Let $f$ be bounded and uniformly continuous. $K_n \in L^1(R)$ has the following properties: $$\|K_n\|_1\leq M$$ $$\int_RK_n \to 1$$ $$\int_{|x|>\delta}|K_n|\to 0 (\delta >0)$$ I am to show that $K_n*f\to f$ uniformly.
Observation: I am not sure if this is important but we know that for each $n$ $K_n*f$ is uniformly continuous as it is a convolution of an $L^1$ and $L^\infty$ function.
My attempt:
$$K_n*f(x)=\int_RK_n(t)f(x-t)dt$$ $$|K_n*f(x)-f(x)|\leq |K_n*f(x)+\int_RK_n(t)f(x)dt-\int_RK_n(t)f(x)dt-f(x)|\leq\int_R|K_n(t)||f(x-t)-f(x)|+|f(x)| |\int_RK_n-1|$$ The second term can be made as small as we want (independently of $x$) as $f(x)$ is bounded and by property 2. There is a $\delta$ s.t if $|x-y|<\delta$ then $|f(x)-f(y)|< \epsilon$. Thus for the first term we split it up as follows:
$$\int_R|K_n(t)||f(x-t)-f(x)|=\int_{[-\delta,\delta]}|K_n(t)||f(x-t)-f(x)|+\int_{[-\delta,\delta]^C}|K_n(t)||f(x-t)-f(x)|$$Now the second term is again as small as we want as $f$ is bounded and by property 3. The first term can also is made as small as we by holders, and the fact that in our interval $[-\delta,\delta]$ the uniform continuity kick in.
Is this correct?