ Thank you for the help! I made an edit to the second part of my proof using what you said, does it look good now? $\endgroup$ Apr 13, 2020 at 2:09
  • $\begingroup$ The proof in higher dimensions, where the balls are actually not just intervals with simple geometry, is extremely similar, but it doesn't come out to the simple analysis in the line $L_2-L_1-2\epsilon>\epsilon$. $\endgroup$
    – ProfOak
    Apr 13, 2020 at 2:49
  • $\begingroup$ Here's an easy way to get the necessary $\epsilon$. We want $\epsilon$ small so that $d(B(L_1;\epsilon),B(L_2;\epsilon))>\epsilon$. Let $F(\epsilon)=d(B(L_1;\epsilon),B(L_2;\epsilon))-\epsilon$. Then $F(0)=||L_2-l_1||>0$. Hence, for $\epsilon$ sufficiently small, $F(\epsilon)>0$. Also, you may want to write my idea up cleaner. Do you understand the proof and can picture it holding in higher dimensions? $\endgroup$
    – ProfOak
    Apr 13, 2020 at 2:52
  • $\begingroup$ So I understand the proof prior to the higher dimension case. Doesn't the same application of the line case apply to the higher dimension case? We essentially want to be able to construct sufficiently small balls such that their distance is greater than $\epsilon$? $\endgroup$ Apr 13, 2020 at 3:47
  • $\begingroup$ See the comment I made above. The function $d(B(L_1;\epsilon),B(L_2;\epsilon))$ is the length of the "shortest" (infinimum, really) line connecting the spaces. The formula is messier than in the one dimensional case where one can easily see a formula for the distance between two intervals. The concept is the same though. $\endgroup$
    – ProfOak
    Apr 13, 2020 at 14:33

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