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Let $\|(x,y)\| _{X\times Y} := \sqrt{\|x\|_X^2+\|y\|_Y^2}$ be the $l^2$ product norm of the banach spaces $X,Y$.

I somehow struggle with proving that this norm satisfies the triangle inequality.

EDIT: Using the TI of the norms in $X$ and $Y$ I get

$\|v+u\|_{X\times Y} =(\|v_x +u_x\|_X^2 +\|v_y +u_y\|_Y^2)^{1/2} \le \left ( (\|v_x\|_X+\|u_x\|_X)^2 + (\|v_y\|_Y+\|u_y\|_Y)^2\right )^{1/2}$

I then tried to use variations of $\sqrt{a_1 +a_2}\le \sqrt{a_1}+\sqrt{a_2}$, but did not succeed.

EpsilonDelta
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1 Answers1

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Now you have to use $$\sqrt{(x_1+x_2)^2+(y_1+y_2)^2}\leq\sqrt{x_1^2+x_2^2}+\sqrt{y_1^2+y_2^2} ,$$ which is the triangle inequality in $\mathbb R^2$.

Martin Argerami
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