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Here I am using $l^2$ to define the set of complex-sequences which are square summable and $l^{\infty}$ to define the set of bounded complex sequences.

I think not. If we take $x(n) = 1 \;\; n\in \mathbb{N}$ ,which is bounded sequence. Then for any $y\in l^2$ must be a null sequence, so

$$\|x-y\|_{\infty} = \sup|1- y(n)| \geq 1$$

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  • $\begingroup$ You are correct about it not being dense, but i don't know why you said any $y \in l^2$ must be a null sequence. It doesn't have to be null itself, rather the terms converge to $0$ $\endgroup$ – mathworker21 May 13 '17 at 19:17
  • $\begingroup$ @mathworker21 A null sequence is defined to be a sequence converging to zero. $\endgroup$ – Paul K May 13 '17 at 19:20
  • $\begingroup$ Yup, that's correct. And the closure of $\ell^2$ in $\ell^{\infty}$ is $c_0$, the subspace of sequences converging to $0$. $\endgroup$ – Daniel Fischer May 13 '17 at 19:23
  • $\begingroup$ Thanks Daniel. I should have specified what I refer to as a null sequence! $\endgroup$ – user197848 May 13 '17 at 19:42
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Here is similar way, the function $\phi(x) = \limsup_n |x_n|$ is a continuous function on $l_\infty$ ( since $|\phi(x)-\phi(y)| \le \|x-y\|$) and $\phi((1,1,...)) = 1$, but $\phi(x) = 0$ for all $x \in l_2$.

If $l_2$ was dense in $l_\infty$, then we would have $x_n \in l_2$ such that $x_n \to (1,1,...)$ (in $l_\infty$) and then $\phi(x_n) \to 1$ which would be a contradiction.

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