5

As stated in my notes:

Remark: Suppose $f: E \to \mathbb{R}$, $E \subseteq \mathbb{R}^n$, and $p \in E$. Also, suppose that $D_if$ exists in some neighborhood of $p$, say, $N(p, h)$ where $h>0$. If all partial derivatives of $f$ are bounded, then $f$ is continuous on $E$.

I found a sketch of the proof here. I'm wondering if I can adapt this proof as follows:

$f(x_1+h_1,...,x_n+h_n)-f(x_1,...,x_n)=f(x_1+h_1,...,x_n+h_n)-f(x_1,x_2+h_2,...,x_n+h_n)-...-f(x_1,x_2,...,x_{n-1}+h_{n-1},x_n+h_n)-f(x_1,...,x_{n-1},x_n+h_n)-f(x_1,...,x_n)$

However, I'm not sure how to apply the contraction principle to finish off the proof. Is there a more efficient way to prove the above remark?

JamieUSF
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    University of South Florida? It looks like you have a typo. Do you mean $f(x_1 +h,\ldots,x_n +h)$ to begin with? I don't think you can use the same $h$ for all the coordinates. I should think using the triangle inequality and the Mean Value Theorem would work. Begin by expressing $f(x_1 + h_1, \ldots, x_n + h_n)-f(x_1,\ldots,x_n)$ as the sum of $n$ differences. I didn't look at the link you provided. – Stefan Smith Apr 04 '13 at 21:58
  • I sort of provided it as $n$ differences...just with a fixed $h$. That's where I'm confused. I'm not sure I see how the MVT quickly applies. – JamieUSF Apr 04 '13 at 21:59
  • You need $f(\mathbf{x}+\mathbf{h})$ to be close to $f(\mathbf{x})$ for all $\mathbf{h}$ small enough, not just $\mathbf{h}$ with equal components. – Stefan Smith Apr 04 '13 at 22:00
  • Each of the $n$ differences can be bounded using the Mean Value Theorem, applied to $f$ along a line parallel to a coordinate axis and using a bound on a partial derivative. – Stefan Smith Apr 04 '13 at 22:02
  • I don't think the equation you entered is correct. You need a finite telescoping series with as many plusses as minuses. – Stefan Smith Apr 04 '13 at 22:07
  • I looked at the link you provided and it looks correct. They didn't explicitly mention the triangle inequality, which they used. I'm not sure what "contraction principle" you're talking about. – Stefan Smith Apr 06 '13 at 02:18

2 Answers2

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The proof is a combination of two facts:

  1. A function of one real variable with a bounded derivative is Lipschitz.

  2. Let $Q\subset \mathbb R^n$ be a cube aligned to coordinate axes. If a function $f:Q\to\mathbb R$ is Lipschitz in each variable separately, then it is Lipschitz.

The proof of 2 involves a telescoping sum such as $$\begin{split} f(x,y,z)-f(x',y',z')&= f(x,y,z)-f(x',y,z) \\ & + f(x',y,z)-f(x',y',z)\\&+f(x',y',z)-f(x',y',z') \end{split}$$

Community
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ˈjuː.zɚ79365
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0

Can we use the increment theorem for the proof ? $\delta z= f_{x}\delta x+f_{y}\delta y+\epsilon_{1}\delta x+\epsilon_{2}\delta y$ where $\epsilon_{1}$ and$\epsilon_{2}$ tends to $0$ as $\delta x,\delta y$ tends to $0$. But $\delta z =f(x+\delta x,y+\delta y)-f(x,y)$. Applying boundedness of partial derivatives $f_{x}$ and $f_{y}$ we get as $\lim f(x+\delta x,y+\delta y)=f(x,y) $ as $\delta x,\delta y$ tends to $0$

Maths Rahul
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Madhu
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