Let $\theta:\mathbb{R}\to\mathbb{R}$ be a measurable bounded function with bounded support such that $\int_\mathbb{R} \theta(x)dx = 1$ and $\theta\ge0$. Also let $f:\mathbb{R}\to\mathbb{R}$ be a summable function.
Define $\theta_t$(x) = $\frac{1}{t} \theta\left(\frac{x}{t}\right)$
Show that $\lim\limits_{t\to0^+} ||f*\theta_t-f||_1 = 0$
I understand intuitively why it's true, but I didn't manage to prove it. Any help would be appreciated.
$$\begin{align} |f\ast\theta_t(x)-f(x)|&=\Bigl|\int_\mathbb{R}\theta_t(y)(f(x-y)-f(x))\,dy\Bigr|\\ &\le\int_\mathbb{R}\theta_t(y)|f(x-y)-f(x)|\,dy\\ &=\int_\mathbb{R}\theta(y)|f(x-t\,y)-f(x)|\,dy. \end{align}$$ $$\begin{align} \int_\mathbb{R}|f\ast\theta_t(x)-f(x)|\,dx&\le\int_\mathbb{R}\int_\mathbb{R}\theta(y)|f(x-t\,y)-f(x)|\,dy\,dx\\ &=\int_\mathbb{R}\theta(y)\int_\mathbb{R}|f(x-t\,y)-f(x)|\,dx\,dy. \end{align}$$ Now $$ \int_\mathbb{R}|f(x-t\,y)-f(x)|\,dx\le2\,\|f\|_1 $$ and $$ \lim_{t\to0}\int_\mathbb{R}|f(x-t\,y)-f(x)|\,dx=0\quad\forall y\in\mathbb{R}. $$ The dominated convergence theorem implies the result.