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Problem:True/False?


If every differentiable function on a subset \(X\subset \mathbb{R}^n \) (i.e, restriction of a differentiable function on a neighbourhood of \(X\) ) is bounded, then \(X\) is compact.


Hint:


Recall that  \(X\subset \mathbb{R}^n \) is compact if and only if it is closed and bounded. Find suitable functions to show bounded and closed respectively.


Discussion:


Let \(f(x)=(||x||^2,||x||^2,…,||x||^2)\), then \(f\) is differentiable. Hence \(||x||\) must be bounded on \(X\), which is the same as saying \(X\) is bounded.

Note, \(||x||\) is the standard euclidean norm. And \(f\) is differentiable because each component of \(f\) is differentiable.

Now all we need is a function which will show that \(X\) is closed as well.

Let \(p\in \mathbb{R}^n \) be a point which is a limit point of \(X\). Suppose that \(p\notin X \). Then \(||x-p|| \ne 0 \) for any \(x\in X\). Hence the function \(g(x)=(\frac{1}{||x-p||},\frac{1}{||x-p||},…,\frac{1}{||x-p||} )  \) is a well-defined and differentiable function. Again, the differentiation part follows from the fact that it is differentiable in each component. And that it is differentiable in each component is easy to check. (Hint: Take the partial derivatives).

Now that we have \(g\), a differentiable function, we know that \(g\) must be bounded on \(X\). But as \(x \to p \) we see that \(||g(x)|| \to \infty \), i.e, g is not bounded. The only assumption we have made so far is that \(p\) is not in \(X\). This must be false, so \(p\in X\). So in fact, every limit point of \(X\) is in \(X\). This shows that \(X\) is closed.


Helpdesk

  • What is this topic: Real Analysis
  • What are some of the associated concept: Standard euclidean norm, differentiable function, Bounded Set
  • Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert