If $S$ is bounded with respect to one norm, it will be bounded with respect to all norms equivalent to that norm. I am trying to formulate an explicit proof to show that if a set $S$ is bounded with respect to $∥·∥_1$ then it is bounded in the other norm $∥·∥_2$ if and only if it $∥·∥_2$ is equivalent to $∥·∥_2$.
I have tried to show that if there is sequence $\{x_n\}$ in $S$.  Then $\{x_n\}$ converges to a point $x$ with respect to $∥·∥_1$ if and only if it converges to the same point $x$ with respect to $∥·∥_2$.  However, I am unable to show how the norms are equivalent. 
 A: Since your question is formulated a little vague, I assume you want to prove the following:
Let $X$ be a normed space with two norms $\|{\cdot}\|_1$ and $\|{\cdot}\|_2$. Then the following statements are equivalent.


*

*Every set $S$ is bounded w.r.t. $\|{\cdot}\|_1$, if and only if it is bounded w.r.t. $\|{\cdot}\|_2$.

*The norms $\|{\cdot}\|_1$ and $\|{\cdot}\|_2$ are equivalent.


Proof.
"$(1)\Rightarrow(2)$": Assume for a contradiction that there is no constant $M>0$ such that $\|x\|_2\leq M\|x\|_1$ for all $x\in X$. Then, for each $n\in\mathbb N$ there is some $x_n\in X$ such that $\|x_n\|_2>n\|x_n\|_1$; in particular, $x_n\neq0$. The set $S:=\{x_n/\|x_n\|_1\ |\ n\in\mathbb N\}$ is bounded w.r.t. $\|{\cdot}\|_1$, since each point in $S$ is normalized. Thus, by (1), there is some $C>0$ such that $||x_n/\|x_n\|_1||_2\leq C$, which implies $\|x_n\|_2\leq C\|x_n\|_1$, contradiction. The other inequality is shown similarly.
"$(2)\Rightarrow(1)$": Assume that $\|{\cdot}\|_1$ is equivalent to $\|{\cdot}\|_2$, i.e. there are some constants $m,M>0$ such that $m\|x\|_1\leq \|x\|_2\leq M\|x\|_1$ for all $x\in X$. Let $S\subset X$ be a set bounded w.r.t. $\|{\cdot }\|_1$, i.e. there is some $C>0$ such that $\|x\|_1\leq C$ for all $x\in S$. Then $\|x\|_2\leq MC$ for all $x\in S$, so $S$ is also bounded w.r.t. $\|{\cdot}\|_2$. The remaining part is shown similarly.
Remark. Note that, in order to prove "$(1)\Rightarrow(2)$", it is NOT sufficient to assume that every set which is bounded w.r.t. $\|{\cdot}\|_1$ is also bounded w.r.t. $\|{\cdot}\|_2$. Take, for instance, $X=C^1([0,2\pi],\mathbb R)$ endowed with $\|x\|_2:=\sup |x|([0,2\pi])$ and $\|x\|_1:=\|x\|_2+||x'||_2$. Then clearly $\|{\cdot}\|_2\leq\|{\cdot}\|_1$ (and hence each set which is bounded w.r.t. $\|{\cdot}\|_1$ is also bounded w.r.t. $\|{\cdot}\|_2$), but the two norms are not equivalent, as the sequence $x_n(t)=\sin(nt)$ shows.
