$M \leq X$. Is it true that $M^*$ is a subspace of $X^*$ where $*$ denotes the dual space?

Let $ X$ be a Banach space and $ M$ be a closed subspace i.e. $ M \leq X$ .

Is it true that $ M^*$ is a subspace of $ X^*$ where $ *$ denotes the dual space?

What I have tried:

I know by Hahn Banach theorem that every $ f \in M^*$ has an extension $ \hat f \in X^*$ such that $ \hat f |_M =f$ and the operator norm doesn’t change i.e. $ \|\hat f\| =\|f\|$ .

I found that such an extension can be chosen to be linear if $ X$ is a Hilbert space here, hence my claim might not hold in the general case.

Any help is appreciated.

If $f$ belongs to $M^{+} $ and $c \ge 0$ then $cf$ belongs to $M^{+}$ and $ \int cf = c\int f$

If $ f$ belongs to $ M^{+} $ and $ c \ge 0$ then $ cf$ belongs to $ M^{+}$ and $ \int cf = c\int f$ .

I need to proove that, using the following observation:

if $ f\in M^{+}$ and $ c>0 $ , then the mapping $ \varphi \rightarrow \psi = c\varphi$ is a one-toone mapping between simple function $ \varphi \in M^{+}$ with $ \varphi \le f $ and simple functions $ \varphi$ in $ M^{+} $ with $ \psi \le cf $ .

I know that this question is already answer here:One-to-one mapping of simple functions $ \phi \to \psi = c\,\phi$ implies $ \int cf\,d\mu = c \int f\,d\mu$ ?

But I can’t follow the verbal explanation.

My original idea was to proove $ $ c \int f \le \int cf \le c\int f $ $ But I can’t… some idea?

$Whether\ the\ CTL\ formula\ M^{‘},\ s_0 \models EG\ AF\ p\ is\ established\ or\ not\ in\ this\ model\ M?$

I have been learning Verification by model checking recently and I get the following question:

$ Whether\ the\ CTL\ formula\ M^{‘},\ s_{0} \models EG\ AF\ p\ is\ established\ or\ not\ in\ this\ model\ M\ ?$

And there is the description of model M.Model

I think it is incorrect because there is a deadlock or infinite loop about $ s_0$ after the state is starting from $ s_0$ , which make it invalid under $ EG\ AF\ $ condition.

if not mind, could anyone tell me my thought is right or wrong? And prove it right or give a counterexample.

Thanks in advances.