Learning starts with a question
First time here? Checkout the FAQs!
x

*Math Image Search only works best with zoomed in and well cropped math screenshots. Check DEMO

0 like 0 dislike
53 views
If \(f\) is a real function defined in a convex open set \(E \subset R^{n}\), such that \(\left(D_{1} f\right)(\mathbf{x})=0\) for every \(\mathbf{x} \in E\), prove that \(f(\mathbf{x})\) depends only on \(x_{2}, \ldots, x_{n}\).
Show that the convexity of \(E\) can be replaced by a weaker condition, but that some condition is required. For example, if \(n=2\) and \(E\) is shaped like a horseshoe, the statement may be false.
in Mathematics by Diamond (66,887 points) | 53 views

1 Answer

0 like 0 dislike
Best answer
We need to show that \(f\left(x_{1}^{0}, x_{2}, \ldots, x_{n}\right)=f\left(x_{1}^{1}, x_{2}, \ldots, x_{n}\right)\) whenever \(\mathbf{x}^{0}=\left(x_{1}^{0}, x_{2}, \ldots, x_{n}\right)\) and \(\mathbf{x}^{1}=\left(x_{1}^{1}, x_{2}, \ldots, x_{n}\right)\) both belong to \(E\). Since \(E\) is convex, the line segment joining \(\mathrm{x}^{0}\) and \(\mathrm{x}^{1}\) is contained in \(E\). The meanvalue theorem applies on this line segment, showing that \(f\left(\mathbf{x}^{0}\right)-f\left(\mathbf{x}^{1}\right)=\left(x_{1}^{0}-\right.\) \(\left.x_{1}^{1}\right)\left(D_{1} f\right)(\mathbf{x})\) for some point \(\mathbf{x}\) on this interval. Hence the result now follows from the hypothesis.

Note that convexity is needed only on each line segment through \(E\) parallel to the \(x_{1}\)-axis. Thus if the intersection of \(E\) with each line parallel to the \(x_{1}\)-axis is an interval and \(\left(D_{1} f\right)(\mathbf{x})=0\) for all \(\mathbf{x} \in E\), then \(f\) is independent of \(x_{1}\).
If we define \(f(x, y)\) on all of \(R^{2}\) except the nonnegative portion of the \(y\)-axis by specifying
\[
f(x, y)= \begin{cases}0 & \text { if } y<0 \text { or } x<0, \\ y^{2} & \text { if } y \geq 0 \text { and } x>0,\end{cases}
\]
then \(f(x, y)\) is continuously differentiable on its domain, \(\left(D_{1} f\right)(x, y)=0\) everywhere on that domain, yet \(f(-1,1)=0 \neq 1=f(1,1)\), so that \(f\) is not independent of \(x\).
by Diamond (66,887 points)

Related questions

0 like 0 dislike
1 answer

Join the MathsGee Q&A forum where you get STEM education support to succeed from our community. Connect and Learn.


On the MathsGee Q&A Forum, you can:


1. Ask questions


2. Answer questions


3. Vote on questions and answers


4. Start a fundraiser


5. Tip your favorite community members


6. Create Live Video Tutorials (Paid/Free)


7. Join Live Video Tutorials (Paid/Free)


8. Earn points by participating



MathsGee Q&A forum post


1. Remember the human


2. Act like you would in real life


3. Find original source of content


4. Check for duplicates before publishing


5. Read the community guidelines




FORUM RULES


1. Answers to questions will be posted immediately after moderation


2. Questions will be queued for posting immediately after moderation


3. Depending on the number of messages we receive, you could wait up to 24 hours for your message to appear. But be patient as posts will appear after passing our moderation.




USEFUL LINKS


Acalytica | Web Analytics | SEO Reports | Social Proof Tool | Email Marketing


MathsGee Android Q&A