How do I compute $=\inf_{x:x\delta_1+(1-x)\delta_0 =y} I_p(x))$ ? ( using the contraction principle to get the rate funct. of the L.D.P in an example)

Question

Consider the following example,

Also consider

and just in case the contraction principle is quoted below

I don't get how they apply the contraction principle to get $H(\alpha)=I_p(s)$

According to the theorem below, I have to compute for $y \in\mathscr Y=\mathscr M_1({0,1})$$ this expression:

$\tilde H(y)=\inf_{x:f(x)=y} I_p(x)=\inf_{x:f(x)=y} x\log(\frac{x}{p})+(1-x)\log(\frac{1-x}{1-p})$

$=\inf_{x:x\delta_1+(1-x)\delta_0 =y} x\log(\frac{x}{p})+(1-x)\log(\frac{1-x}{1-p})$.

After that, I have to compute its lower semicontinuous regularization to get $H=\tilde H_\text {lsc}$ that is

using the definition of lsc regularization:

$H(\alpha)=\tilde H_\text{lsc}(\alpha)=\sup\{\inf_{\{G\in y\}} \tilde H(y): \alpha \in G, G \text{ open}\}$ with the suggestion in the comments $\tilde H(y)=Ip(s)$ if $y= s\delta_1+(1-s)\delta_0$

I am clueless about this computation, How do I do this?

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Contraction principle