Why is every positive linear map between $C^*$-algebras bounded? We know that every positive linear functional on a $C^*$-algebra is bounded.
How can we prove every positive linear map between $C^*$-algebras is bounded?
 A: Let $\phi:A\longrightarrow B$ be a positive linear map between two $C^*$-algebras. We will show that $\phi$ is automatically bounded. We can assume that $B$ is unital without loss of generality, but this does not change anything. The problem is with $A$. Here is a proof I learned in Blackadar's book which works whether $A$ is unital or not, for positive linear functionals like for positive linear maps in general. We will eventually use the partial order on self-adjoint elements $x\leq y$ if $y-x$ is positive. By assumption, $x\leq y$ implies $\phi(x)\leq \phi(y)$. Note that for $z$ self-adjoint in $B$ unital, we have $\|z\|\leq M$ if and only if $-M1\leq z\leq M1$.
1) Every $x\in A$ can be written  $x=h+ik$ with $h,k$ self-adjoint, $\|h\|\leq \|x\|$ and $\|k\|\leq \|x\|$. Indeed, set $h:=\frac{x+x^*}{2}$ and $k:=\frac{x-x^*}{2i}$. So it suffices to prove that $\phi$ is bounded on self-adjoint elements.
2) Every self-adjoint $x=x^*\in A$ can be written $x=x_+-x_-$ with $x_+,x_-$ positive and $\|x_+\|\leq \|x\|$, $\|x_-\|\leq \|x\|$. Indeed, by functional calculus, take $f_+(t)=\max(t,0)=\frac{|t|+t}{2}$ and $f_-(t)=\max(-t,0)=\frac{|t|-t}{2}$. Then set $x_+:=f_+(x)$ and $x_-:=f_-(x)$. So it suffices to prove that $\phi$ is bounded on positive elements.
3) Assume for a contradiction that $\phi$ is not bounded on positive elements and take $x_n$ a sequence of positive elements such that $\|x_n\|=1$ and $\|\phi(x_n)\|\geq 4^n$. Now set $x:=\sum_{n\geq 1}2^{-n}x_n\in A$. For every $n\geq 1$, $x\geq 2^{-n}x_n$ whence $\phi(x)\geq \phi(2^{-n}x_n)$ and therefore $\|\phi(x)\|\geq \|2^{-n}\phi(x_n)\|\geq 2^{-n}4^n=2^n$. Contradiction since $\|\phi(x)\|<\infty$.
Note: with more work, we can prove that if $A,B$ are unital and if $\phi:A\longrightarrow B$ is a unital linear map, then $\phi$ is positive if and only if $\|\phi\|=1$.
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Suppose $\varphi:A\to B$ is a positive linear map between $C^*$ algebra, then for all positive linear functional $\tau$ on $B$, $\tau\circ\varphi $ is a positive linear functional on $A$ and thus bounded, i.e.$$\lVert\tau\circ\varphi(x)\rVert\leq M_\tau ~~ (\forall x\in A \text{ satisfying } \l{x}\leq 1)$$ for some positive constant $M_\tau$. 
Every bounded linear functional on $B$ is a linear combination of 4 positive linear functionals on $B$ by Jordan Decomposition, thus for all $b^*\in B^*,$ $$\l{b^*(\varphi(x))}\leq M_{b^*} ~~( \forall x\in A \text{ satisfying } \l{x}\leq 1)$$for some positive constant $M_{b^*}$. Hence $\l{\varphi(x)}\leq M$ for some positive constant $M$ by the Principle of Uniform Boundedness, i.e. $\varphi$ is bounded.
