If $\ $ $yx_n\to 0 $ for all $y$ in the C$^*$-algebra A, Is it true that $x_n$ is weakly convergent to $0$? 
$A$ is a C$^*\! $-algebra and $(x_n)_{n\in \mathbb{N}} \subseteq A $.
   If $\ $ $yx_n\to 0 $  for all  $y\in A$, Is it true that $x_n$ is weakly
  convergent to $0$  ?

For unitals this is trivial but I didn't find any idea about nonunitals.
 A: The answer is yes. The proof will use the following propositions.
Proposition 1: Let $A$ be a $C^*$-algebra with $a,b,e\in A$ such that $e\geq0$ and $\|e\|\leq1$. If $\|ae-a\|\leq\varepsilon$ and $\|be\|\leq\varepsilon$, then $\|a+b\|\leq\sqrt{\|a\|^2+\|b\|^2}+2\varepsilon$.
Proof: Is $\|a+b\|\leq\sqrt{\|a\|^2+\|b\|^2}+2\varepsilon$ when $\|ae-a\|\leq\varepsilon$ and $\|be\|\leq\varepsilon$ in a $C^*$-algebra?
Corollary 2: Let $A$ be a $C^*$-algebra with $a,b,e\in A$ such that $e\geq0$ and $\|e\|\leq1$. If $\|ea-a\|\leq\varepsilon$ and $\|eb\|\leq\varepsilon$, then $\|a+b\|\leq\sqrt{\|a\|^2+\|b\|^2}+2\varepsilon$.
Proof: Note the only thing that changed is the order of the multiplication. However, reversing the order of multiplication of a $C^*$-algebra just gives you another $C^*$-algebra, so the proposition still holds.
Proposition 3: Let $A$ be a $C^*$-algebra with $a\in A$ and let $\varepsilon>0$. Then there exists $e\in A$ with $e\geq0$ and $\|e\|\leq1$ such that $\|ea-a\|\leq\varepsilon.$
Proof: This is by the principle of an approximate identity.
Before proving the main result, we need to prove the following lemma.
Lemma 4: Let $A$ be a $C^*$-algebra and let $\{x_n\}$ be a sequence in $A$ such that $yx_n\to0$ for all $y\in A$. Then the sequence $\{x_n\}$ is bounded.
Proof: The proof is by contradiction. We assume for the contrary that $\{x_n\}$ is unbounded.
The idea is to recursively define a sequence of elements $\{y_n\}$ and a sequence of finite index sets $\{I_n\}$ with $|I_n|=n$ such that $\|y_{n+1}-y_n\|\leq2^{-n-1}$ and $\|y_nx_i\|\geq1+2^{-n}$ for all $i\in I_n$. Then the sequence $\{y_n\}$ is Cauchy, so it has a limit $y$. We find $yx_n\not\to0$, which gives our contradiction.
We first define $y_0=0$ and $I_0=\emptyset$. Then, if $y_n$ and $I_n$ are defined, we use that $\{x_n\}$ is unbounded and that $yx_m\to0$ as $m\to\infty$ for all $y\in A$. We find some $i\in\mathbb{N}\setminus I_n$ such that $\|x_i\|\geq2^{n+1}(1+2^{-n})$ and $\|y_nx_i\|\leq2^{-n-1}$ and $\|x_j^*x_i\|\leq1$ for all $j\in I_n$.
We define $y_{n+1}=y_n+2^{-n-1}\|x_i^*\|^{-1}x_i^*$ and $I_{n+1}=I_n\cup\{i\}$. For $j\in I_n$ we get $$\|y_{n+1}x_j\|\geq\|y_nx_j\|-2^{-n-1}\|x_i^*\|^{-1}\|x_i^*x_j\|\geq1+2^{-n}-2^{-n-1}=1+2^{-n-1}.$$ Note that we use $\|x_i^*\|^{-1}\leq1$ and $\|x_i^*x_j\|=\|x_j^*x_i\|\leq1$. Finally, we have $$\|y_{n+1}x_i\|\geq2^{-n-1}\|x_i^*\|^{-1}\|x_i^*x_i\|-\|y_nx_i\|\geq2^{-n-1}\|x_i\|-2^{-n-1}\geq1+2^{-n-1}.$$ Here we use $\|x_i^*x_i\|=\|x_i^*\|\|x_i\|$ and $\|x_i\|\geq2^{n+1}(1+2^{-n})$.
Proof of main result: The proof is by contraposition. Let $A$ be a $C^*$-algebra and let $\{x_n\}$ be a sequence in $A$ such that $yx_n\to0$ for all $y\in A$. Note that by Lemma 4 we can assume without loss of generality that $\|x_n\|\leq1$ for all $n$. Let $\phi$ be a linear functional on $A$ such that $\phi(x_n)\not\to0$. The goal is to show that $\phi$ is unbounded.
Since $\phi(x_n)\not\to0$, passing to a subsequence we may assume without loss of generality that there exists $\varepsilon>0$ such that $|\phi(x_n)|\geq\varepsilon$ holds for all $n$. Again, passing to a subsequence, we can assume without loss of generality that $\phi(x_n)$ has a constant sign, which we can then assume without loss of generality to be positive. So $\phi(x_n)\geq\varepsilon$ holds for all $n$.
We define a sequence $a_n$ in $A$ recursively, such that $\|a_n\|\leq\sqrt{n+1}+(1-2^{-n})\varepsilon$ and $\phi(a_n)\geq(n+1)\varepsilon$. Notice that this indeed implies that $\phi$ is unbounded.
We first define $a_0=x_0$. Then, if $a_n$ is defined, we use Proposition 3. We find some $e\in A$ with $e\geq0$ and $\|e\|\leq1$ such that $\|ea_n-a_n\|\leq2^{-n-2}\varepsilon$.
By hypothesis, we get $ex_m\to0$ as $m\to\infty$, so we find $i$ such that $\|ex_i\|\leq2^{-n-2}\varepsilon$. By Corollary 2, we get
\begin{align}
\|a_n+x_i\|
&\leq\sqrt{\|a_n\|^2+\|x_i\|^2}+2^{-n-1}\varepsilon
\\&\leq\sqrt{n+2}+(1-2^{-n-1})\varepsilon.
\end{align}
Here we use that for $c\geq0$ the inequality $$\sqrt{(\sqrt{n}+c)^2+1}\leq\sqrt{n+1}+c$$ holds, which can be verified by squaring both sides. Furthermore, by linearity of $\phi$ we have $\phi(a_n+x_i)\geq(n+2)\varepsilon$, so $a_{n+1}=a_n+x_i$ is a sufficient recursive definition.
