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Suppose we have two equivalent norms, $\|\cdot \| \sim \left\vert\!\left\vert\!\left\vert \cdot\right\vert\!\right\vert\!\right\vert$. We assume that $x_n \rightarrow x $ in $(X,\|\cdot \|)$. Show that $x_n \rightarrow x $ in $(X,\left\vert\!\left\vert\!\left\vert \cdot\right\vert\!\right\vert\!\right\vert)$

We observe since the norms are equivalent, there exits constants $\alpha, \beta$, such that $\alpha \|\cdot \| \leq \left\vert\!\left\vert\!\left\vert \cdot\right\vert\!\right\vert\!\right\vert \leq \beta \|\cdot \|$, with $0 \lt \alpha \leq \beta$

Since, $x_n \rightarrow x$ in $(X,\|\cdot \|$) we know that for all $\epsilon>0, \exists K(\epsilon)$ such that for n>$K(\epsilon),$ we have $\|x_n -x\| \leq \epsilon$

Since, we know $\beta \gt 0$ , then $ \beta \|x_n -x\| \leq \beta \epsilon$, Thus we see that $\left\vert\!\left\vert\!\left\vert x_n-x\right\vert\!\right\vert\!\right\vert\leq \beta \epsilon$.

This is how far I've gotten, our goal to is to find a a $J(\epsilon)$ such that for all $\epsilon>0, \exists J(\epsilon)$ such that for n>$J(\epsilon),$ we have $\left\vert\!\left\vert\!\left\vert x_n-x \right\vert\!\right\vert\!\right\vert \leq \epsilon$

My candidate for $J(\epsilon$) is $J(\epsilon) = \frac{K(\epsilon)}{\beta}$ from my above argument, but I'm not exactly sure how to validate this. If someone could help me on the proof, it would be very appreciated. I believe I'm pretty close, just missing something small.

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I think you mean $J(\epsilon) = K(\epsilon/\beta)$.

For all $n \ge J(\epsilon)=K(\epsilon/\beta)$, we have $\|x_n-x\| \le \epsilon/\beta$, so $$||| x_n-x||| \le \beta \|x_n-x\| \le \beta \epsilon/\beta=\epsilon.$$

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  • $\begingroup$ Perfect, this is exactly what I was looking for to get me to the end. $\endgroup$ – Vogtster Aug 28 '16 at 22:19
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It is far easier to show it this way:

$x_n \rightarrow x $ in $(X,||*||)$, implies that $||x_n - x|| \to 0$ as $n \to \infty $.

Due to norm equivalence, there exist constants $\alpha, \beta$, such that $$\alpha ||x_n - x|| \leq |||x_n - x||| \leq \beta ||x_n - x||,$$

so if both the LHS and RHS goes to zero, then so does the middle term and thus we have shown convergence in the second norm.

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  • $\begingroup$ Thanks, this is actually a lot more simple than my argument. $\endgroup$ – Vogtster Aug 28 '16 at 22:17

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