Let's focus on a European call option for the sake of the argument. Assume deterministic rates to keep notations uncluttered. Define $\Bbb{Q}$ as the probability measure associated to the money market numéraire $B_t$.
$$ C(K,T) = \frac{1}{B_T} \Bbb{E}^\Bbb{Q} \left[ (S_T-K)^+ \right] = \frac{1}{B_T} \int_K^\infty (S - K) q(S) dS $$
Whence (Leibniz rule)
$$ \frac{\partial C}{\partial K}(K,T) = - \frac{1}{B_T} \int_K^\infty q(S) dS = -\frac{1}{B_T}\Bbb{Q}(S_T \geq K)$$
since $B_T$ is a numéraire (traded asset with positive value at all times) and since by definition of a probability
$$\Bbb{Q}(\omega) \geq 0 , \forall \omega \in \Omega$$
the resulting no static arbitrage condition is indeed $ \frac{\partial C}{\partial K}(K,T) \leq 0 $
Similarly,
$$ \frac{\partial^2 C}{\partial K^2}(K,T) = \frac{1}{B_T}q(K) $$
where the same argument applies for the sign of $B_T$ and again by definition of a p.d.f.
$$ q(\cdot) \geq 0 $$
so that we we get $ \frac{\partial^2 C}{\partial K^2}(K,T) \geq 0 $
As you can see the weak inequalities directly come from the positivity on the c.d.f. and p.d.f.
As far as calendar arbitrage is concerned, the relation you wrote holds for American options and is a direct consequence of the latter's definition
$$ C^{AM}(K,T) = \sup_{\tau \in [0,T]} \Bbb{E}^\Bbb{Q} \left[ \frac{1}{B_\tau}(S_{\tau} - K)^+ \right] $$
where $\tau$ is a stopping time and obviously
$$ \sup_{\tau \in [0,T_2]} \Bbb{E}^\Bbb{Q} \left[ \frac{1}{B_\tau}(S_{\tau} - K)^+ \right] \geq \sup_{\tau \in [0,T_1]} \Bbb{E}^\Bbb{Q} \left[ \frac{1}{B_\tau}(S_{\tau} - K)^+ \right] $$
for any $T_2 \geq T_1$.
You have a similar inequality for the absence of calendar arbitrage for European options though, see this related question for further info.
