Convolution of a sequence of L1 function has uniformly convergence. 
Problem: Let $f: \mathbb{R} \to \mathbb{R}$ be bounded uniformly continuous function. And $(K_{n})_{n=1}^{\infty}$ is a sequence of $L^{1}(\mathbb{R})$ function such that 
  
  
*
  
*$\left\|K_{n}\right\|_{L^{1}} \leq M < +\infty$
  
*$\int_{-\infty}^{\infty}K_{n}(x)dx \to 1$ as $n \to \infty$
  
*$\int_{x:|x|> \delta} |K_{n}(x)| \to 0$ as $n \to \infty$.
  
  
  Then, $K_{n} *f \to f$ uniformly, where $*$ means a convolution.

I think I solved this problem, however, I didn't use all conditions I have, especially the condition that $\left\|K_{n}\right\|_{L^{1}}$ is uniformly bounded. Could you check that whether my attempt is right or not?

My attempt: Suppose $f\geq 0$. And $||f||_{\infty} <+\infty$ Note that for any $\delta >0$ and any $n \in \mathbb{N}$, $$\int_{-\infty}^{\infty}K_{n}(x)dx = \int_{|x|\leq \delta}K_{n}(x)dx+\int_{x:|x|> \delta} K_{n}(x) $$
  and from the fact $|\int_{x:|x|> \delta} K_{n}(x)| \leq \int_{x:|x|> \delta} |K_{n}(x)|,$ $\int_{x:|x|> \delta} K_{n}(x)$ also converges to $0$, hence
  $$\int_{|x|\leq \delta}K_{n}(x)dx \to 1 \textrm{ as } n \to \infty , \forall \delta>0. $$
  Now from uniform continuity of $f$ and $f\geq 0$, for given $\epsilon>0$, $\exists \delta>0$ such that $$|x-y|<\delta \implies |f(y)-f(x)|<\epsilon \implies -\epsilon +f(x) < f(y) < \epsilon +f(x).$$
  Also, take $N_{1}$ such that $\forall n > N_{1}$, 
   $$\left|\int_{|x|\leq \delta}K_{n}(x)dx - 1\right|< \epsilon \implies 1-\epsilon< \int_{|x|\leq \delta}K_{n}(x)dx<1+\epsilon.$$
  Also, take $N_{2}$ such that $\forall n > N_{2}$,
   $$\left|\int_{x:|x|> \delta} K_{n}(x)\right|< \epsilon.$$
  Now let $N = \max(N_{1}, N_{2})$ Then, for any $x \in \mathbb{R}$ and $\forall n>N$,
  \begin{align*}
|K_{n}*f(x) - f(x)| &=  \left|\int_{|y|\leq \delta}K_{n}(y)f(x-y)dy+\int_{y:|y|> \delta} K_{n}(y)dy - f(x)\right| \\
& \leq \left|\int_{|y|\leq \delta}K_{n}(y)f(x-y)dx - f(x)\right| + \left|\int_{y:|y|> \delta} K_{n}(y)dy\right| \\
& \leq \left|(f(x)+\epsilon)(1+\epsilon) - f(x)\right| + \epsilon \\
& \leq \left|\epsilon f(x)+\epsilon(1+\epsilon)\right| + \epsilon \\
& \leq \epsilon(||f||_{\infty}+2+\epsilon)
\end{align*}
  Hence, by letting $\epsilon \to 0$, we can conclude that $K_{n}*f \to f$ uniformly. 

