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The answer is no. Indeed, your question can be restated as follows: Is it true that for some constant $c>0$ we have $$\int P\ln\frac PQ\,d\mu\ge c\int(\ln P-\ln Q)^2\,d\mu, \tag{1} $$ where $P$ and $Q$ are probability densities with respect to a measure $\mu$?

Let $\mu$ be the counting measure on the set $\{0,1\}$, and let $P(0)=p\in(0,1)$ and $Q(0)=q\in(0,1)$. Then (1) will become $$p\ln\frac pq+(1-p)\ln\frac{1-p}{1-q}\ge c\Big[\Big(\ln\frac pq\Big)^2+\Big(\ln\frac{1-p}{1-q}\Big)^2\Big]. \tag{2} $$ Letting now $p\to0$, we get a contradiction, because the left-hand side of (2) will go to $\ln\frac1{1-q}<\infty$, but the right-hand side of (2) will go to $\infty$.

Added: You have changed your original question quite a bit, without any further comment, thus just invalidating my answer. Anyhow, the answer now changes from "no" to "obviously no". Indeed, to disprove your latest conjecture, just take any $f,f',g$ such that $f'=f\ne g$.

Added more: The second version of the question, very different from the first version, has again been changed by the OP. It is hardly possible to keep up with all these changes.

The answer is no. Indeed, your question can be restated as follows: Is it true that for some constant $c>0$ we have $$\int P\ln\frac PQ\,d\mu\ge c\int(\ln P-\ln Q)^2\,d\mu, \tag{1} $$ where $P$ and $Q$ are probability densities with respect to a measure $\mu$?

Let $\mu$ be the counting measure on the set $\{0,1\}$, and let $P(0)=p\in(0,1)$ and $Q(0)=q\in(0,1)$. Then (1) will become $$p\ln\frac pq+(1-p)\ln\frac{1-p}{1-q}\ge c\Big[\Big(\ln\frac pq\Big)^2+\Big(\ln\frac{1-p}{1-q}\Big)^2\Big]. \tag{2} $$ Letting now $p\to0$, we get a contradiction, because the left-hand side of (2) will go to $\ln\frac1{1-q}<\infty$, but the right-hand side of (2) will go to $\infty$.

The answer is no. Indeed, your question can be restated as follows: Is it true that for some constant $c>0$ we have $$\int P\ln\frac PQ\,d\mu\ge c\int(\ln P-\ln Q)^2\,d\mu, \tag{1} $$ where $P$ and $Q$ are probability densities with respect to a measure $\mu$?

Let $\mu$ be the counting measure on the set $\{0,1\}$, and let $P(0)=p\in(0,1)$ and $Q(0)=q\in(0,1)$. Then (1) will become $$p\ln\frac pq+(1-p)\ln\frac{1-p}{1-q}\ge c\Big[\Big(\ln\frac pq\Big)^2+\Big(\ln\frac{1-p}{1-q}\Big)^2\Big]. \tag{2} $$ Letting now $p\to0$, we get a contradiction, because the left-hand side of (2) will go to $\ln\frac1{1-q}<\infty$, but the right-hand side of (2) will go to $\infty$.

Added: You have changed your original question quite a bit, without any further comment, thus just invalidating my answer. Anyhow, the answer now changes from "no" to "obviously no". Indeed, to disprove your latest conjecture, just take any $f,f',g$ such that $f'=f\ne g$.

The answer is no. Indeed, your question can be restated as follows: Is it true that for some constant $c>0$ we have $$\int P\ln\frac PQ\,d\mu\ge c\int(\ln P-\ln Q)^2\,d\mu, \tag{1} $$ where $P$ and $Q$ are probability densities with respect to a measure $\mu$?

Let $\mu$ be the counting measure on the set $\{0,1\}$, and let $P(0)=p\in(0,1)$ and $Q(0)=q\in(0,1)$. Then (1) will become $$p\ln\frac pq+(1-p)\ln\frac{1-p}{1-q}\ge c\Big[\Big(\ln\frac pq\Big)^2+\Big(\ln\frac{1-p}{1-q}\Big)^2\Big]. \tag{2} $$ Letting now $p\to0$, we get a contradiction, because the left-hand side of (2) will go to $\ln\frac1{1-q}<\infty$, but the right-hand side of (2) will go to $\infty$.

Added: You have changed your original question quite a bit, without any further comment, thus just invalidating my answer. Anyhow, the answer now changes from "no" to "obviously no". Indeed, to disprove your latest conjecture, just take any $f,f',g$ such that $f'=f\ne g$.

Added more: The second version of the question, very different from the first version, has again been changed by the OP. It is hardly possible to keep up with all these changes.

The answer is no. Indeed, your question can be restated as follows: Is it true that for some constant $c>0$ we have $$\int P\ln\frac PQ\,d\mu\ge c\int(\ln P-\ln Q)^2\,d\mu, \tag{1} $$ where $P$ and $Q$ are probability densities with respect to a measure $\mu$?

Let $\mu$ be the counting measure on the set $\{0,1\}$, and let $P(0)=p\in(0,1)$ and $Q(0)=q\in(0,1)$. Then (1) will become $$p\ln\frac pq+(1-p)\ln\frac{1-p}{1-q}\ge c\Big[\Big(\ln\frac pq\Big)^2+\Big(\ln\frac{1-p}{1-q}\Big)^2\Big]. \tag{2} $$ Letting now $p\to0$, we get a contradiction, because the left-hand side of (2) will go to $\ln\frac1{1-q}<\infty$, but the right-hand side of (2) will go to $\infty$.

The answer is no. Indeed, your question can be restated as follows: Is it true that for some constant $c>0$ we have $$\int P\ln\frac PQ\,d\mu\ge c\int(\ln P-\ln Q)^2\,d\mu, \tag{1} $$ where $P$ and $Q$ are probability densities with respect to a measure $\mu$?

Let $\mu$ be the counting measure on the set $\{0,1\}$, and let $P(0)=p\in(0,1)$ and $Q(0)=q\in(0,1)$. Then (1) will become $$p\ln\frac pq+(1-p)\ln\frac{1-p}{1-q}\ge c\Big[\Big(\ln\frac pq\Big)^2+\Big(\ln\frac{1-p}{1-q}\Big)^2\Big]. \tag{2} $$ Letting now $p\to0$, we get a contradiction, because the left-hand side of (2) will go to $\ln\frac1{1-q}<\infty$, but the right-hand side of (2) will go to $\infty$.

Added: You have changed your original question quite a bit, without any further comment, thus just invalidating my answer. Anyhow, the answer now changes from "no" to "obviously no". Indeed, to disprove your latest conjecture, just take any $f,f',g$ such that $f'=f\ne g$.