Let $ K(x) = \left\{ \begin{array}{ll} \frac{e^{-1/1-x^2}}{C} & {|x|<1}, \\ 0 &{|x|\geq 1} \end{array} \right.$, where $C = \int^{1}_{-1}e^{-1/(1-x^2)}$, and let $K_{t}(x) = \frac{1}{t}K(x/t).$
Prove that the family of operators
$$A_{t}:L^{1}(\mathbb{R}) \rightarrow L^{1}(\mathbb{R}), A_{t}(f) = f*K_{t},$$
is uniformly bounded.
Could anyone how can I solve this, if the only given information that I have is the following theorem:
Or this theorem is not needed for solving this excercise?
This is simpler than that. There is a named result called Young's inequality that tells you that if $1 \leq p,q,r \leq \infty$ satisfy $\frac{1}{p} + \frac{1}{q} = \frac{1}{r} + 1$ and $f \in L^p$, then the operator $g \mapsto f*g$ is bounded as an operator from $L^q$ to $L^r$. It also gives you a bound on the operator norm, which as you might expect is $\| f \|_{L^p}$. Since this does your problem in its entirety, you should probably figure out how to prove this yourself rather than just citing it.