Cramer's theorem is as follows:
Let $(Y_i)_{i\geq 1}$ be a sequence of i.i.d. random variables, ${S_n=\frac{1}{n}\sum_{i=1}^n Y_i}$ be their average sum and $M_{Y_1}(u):=\mathrm{E}[e^{uY_1}]<\infty$ be the moment generating function of the r.v. $Y_1$. Then, for all $t>\mathrm{E}[Y_1]$ \begin{equation} \lim_{n\rightarrow \infty}\frac{1}{n}\ln P(S_n\geq t)=-I(t)\quad\quad\quad (1) \end{equation} where the rate function $I$ is defined by \begin{equation*} I(t):=\sup_{u}\left(tu-\ln M_{Y_1}(u)\right) \end{equation*} which is the Legendre transform of the log moment generating function.
I showed that Cramers theorem implies
\begin{equation*} \lim_{n\rightarrow \infty}\frac{1}{n}\ln P(S_n< t)=-I(t)\quad\quad\quad(2) \end{equation*} for all $t<\mathrm{E}[Y_1]$.
Later on, I received a comment saying that if $(1)$ is true $(2)$ should not be true and I am confused. Of course there is a difference between two cases. In the first $t>\mathrm{E}[Y_1]$ and in the second $t<\mathrm{E}[Y_1]$. However $(1)$ and $(2)$ do not seem to be compatible.
Question: What is wrong here (if there is any)? how can it be corrected?
Let me tell you how I got $(2)$ from Cramer's theorem:
Let $X_i=-Y_i$ and $t^{'}=-t$. Applying Cramers theorem to $(X_i)_{i\geq 1}$ and the threshold $t^{'}=-t$, the Equation $(1)$ becomes
$$\lim_{n\rightarrow \infty}\frac{1}{n}\ln P(\frac{1}{n}\sum_{i=1}^n X_i\geq t^{'})=\lim_{n\rightarrow \infty}\frac{1}{n}\ln P(-\frac{1}{n}\sum_{i=1}^n Y_i\geq -t)=\lim_{n\rightarrow \infty}\frac{1}{n}\ln P(\frac{1}{n}\sum_{i=1}^n Y_i< t)=-I(t)$$
which is the same with $(2)$. With $t^{'}=-t$ and $X_1$, $$t^{'}>\mathrm{E}[X_1]\Longrightarrow -t>\mathrm{E}[-Y_1]\Longrightarrow t<\mathrm{E}[Y_1]$$
so the other condition is also true. Simlarly, $$M_{X_1}(u):=\mathrm{E}[e^{uX_1}]=\mathrm{E}[e^{-uY_1}]=M_{Y_1}(-u)$$ Applying this result here:
$$I(t):=\sup_{u}\left(t^{'}u-\ln M_{X_1}(u)\right)=\sup_{u}\left(-tu-\ln M_{Y_1}(-u)\right)=\sup_{u}\left(tu-\ln M_{Y_1}(u)\right)$$
Namely, $(1)$ and $(2)$ have the same rate function $I$.
Who is telling you that $(2)$ is wrong? The presentation in your proof could be better - you should not be using $I$ for both rate functions, for example, and in fact we should end up with $I_{X_1}(t)=I_{Y_1}(-t)$ - but the idea is essentially correct, and the conclusion is certainly correct.
In fact, the full version of Cramer's theorem states that, for any closed set $C$ and open set $G$, we have
\begin{align*} \limsup_{n\to\infty}\frac1n\log P(S_n\in C)&\le-\inf_{x\in C}I(x),\\ \liminf_{n\to\infty}\frac1n\log P(S_n\in G)&\ge-\inf_{x\in G}I(x). \end{align*}
To derive your version, put $C=[t,\infty)$ and $G=(t,\infty)$. Since $I$ is convex with minimum at $\mathrm E[X_1]$, $\inf_{x\in C}I(x)=\inf_{x\in G}I(x)=I(t)$. Also $C\supset G$, so $P(S_n\in C)\ge P(S_n\in G)=P(S_n\ge t)$. Thus
$$-I(t)\le\liminf_{n\to\infty}\frac1n\log P(S_n\ge t)\le\limsup_{n\to\infty}\frac1n\log P(S_n\ge t)\le-I(t).$$
Of course, the above argument would work just as well if we let $C=(-\infty,t]$ and $G=(-\infty,t)$, with $t<\mathrm E[X_1]$. Again by convexity, $\inf_{x\in C}I(x)=\inf_{x\in G}I(x)=I(t)$, so the conclusion follows.