For general $f$, consider $f=f^{+}-f^{-}$, then each $f^{+}, f^{-}$ is also bounded uniformly continuous function, with the triangle inequality 
$$|K_{n}*f - f| \leq|K_{n}*f^{+} - f^{+}|+|K_{n}*f^{-} - f^{-}| $$
gives the desired result.
 A: As noted in Daniel Fischer's answer, the result is false without condition (1), so a proof that doesn't use (1) cannot be right. Regarding where the error is, I really don't follow anything in your chain of inequalities. I mean the very first step, $$|K_{n}*f(x) - f(x)| =  \left|\int_{|y|\leq \delta}K_{n}(y)f(x-y)dy+\int_{y:|y|> \delta} K_{n}(y)dy - f(x)\right|,$$is wrong - there's no reason to think that $$K_{n}*f(x) - f(x)=  \int_{|y|\leq \delta}K_{n}(y)f(x-y)dy+\int_{y:|y|> \delta} K_{n}(y)dy - f(x).$$
Hint to get you started  towards a correct solution:
Since $\int K_n\to1$ there exists $N$ so $\int K_n\ne0$ for $n>N$. Define $$c_n=\frac1{\int K_n}\quad(n>N).$$
Since $$||K_n*f-c_nK_n*f||_\infty\le||K_n-c_nK_n||_1||f||_\infty
=|1-c_n|||K_n||_1||f||_\infty\to0,$$it's enough to show that $c_nK_n*f\to  f$ uniformly. So we may as well simplfy the notation and assume $$\int K_n=1.$$
Main Trick, that you more or less always use when proving this sort of thing: Since $\int K_n=1$, $$K_n*f(x)-f(x)=\int K_n(y)(f(x-y)-f(x))\,dy,$$hence $$|K_n*f(x)-f(x)|\le\int |K_n(y)(f(x-y)-f(x))|\,dy.$$
A: Let $f(x) = \min \{ 1, \lvert x\rvert\}$, and define
$$K_n(x) = \begin{cases}C_n + n &\text{if } \lvert x\rvert < \frac{1}{2n}\\ \;\;-C_n &\text{if } \frac{1}{2n} \leqslant \lvert x\rvert < \frac{1}{n}\\ \quad 0 &\text{if } \lvert x\rvert \geqslant \frac{1}{n}\end{cases}$$
where $C_n > 0$ is to be chosen later. It's easy to see that $(K_n)$ satisfies conditions 2 and 3, and that $\lVert K_n\rVert = 1 + 2C_n$, so $(K_n)$ satisfies condition 1 if and only if $(C_n)$ is bounded.
When $n \geqslant \frac{1}{\delta}$, we compute
\begin{align}
(K_n \ast f)(0) &= \int_{-\delta}^{\delta} K_n(y)f(-y)\,dy \\
&= 2\int_0^{\frac{1}{2n}}(C_n+n) y\,dy - 2\int_{\frac{1}{2n}}^{\frac{1}{n}}C_n y\,dy \\
&= \frac{C_n + n}{4n^2} - \frac{3C_n}{4n^2} \\
&= \frac{1}{4n} - \frac{C_n}{2n^2}\,.
\end{align}
Thus $K_n\ast f$ doesn't even converge pointwise to $f$ if $(C_n)$ grows sufficiently fast.
If we split the integral giving the convolution,
\begin{align}
(K_n\ast f)(x) - f(x) &= \int_{\mathbb{R}} K_n(y) \bigl(f(x-y) - f(x)\bigr)\,dy \\
&= \int_{\lvert y\rvert \leqslant \delta} K_n(y)\bigl(f(x-y)-f(x)\bigr)\,dy + \int_{\lvert y\rvert > \delta} K_n(y)\bigl(f(x-y)-f(x)\bigr)\,dy\,,
\end{align}
we note that the second summand is easily controlled, since
$$\Biggl\lvert \int_{\lvert y\rvert > \delta} K_n(y)\bigl(f(x-y)-f(x)\bigr)\,dy\Biggr\rvert \leqslant 2\lVert f\rVert_{\infty} \int_{\lvert y\rvert > \delta} \lvert K_n(y)\rvert\,dy\,$$
and by condition 3, the right hand side tends to $0$ and is independent of $x$.
It is to control the first summand that condition 1 is used. Pulling the modulus inside the integral, we get
$$\Biggl\lvert \int_{\lvert y\rvert \leqslant \delta} K_n(y)\bigl(f(x-y) - f(x)\bigr)\,dy\Biggr\rvert \leqslant \int_{\lvert y\rvert \leqslant \delta} \lvert K_n(y)\rvert\,\underbrace{\bigl\lvert f(x-y) - f(x)\bigr\rvert}_{\leqslant \varepsilon}\,dy \leqslant \varepsilon \int_{\lvert y\rvert \leqslant \delta}\lvert K_n(y)\rvert\,dy\,,$$
and if we have condition 1, this gives a uniform bound of $\varepsilon M$.
Without condition 1, it is possible that (at least for some $x$) the values of $f(x-y) - f(y)$ and of $K_n(y)$ "conspire" to blow up that part of the integral. For example if $\lvert f(x-y) - f(x)\rvert$ is relatively large and of constant sign where $K_n$ is negative, and $\lvert f(x-y) - f(x)\rvert$ is relatively small where $K_n$ is positive, as we saw above.
