# Value of European Call equals Value of American Call, Question on Explanation/Proof

I am reading S. Shreve, Stochastic Calculus for Finance, Vol. I. There he proves that American Call Options have the same value as European Call Options. In the proof he uses that for a Call option the payoff function $g(s) = (s - K)^+$ is convex, and then shows that $$\frac{1}{(1+r)^n} g(S_n)$$ is a submartingale (he calles this process the discounted intrinsic value process, where the intrinsic value process is the process which gives the immediate payout at each time and for each event). Because it is a submartingale under the risk-neutral measure by the optional sampling theorem for each stopping time $\tau : \Omega \to \mathbb N \cup \{\infty\}$ the stopped process considered at the $N$-th time step $$\frac{1}{(1+r)^{N \wedge \tau}} g(S_{N\wedge \tau})$$ has a lower expectation, i.e. we can not get better by stopping some time before $N$, which shows the claim.

After the proof, there are some explanations I do not understand:

[The Theorem] shows that the early exercise feature of the American call contributes nothing to its value. An examination of the proof of the theorem indicates that this is because the discounted intrinsic value of the call is a submartingale (i.e. has a tendency to rise) under the risk-neutral probabilities. The discounted intrinsic value of an American put is not a submartingale. If $g(s) = (K - s)^+$, then the Jensen inequality still holds but [here he refers to an equation in the proof] does not. Jensen's inequality says that the convex payoff of the put imparts to the discounted intrinsic value of the put a tendency to rise over time, but this may be overcome by a second effect. Because the owner of the put receives $K$ upon exercise, she may exercise early in order to prevent the value of this payment from being discounted away. For low stock prices, this second effect becomes more important than the convexity, and early exercise becomes optimal.

In his explanations he seems to say that $(K - s)^+$ is convex too ("[...] Jensen's inequality says that the convex payoff of the put [...]"), but it is not (indeed it is concave...) and so his explanations and Jensen's inequality does not apply, so I can not follows his explanations, maybe I have understood something wrong? Could someone please explain?

EDIT: The relevant part could be accessed with google books.

• I'm not sure where you get that the payoff of a put is concave, because it is most definitely convex. Jun 11, 2015 at 19:01
• @ocstl It certainly actually depends on whether you are long or short... Jun 11, 2015 at 22:02

So the author is correct in saying that $(K-s)^+$ is convex.