Let $A$ be a Banach algebra. Some textbooks define a (left ) multiplier as a map $T:A\rightarrow A$ satisfying $T(ab)=T(a)b$ for all $a,b\in A$ and assume that $A$ needs to be a without order Banach algebra, that is, if $xA=(0)$ for some $x\in A$, then $x=0$. (right multiplier is defined in a similar way). However, some other define a (left) multiplier as a linear map (even continuous) on a (general) Banach algebra satisfying the above condition. Am I missing something here? I know the first definition does not assume that $T$ is linear however it follows from that $A$ is without order. Is a multiplier assumed to be continuous by definition?
I would be interested to know exactly what sources you are using, as any standard textbook does prove the following.
I actually don't know what happens for just a left multiplier (though also do not know any counter-examples off the top of my head), but suppose we have a double multiplier $(L,R)$, that is, $L$ is a left multiplier, and $R$ is a right multiplier, and $R(a)b = aL(b)$ for all $a,b\in A$.
Notice that being "without order" is equivalent to:
For $a,b\in A$, if $ac=bc$ for all $c\in A$, then $a=b$.
There is an obvious equivalent definition on the other side. I will assume both these conditions hold for $A$.
Using this, we find that for $a,b\in A$ and $\lambda$ a scalar, $$ c(L(a)+\lambda L(b)) = cL(a) + c (\lambda L(b)) = R(c)a + R(c) (\lambda b) = R(c)(a+\lambda b) = cL(a+\lambda b), $$ for any $c\in A$, and thus $L$ is linear. Similarly $R$ is linear. In fact, it is enough that $R(a)b=aL(b)$ for $a,b\in A$; then a similar argument shows that $L$ is automatically a left multiplier, and $R$ a right multiplier.
We don't assume that $L$ and $R$ are bounded. However, we can use the closed graph theorem in the following way: if $(a_n)\rightarrow 0$ and $L(a_n)\rightarrow a$, then $$ ba = \lim_n bL(a_n) = \lim_n R(b) a_n = 0 $$ for any $b$, and so $a=0$, and thus $L$ is continuous; similarly $R$ is continuous.