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Prove that $f: \mathbb{R^3} \to \mathbb{R^3}, f(x_1,x_2,x_3) = (x_1+x_2+x_3, 2x_1-3x_2+x_3, 2x_1+x_2-x_3)$ is continuous on $\mathbb{R^3}$ with the Euclidean norm.

Here is my attempt:

It suffices to show that $f$ is continuous at the zero vector $(0,0,0)$. Let $\varepsilon >0$, choose $\delta = $ (tbd) and suppose $||(x_1,x_2,x_3) - (0,0,0)|| = ||(x_1,x_2,x_3)|| < \delta$. Then

$$||f(x_1,x_2,x_3) - f(0,0,0)|| = ||(x_1+x_2+x_3, 2x_1-3x_2+x_3, 2x_1+x_2-x_3) - (0,0,0)|| = ||(x_1+x_2+x_3, 2x_1-3x_2+x_3, 2x_1+x_2-x_3)|| < \varepsilon.$$

I got stuck here. Any hints on how i could use the assumption $||(x_1,x_2,x_3)|| < \delta$?

Calvin Khor
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860009898987
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2 Answers2

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Define $x=(x_1,x_2,x_3)$, $y=(y_1,y_2,y_3)$ and $A =\begin{pmatrix} 1 & 1 & 1 \\ 2 & -3 & 1 \\ 2 & 1 & -1 \end{pmatrix}$ then $$\|f(x)-f(y)\| = \|A(x-y)\| \leq \|A\|\|x-y\| < \delta\|A\|$$ where $\|A\|=\sqrt{\sum_{ij}A_{ij}^2} = \sqrt{23}$ then choosing $\delta = \varepsilon /\|A\| $ you will get the continuity.

Pablo Herrera
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Use that all norms of $\mathbb{R}^n$ are equivalent: Note that $$ ||(x_1+x_2+x_3, 2x_1-3x_2+x_3, 2x_1+x_2-x_3)|| = ||x_1(1,2,2)+x_2(1,-3,1)+x_3(1,1,-1)||\leq 3|x_1|+ \sqrt{11}|x_2|+\sqrt{3}|x_3| \leq \sqrt{11}(|x_1|+|x_2|+|x_3|) \leq \sqrt{11}c||x_1,x_2,x_3)||$$ Now take $\delta = \dfrac{\varepsilon}{\sqrt{11}c}$.

remarks: above I used triangular inequality, and in the penultimate inequality I used the equivalence of the sum norm with the Euclidean norm.

Ilovemath
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  • where did $c$ come from? – 860009898987 Nov 01 '21 at 14:54
  • @860009898987 $\vert x_1 \vert + \vert x_2 \vert +\vert x_3 \vert$ is a norm on $\mathbb{R}^3$ so it is equivalent to the usual norm $\Vert (x_1,x_2,x_3)\Vert$ so such a $c$ exists by definition of equivalent norms. Alternatively we have $\vert x_i\vert \leq \Vert (x_1,x_2,x_3)\Vert$ for $i = 1,2,3$ so you can just take $c = 3$. – Digitallis Nov 01 '21 at 16:17