How to get this inequality in Santambrogio's book about optimal transport?

Question

Let $\hat{\varrho}, \tilde{\varrho}$ be probability density functions on $\mathbb R^d$ where $\tilde{\varrho} \in L^{\infty} (\mathbb R^d)$. For $\varepsilon \in [0, 1]$, we define $\varrho_{\varepsilon}:=(1-\varepsilon) \hat{\varrho}+\varepsilon \tilde{\varrho}$. Set $M=\|\tilde{\varrho}\|_{L^{\infty}}$ and look at $\varepsilon \mapsto \int \varrho_{\varepsilon} \log \varrho_{\varepsilon}$.

It is mentioned in the proof of Proposition 8.7 in Santambrogio's Optimal transport for applied mathematicians that

The integrand can be differentiated in $\varepsilon$ pointwisely, thus getting $\left(1+\log \varrho_{\varepsilon}\right)(\tilde{\varrho}-\hat{\varrho})$. For $\varepsilon<1 / 2$, we can check that these functions are dominated by $(\hat{\varrho}+M)(|\log \hat{\varrho}|+\log M+1)$.

Could you explain how we get the inequality $$ |\left(1+\log \varrho_{\varepsilon}\right)(\tilde{\varrho}-\hat{\varrho})| \le (\hat{\varrho}+M)(|\log \hat{\varrho}|+\log M+1), \quad \forall \varepsilon \in (0, \frac{1}{2}) $$ ?

Thank you so much for your elaboration!

Answer 1

$\newcommand\R{\mathbb R}\newcommand\b{\hat\rho(x)}\newcommand\a{\tilde\rho(x)}$This inequality is false in general.

For instance, if $\varepsilon=1/10$, $M=1/10$, and for some $x\in\R^d$ we have $\a=1/10$ and $\b=10$, then the values at $x$ of the left- and right-hand sides of the inequality in question are $31.66\ldots$ and $10.1$, respectively.

Answer 2

Iosif already pointed out the trivial typo. For the purposes of the argument in that proof (incidentally, it requires that the objects be probability densities on a compact set $\Omega$, not on whole of $\mathbb{R}^d$, as otherwise the RHS of the inequality below will not be $L^1$), it suffices that the following inequality is true: there exists a universal constant $C$ such that

$$ |\left(1+\log \varrho_{\varepsilon}\right)(\tilde{\varrho}-\hat{\varrho})| \le C (\hat{\varrho}+M)(|\log \hat{\varrho}|+|\log M|+1) $$

Log term

For $\varepsilon \in (0,\frac12)$ we have $$ \frac12 \hat{\varrho} \leq \varrho_{\varepsilon} \leq \hat{\varrho} + \frac12 M $$ and since $\log$ is monotonic, we have $$ |\log \varrho_{\varepsilon} | \leq \max\big( |\log (\frac12 \hat{\varrho})|, |\log(\hat{\varrho} + M)|\big) $$ Noting that $\hat{\varrho}+M$ is between $2\hat{\varrho}$ and $2M$, we find $$ |\log \varrho_{\varepsilon} | \leq \log 2 + \max\big( |\log\hat{\varrho}|,|\log M|\big) \leq \log 2 ( 1 + |\log\hat{\varrho}| + |\log M|).$$

Other term

By triangle inequality we have $|\hat{\varrho} - \tilde{\varrho}| \leq M + |\hat{\varrho}|$

Combining

The desired inequality holds with $C = \log 2$.