Problem With Lower And Upper No-Arbitrage Barriers And Inequalities-Financial Mathematics EXERCISE

We consider a financial market of one period with two stocks $(S_0,S_1)$.
  We further consider a purchase option with repayment $C = (S_1^{1} - K)^+$
  with a value of $K> 0$ and maturing time $T = 1$. Let $\underline{π}(C)$,$\overline{π}(C)$ the lower and upper no-arbitrage barriers for the value of $C$.
a)Show that the following inequality applies
  $$\underline{π}(C)\geq \bigg(S_1^{1} - \dfrac{k}{1+r}\bigg)^+$$
Please also show that the following equality applies
  $$\underline{π}(C)= \bigg(S_1^{1} - \dfrac {k}{1+r}\bigg)^+$$
if it is further assumed that:
  $$P[S_1^{1}=S_0^{1} \cdot(1+r)]> 0$$
Hint : Use the probability measure $$Q[·]: = P [·|S_1^{1} =S_0^{1}\cdot (1+r)]$$
b) Also show that the following inequality applies
  $$\underline {π}(C)\leq S_0^{1}$$
and that the following equality applies
  $$\overline{π}(C)=S_0^{1}$$
assuming that $\rm ess \sup S_1^{1}=\infty $ and that $\rm ess \inf  S_1^{1}=0$.

ATTEMPT
a)We have a financial market with no-arbitrage so we have the form:
$$π(C)=E_Q\bigg[\dfrac{c}{1+r}\bigg]<\infty$$ for $Q\subset P$
So,$$\underline{π}(C)\geq \inf_{Q \in P}E_Q\bigg[\dfrac{c}{1+r}\bigg]=\inf_{Q \in P}E_Q\bigg[\dfrac{S_1^{1}-k}{1+r}\bigg]=\inf_{Q \in P}\dfrac{1} {1+r}E_Q[S_1^{1}-k]$$
UPDATE
Ι think that i have to use put-call-parity equation:
$$C_t-P_t=S_t-kB(t,T)$$
with $B(t,T)=\dfrac{S_0^{0}} {S_0^{1}}=\dfrac{1}{1+r}$
Then we rearrange the equation so that C is on the left hand side and use P_t>0 (using 1/(1+r) for the discount factor).So,we have $$C_t=P_t+S_t-k \dfrac{1}{1+r}$$
Can anyone help me from here with a thorough solution so to find the question a)?
I don't know if this is right but i want to know if I am in the right way to solve the problem.
Also,I would like to ask if $\min=\inf$ in this problem
For question b) i know that i have to take this theorem

and this remark but i don't know even how to start!

I would really appreciate any hints/thorough solution because it's the first problem that I have to solve and I don't have any experience in this type of exercise and generally it's my first try in the field of financial mathematics
Thanks, in advance!
 A: (a) Use put-call parity and the fact that the put price must be +ve. (b) Call price cannot exceed the value of the stock, as otherwise buying the stock and selling the call would generate an arbitrage profit.
A: $\mathbf{(a)}$ From the extension of the Theorem hinted, the lower no-arbitrage bound is :
$$\underline{\pi}(C) = \inf_{\mathbb{Q} \in \mathcal{P}}\mathbb{E}_\mathbb{Q}\bigg[\frac{C}{1+r}\bigg]=\inf_{\mathbb{Q} \in \mathcal{P}}\mathbb{E}_\mathbb{Q}\bigg[\frac{(S_1^1 - K)^+}{1+r}\bigg]$$
The Jenses inequality in context of probability theory, says thati f $X$ is a random variable and $\varphi$ is a convex function, then it is :  $\varphi\big(\mathbb{E}[X]\big) \leq \mathbb{E}[\varphi(X)]$. The function $f(x) = (x-K)^+, \; K>0$ is convex, thus :
$$\mathbb{E}_\mathbb{Q}\bigg[\frac{(S_1^1 - K)^+}{1+r}\bigg] \geq\frac{1}{1+r}\big[\mathbb{E}_\mathbb{Q}(S_1^1 - K)\big]^+ $$
$$= \frac{1}{1+r}\big[\mathbb{E}_\mathbb{Q}(S_1^1) - \mathbb{E}_\mathbb{Q}(K)\big]^+= \frac{1}{1+r}\bigg[S_0^1(1+r) - K\bigg]^+$$
$$\implies$$
$$\underline{\pi}(C) \geq \bigg(S_0^1 - \frac{K}{1+r}\bigg)^+$$
where the probability measure $\mathbb{Q}[\cdot] := \mathbb{P}[\cdot | S_1^1 = S_0^1(1+r)]$ was used.
Now if $\mathbb{P}[S_1^1 = S_0^1(1+r)] >0$, this means that at some point it will be $S_1^1 = S_0^1(1+r)$ and thus :
$$\mathbb{E}_\mathbb{Q}\bigg[\frac{(S_1^1 - K)^+}{1+r}\bigg] = \mathbb{E}_\mathbb{Q}\bigg[\frac{(S_0^1(1+r) - K)^+}{1+r}\bigg] = \mathbb{E}_\mathbb{Q}\bigg[\bigg(S_0^1 - \frac{K}{1+r}\bigg)^+\bigg]$$
which means that the equality will hold, since $S_0^1$ is just a constant $>0$ :
$$\underline{\pi}(C) = \bigg(S_0^1 - \frac{K}{1+r}\bigg)^+$$
For part $\mathbf{(b)}$, again from the extension of the Theorem hinted, we have :
$$\overline{\pi}(C) = \sup_{\mathbb{Q} \in \mathcal{P}}\mathbb{E}_\mathbb{Q}\bigg[\frac{C}{1+r}\bigg]=\sup_{\mathbb{Q} \in \mathcal{P}}\mathbb{E}_\mathbb{Q}\bigg[\frac{(S_1^1 - K)^+}{1+r}\bigg]$$
But, note that simply it is :
$$\mathbb{E}_\mathbb{Q}\bigg[\frac{(S_1^1 - K)^+}{1+r}\bigg] \leq \mathbb{E}_\mathbb{Q}\bigg[\frac{S_1^1}{1+r}\bigg] = \frac{\mathbb{E}_\mathbb{Q}(S_1^1)}{1+r} = S_0^1 $$
$$\implies$$
$$\overline{\pi}(C) \leq S_0^1 $$
where the probability measure $\mathbb{Q}[\cdot] := \mathbb{P}[\cdot | S_1^1 = S_0^1(1+r)]$ was used.
Can you figure out a similar approach for the last part now ?
