Convergence in $L^1$ and boundedness in $L^{\infty}$ imply strong convergence in $L^p$ for $1<p<\infty$?

Question

Let $f_n$ be a sequence of real functions and $f\in L^1\cap L^{\infty}(\mathbb{R}^n)$ so that \begin{align} f_n\to f \text{ in } L^1(\mathbb{R}^n) \text{ and } f_n \overset{\ast}{\rightharpoonup} f \text{ in } L^{\infty}(\mathbb{R}^n). \end{align} The weak$^{*}$ convergence immediately implies that $f_n{\rightharpoonup}f$ in $L^p(\mathbb{R}^n)$ for any $1<p<\infty$. Indeed the weak$^{*}$ convergence implies \begin{align} \int_{\mathbb{R}^n} f_n \varphi dx \to \int_{\mathbb{R}^n}f \varphi dx \end{align} for any $\varphi\in C_0^{\infty}(\mathbb{R}^n)$. Hence we have weak convergence in $L^p(\mathbb{R}^n)$ since $f\in L^1\cap L^{\infty}(\mathbb{R}^n)$. Is it true that we even get strong convergence?