$M \leq X$. 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. 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.