I'm trying to prove that the price of an European call option, denoted by $C(T,K),$ where $T$ is the maturity date and $K$ is the strike, is an increasing function of $T.$
I can't see why the map $T\rightarrow C(T,K)$ is an increasing function.I was using the definition of a Call option but I don't find clear the reason of the proposition.
Suppose that $T_1 < T_2$ but $C(T_1, K) > C(T_2,K)$ for a contradiction. Call the option with maturity time $T_i$ option $i$ and write $S_t$ for the price of the underlying asset at time $t$ (e.g. the stock price). I proceed by a no arbitrage argument.
We begin by buying the cheap option and writing the expensive one. That is we buy option $2$ and we sell option $1$ giving a profit of $C(T_1,K) - C(T_2,K) > 0$ which we place in a risk-free asset.
At time $T_1$, we may have $S_{T_1} < K$ in which case the holder of option $1$ does nothing and we can then also do nothing to guarantee the profit from our risk-free asset (though this may be suboptimal).
Otherwise $S_{T_1} \geq K$ and the holder of option $1$ exercises the option so we must short-sell them the stock for the strike price $K$. We are left to close out the short sell, which we do by exercising option $2$ at time $T_2$ to use the $K$ profit from the short-sell to buy the underlying stock leaving us again with the profit from the risk-free asset.
In total this gives a risk-free profit of $[C(T_1,K) - C(T_2,K)] e^{rT_2}$ where $r$ is the risk-free rate so this strategy gives an arbitrage. Therefore by the no-arbitrage assumption we must in fact have $C(T_1,K) \leq C(T_2,K)$.