Let $X$ be vector space over field $F$. Two norm $\Vert\cdot\Vert_1$ and $\Vert\cdot\Vert_2$ on $X$ is equivalent if there exist $k,K\in\mathbb{R}$ such that $$ k\Vert x\Vert_1\leq \Vert x\Vert_2\leq K\Vert x\Vert_1 $$ or $$ K^{-1}\Vert x\Vert_2\leq \Vert x\Vert_1\leq k^{-1}\Vert x\Vert_2 $$ for all $x\in X$.

Now, I want to prove this theorem.

Let $X$ be vector space over field $F$. If $\Vert\cdot\Vert_1$ and $\Vert\cdot\Vert_2$ is equivalent on $X$ with metrics $d_1$ and $d_2$ respectively, then a sequence $\{x_n\}$ convergent to $x$ on metric space $(X,d_1)$ if and only if $\{x_n\}$ convergent to $x$ on metric space $(X,d_2)$.

I know, that a sequence $\{x_n\}$ is convergent on metric space $(X,d_1)$ if $$(\forall \varepsilon>0) (\exists N\in\mathbb{N}) \text{ such that }(\forall n\geq N), d_1(x_n,x)<\varepsilon.$$

Now I confused to associate the definition of $\{x_n\}$ convergent with equivalence of two norms.

Any idea to proof this theorem?


2 Answers 2


$d_1(x,y)=\|x-y\|_1$ so $d_1(x_n,x) \to 0$ iff $\|x_n-x\|_1 \to 0$. Similarly $d_2(x_n,x) \to 0$ iff $\|x_n-x\|_2 \to 0$. Can yo see from the stated inequalities that $\|x_n-x\|_1 \to 0$ iff $\|x_n-x\|_2 \to 0$.


We see that by the definition of equivalence, the the convergence of one sequence implies convergence of the other. Indeed, if we have that $ \{x_n\}_{n\geq 1} $ converging to $ x $, we have that $$||x_n-x||_1 \rightarrow 0$$ If we let $ \varepsilon>0 $, we take the $ K >0 $ such that $$ ||x||_1 \leq K||x||_2 \quad \forall x\in X$$ So, we take $ N $ large such that $ \forall n\geq N $, $$||x_n-x||_1 \leq \frac{\varepsilon}{K} $$ Then we see that for all $ n>N $, $$||x_n-x||_2 \leq K||x_n-x|| < \varepsilon $$ Since $\varepsilon $ was arbitrary, we get convergence in the second norm.

The other direction is basically the same idea.


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