Notations:
$T,K$ are the maturity and the strike of a vanilla call of price $C(T, K)$.
$(S_t)_{t\in [0,T]}$ the price process of the underlying.
$\mathbb{Q}$ is the risk neutral measure.
$x(T,K) = \log(K/S_0) - rT$ is the log-moneyness.
Fix a maturiy $T^* > 0$ and define :
$$C(K) := C^{BS}(T^*, K, \sigma_{imp}(K))$$
S.t the R.H.S is the Black-Schiles price of a vanilla call with strike $K$ and maturity $T^*$
If $\mathbb{E}_{\mathbb{Q}}\left(S_{T^*}^2\right) < \infty$, one can easily show that :
$$\forall \ K > 0 : \quad C(K) \leq e^{-rT^*} \frac{\mathbb{E}_{\mathbb{Q}}\left(S_{T^*}^2\right)}{4K}$$
We will prove the following inequality :
$$ \exists \ \kappa > 0, \ \forall \ x > \kappa : \sigma_{imp}^2(x(K)) \leq 2\frac{x(K)}{T^*}\quad \quad (*)$$
First, notice that by using the bound on the call price we have :
$$\lim_{x\to +\infty} C(K(x)) = 0 \quad \quad (1)$$
Knowing that $\sigma \mapsto C^{BS}(T^*, K, \sigma)$ is continuous and increasing, it suffices to show that :
$$C^{BS}(T^*, K, \sigma_{imp}(x)) \leq C^{BS}(T^*, K, \sqrt{2\frac{x(K)}{T^*}})$$
Now, using l’hôpital’s rule we can prove that :
$$ \lim_{x\to +\infty} C^{BS}(T^*, K, \sqrt{2\frac{x(K)}{T^*}}) = \frac{S_0}{2}$$
So : $$ \exists \ A > 0, \ \forall \ x>A: \quad C^{BS}(T^*, K, \sqrt{2\frac{x(K)}{T^*}}) - \frac{S_0}{2} \geq - \frac{S_0}{4}$$
So that :
$$ \exists \ A > 0, \ \forall \ x>A: \quad C^{BS}(T^*, K, \sqrt{2\frac{x(K)}{T^*}}) \geq \frac{S_0}{4}$$
By (1), the definition of the limit gives: $$\exists \ B > 0, \ \forall \ x>B: \quad C(K(x)) \leq \frac{S_0}{4}$$
Puting $\kappa := \max(A,B)$ completes the proof.
Now expressing (*) in the strike forme rather than log-moneyness proves that :
$$ \sigma_{imp}^2(K)= \underset{K\to +\infty}{\mathcal{O}}(\log(K))$$