Let $\mathcal{A}$ be a unital $C^*$-algebra. Let $\varphi$ be a state on $\mathcal{A}$. Let $\psi$ be a positive linear functional on $\mathcal{A}$ such that $\psi\le \phi$. Let $0\le a \le 1$ be an element in $\mathcal{A}$. Is it true that $\psi(a)\le\varphi(a)\psi\left(1_{\mathcal{A}}\right)$?
My attempt: Since $a$ is non-negative, I can write $$\varphi(a)\psi\left(1_\mathcal{A}\right)-\psi(a)=\left(\varphi(a)^{\frac{1}{2}}\psi\left(1_\mathcal{A}\right)^{\frac{1}{2}}-\psi(a)^{\frac{1}{2}}\right)\left(\varphi(a)^{\frac{1}{2}}\psi\left(1_\mathcal{A}\right)^{\frac{1}{2}}+\psi(a)^{\frac{1}{2}}\right)$$
Clearly, $\left(\varphi(a)^{\frac{1}{2}}\psi\left(1_\mathcal{A}\right)^{\frac{1}{2}}+\psi(a)^{\frac{1}{2}}\right)\ge 0$. All I need to show is that $\left(\varphi(a)^{\frac{1}{2}}\psi\left(1_\mathcal{A}\right)^{\frac{1}{2}}-\psi(a)^{\frac{1}{2}}\right)\ge0$ in order to prove the claim.
The only thing which I know is the Cauchy-Schwartz inequality. So, I resorted to using it. Using it, $$\psi(a)\le \psi(a^2)^{\frac{1}{2}}\psi(1_{\mathcal{A}})^{\frac{1}{2}}\le \varphi(a^2)^{\frac{1}{2}}\psi(1_{\mathcal{A}})^{\frac{1}{2}}$$ Since $0\le a \le 1$, we see that $a^2=a^{\frac{1}{2}}aa^{\frac{1}{2}}\le a^{\frac{1}{2}}1_{\mathcal{A}}a^{\frac{1}{2}}=a$. Using that $\varphi$ is a state, we obtain that $$\psi(a)\le \varphi(a)^{\frac{1}{2}}\psi(1_{\mathcal{A}})^{\frac{1}{2}}.$$
All I am missing is a square root on the left hand side. Is there a way to fix it? Is the inequality even true?
Thank you for your time.
The inequality does not hold. Consider the two dimensional $C^*$-algebra of functions defined on $\{0,1\}.$ Let $\varphi(f)={1\over 2}[f(0)+f(1)]$ and $\psi(f)={1\over 6}f(0)+{1\over 2}f(1).$ Then $\psi\le \varphi$ and $\psi(1)={2\over 3}.$ However ${2\over 3}\varphi(f)={1\over 3}f(0)+{1\over 3}f(1).$ Hence the required inequality does not hold.
In general if $\psi (1)\neq 0$ then $\tilde{\psi}=\psi(1)^{-1}\psi$ is a state. The inequality $\tilde{\psi}\le \varphi$ implies $\tilde{\psi}=\varphi,$ i.e. $\psi $ is a multiple of $\varphi.$