- #1

mathmari

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MHB

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Let $D=\left \{x=(x_1, x_2)\in \mathbb{R}^2: x_2>\frac{1}{x_1}, \ x_1>0\right \}$.

We have the function $f: D\rightarrow \left (0,\frac{\pi}{2}\right )$ with $f(x)=\arctan \left (\frac{x_2}{x_1}\right )$.

I want to show using the mean value theorem in $\mathbb{R}^2$ that for all $x,y\in D$ it holds that $$|f(x)-f(y)|<\frac{1}{\sqrt{2}}\|x-y\|$$ where $\|\cdot \|$ is the euclidean norm in $\mathbb{R}^2$.

Do we have to do the following?

We take an interval $(y,x)\in D$ and then from the mean value theorem we get that $$f'(c)=\frac{f(x)-f(y)}{x-y}$$ where $c\in (x,y)$. Then we have to find an upper bound for $f'(c)$.

(Wondering)