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Suppose that $f: \mathbb{R}^n\setminus\{0\} \to \mathbb{R}$ is differentiable. Assume $\lim\limits_{x \to 0}\frac{\partial f}{\partial x_j}$ exists for each $j \in \{1,2,\dots,n\}$.

Can $f$ be extended to a continuous function on $\mathbb{R}^n$? Furthermore, if we assume continuity at the origin, is $f$ differentiable on $\mathbb{R}^n$?

I can't really come up with counter examples or a proof for either questions. For the second question, I know that if $n =1$, $f$ can be extended to a differentiable function (a L'Hopitals argument), but is this true in general?

INQUISITOR
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  • $f(x)= \frac{1}{x}$ is differentiable on $\mathbb{R}\setminus {0}$. How do you extend $f$ to a continuous function? I don't think its possible. – Uncool May 06 '21 at 18:09
  • @Uncool In your example $\lim_{x\to 0} f'(x)$ does not exist, in the sense it's not a real number – alphaomega May 06 '21 at 18:12
  • @Uncool For the second question, we are assuming that $f$ is continuous at $0$. – INQUISITOR May 06 '21 at 18:13
  • Think about a step function in $n=1$... – Jair Taylor May 06 '21 at 19:16
  • For $n=1$, there is a clear counter example for the first question: $f$ is $1$ on $[0,\infty)$ and $f$ is $-1$ on $(-\infty,0)$. But is there a counter example for $n>1$? – INQUISITOR May 06 '21 at 22:06

2 Answers2

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Yes.

Assume the partial derivatives have a finite limit towards 0. Then, in some neighborhood of 0, their absolute value is bounded by some constant $M$. By breaking it up into segments along each coordinate separately, you can show for any points $x_1$,$x_2$ in this neighborhood, that $|f(x_1)-f(x_2)|\leq M |x_1-x_2|$. You can use this to show that any sequence approaching 0 is Cauchy, so the limit exists.

Eric
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  • But how do we know that the limit of $f(x_n)$ convergence to the same value for any sequence $x_n$ converging to $0$? – INQUISITOR May 06 '21 at 21:43
  • Interleaving the converging series gives the same result. Alternatively, if $a_n$ and $b_n$ converge to 0, then $|a_n-b_n|$ converges to 0, so $M|a_n-b_n|$ converges to 0 so $|f(a_n)-f(b_n)|$ converges to 0. – Eric May 07 '21 at 02:21
  • So what is stopping this from working for $n=1$? In the comment section of my original post, I give a counter example for $n=1$, but what you wrote seems to work for $n=1$. – INQUISITOR May 07 '21 at 19:43
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    Good point. I’m implicitly assuming that the neighborhood of 0 excluding 0 is path connected. Otherwise you can’t use lines parallel to the axis to connect any two points. This is true for $n>1$. – Eric May 07 '21 at 23:18
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I think that I can answer your second question with yes. Let $i \in \lbrace 1, ..., n \rbrace$ be a coordinate direction and for some given $t \in \mathbb{R} \setminus \lbrace 0 \rbrace$ let $g_t:[0, 1] \rightarrow \mathbb{R}$ $$ g_t(s) := f(tse_i), $$ where $e_i$ is the $i$-th unit vector. Then, because of differentiability, we have $g_t'(s) = t\partial_i f(tse_i)$ according to the chain rule. The mean value theorem gives us some $\xi_t \in [0, 1]$ such that ($t \in \mathbb{R} \setminus \lbrace 0 \rbrace$) $$ \frac{f(te_i) - f(0)}{t} = \frac{g_t(1) - g_t(0)}{t} = \frac{g_t'(\xi_t)}{t} = \partial_i f(\xi_t t e_i) $$ whereas the last expression converges to $\lim_{x \rightarrow 0} \partial_i f(x)$ whose existence we assumed. So the partial derivatives $\partial_i f(0)$ exist and are by assumption continuous. So $f$ is differentiable in $0$.

The first question I can not quite answer (yet). But if you wanted to construct a counterexample, you must have strong oscillation near the origin.

Hyperbolic PDE friend
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