Let $\rho_n,\,\rho\in C(\mathbb R^d)\cap L^1(\mathbb R^d)$ positive for every $n\in\mathbb N\,$. Moreover I know
$$\int_{\mathbb R^d}\rho_n(x)\,d x \,=\,1 \;,\quad\, \int_{\mathbb R^d}|\rho_n(x)-\rho(x)|\,d x\xrightarrow[n\to\infty]{}\,0\,.$$
So in particular the sequence $(\rho_n)_{n\in\mathbb N}$ is bounded in $L^1(\mathbb R^d)$. Can I say that $(\rho_n)_{n\in\mathbb N}$ is also uniformly bounded on compact sets?
Precisely for every $K$ compact subset of $\mathbb R^d$ there exists $M=M_K\in[0,\infty)$ such that for every $n\in\mathbb N$, $x\in K$ we have $\rho_n(x)\leq K$ ?
Edit. In a previous version of this question I asserted that $C(\mathbb R^d)\cap L^1(\mathbb R^d)$ is included in $L^\infty(\mathbb R^d)$. This is not true as pointed out in the comments. The question should still make sense, since I am interest in uniformly bounded sequence on compact sets.
Edit 2. A very rough idea of why I think the answer might be positive. $\rho_n$ is a sequence of probability densities. If it is not uniformly bounded on compact sets, the example of an appoximation of identity comes in my mind. But in this case the limit for $n\to\infty$ would be the Dirac delta distribution, which does not belong to $L^1(\mathbb R^d)\,$. Of course this is far from being a proof.
Let $f_n$ be the piecewise linear function defined on $[0,1]$ satisfying $$f(0) = 0,\ f(\frac 1{2n}) = -n,\ f(\frac 1n) = 0,\ f(1 - \frac 1n) = 0,\ f(1-\frac 1{2n}) = n,\ f(1) = 0.$$You can extend $f_n$ to the entire line by setting $f(x) = 0$ outside the interval $[0,1]$. It is easy to verify that $\int f_n \, dx = 0$ for all $n$, that $\int |f_n| \, dx \to 0$, and that $\{f_n\}$ is not uniformly bounded on $[0,1]$.
Now let $\rho$ be any function with $\int \rho \, dx = 1$ and let $\rho_n = f_n + \rho$. Then $\int \rho_n \, dx = 1$ for all $n$, $\int |\rho_n - \rho| \, dx \to 0$, but $\{\rho_n\}$ is not uniformly bounded.
EDIT: The example above missed the requirement that $\rho_n \ge 0$. Another way to proceed is to let $\rho_n$ ($n \ge 2$) be the continuous piecewise linear function on $[0,1]$ satisfying $$\rho_n(0) = 0,\ \rho_n(\frac 1{2n}) = n,\ \rho_n(\frac 1n) = 0,\ \rho_n(\frac 12) = 0,\ \rho_n( \frac 34) = \frac{8n-4}{2n},\ \rho_n(1) = 0.$$ The strange-looking choice of $\rho_n(3/4)$ guarantees that $\int \rho_n \, dx = 1$ for all $n$.
The functions $\rho_n$ converge pointwise on $[0,1]$ to the continuous piecewise linear function $\rho$ satisfying $$\rho(0) = 0,\ \rho(\frac 12) = 0,\ \rho(\frac 34) = 4,\ \rho(1) = 0.$$ Both $\rho_n$ and $\rho$ can be extended to be equal to zero off the interval $[0,1]$. It is routine to check that $$\int |\rho_n - \rho| \, dx \to 0.$$