4
$\begingroup$

I am just trying to find a formal proof for this statement:

Let $f:\mathbb R^n \backslash \ \{0\} \rightarrow \Bbb R$ be a function, and assume that there exits $L\in \Bbb R$ that for every path $$\gamma: (a,b)\rightarrow \Bbb R^n\backslash\{0\} $$ that satisfies $$\lim_{t\rightarrow b}\gamma(t) = 0 $$ the limit $$\lim_{t\rightarrow b} f(\gamma(t)) = L $$ show that : $\lim_{x\rightarrow 0} f(x) =L$

$\endgroup$
2
  • 1
    $\begingroup$ We can just apply it to the path $\gamma(t)=t-b$, unless I'm missing something ? $\endgroup$ – Suzet Feb 13 '20 at 14:07
  • 1
    $\begingroup$ my mistake i wrote it wrong $\endgroup$ – Tair Galili Feb 13 '20 at 14:15
3
$\begingroup$

With the edited question, let us assume towards a contradiction that $\lim_{x\rightarrow 0} f(x) \not = L$ (the limit either doesn't exist or if it does, it isn't $L$).

There exists some positive $\epsilon$ such that for any $n\geq 0$, there is some $x_n\in B^{\star}(0,\frac{1}{n})$ (open ball centered at $0$ with radius $\frac{1}{n}$ and $0$ removed) such that $||f(x_n)-L||\geq \epsilon$.

For $n\geq 0$, let $t_n:=-2^{-n}$ so that $t_0=-1$ and $t_n$ goes to $0$ as $n$ goes to infinity.
Because $B^{\star}(0,\frac{1}{n})$ is path-connected, for every $n\geq 0$ there exists some path $\gamma_n:[t_{n},t_{n+1}] \rightarrow B^{\star}(0,\frac{1}{n})$ with $\gamma(t_{n})=x_{n}$ and $\gamma(t_{n+1})=x_{n+1}$.
I define the path $\gamma : (-1,0)\rightarrow \Bbb R^n\backslash\{0\}$ by letting $\gamma(t)=\gamma_n(t)$ where $n\geq 0$ is the unique nonnegative integer such that $t_n<t\leq t_{n+1}$. Such a path $\gamma$ is well defined, it glues all the $\gamma_n$ together.

Now, by construction I have $\gamma(t)\in B^{\star}(0,\frac{1}{n})$ for every $t\in (t_n,0)$, which implies that $\lim_{t\rightarrow 0}\gamma(t) = 0$. Meanwhile, $f(\gamma(t_n))=f(x_n)$ for every positive $n$, which implies $$||f(\gamma(t_n))-L||\geq \epsilon$$ whence $f(\gamma(t))$ can not converge to $L$ as $t$ goes to $0$, this is the desired contradiction.

$\endgroup$
2
  • $\begingroup$ thanks a lot! I thought that the proof is going to be shorter but this one is really nice though. (just in the first part, $n$ cant be zero... but it really doesnt matter for the convergence) $\endgroup$ – Tair Galili Feb 13 '20 at 14:50
  • 1
    $\begingroup$ @tairgalili It can be made shorter if we skip over the details, constructing the path $\gamma$ is in the end just more difficult to write than to think of ! $\endgroup$ – Suzet Feb 13 '20 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.