I have a question with this test, to give context let's first define the following:
$ \textbf{1.}$ A kernel-function $K:\mathbb{R}\longrightarrow \mathbb{R}_{\geq 0}$ is a continuous function that cancels outside some compact interval and is such that $\int_{\mathbb{R}}K(y)dy=1$
$ \textbf{2.}$ Given a kernel-function $K$, we define for all $k\epsilon \mathbb{N}$, $\mathbb{R}\ni y\mapsto K_k(y):=kK(ky)$
$ \textbf{PROPOSITION:}$
Let $\left\{ K_{k} \right\}_{k\in\mathbb{N}} $ be a sequence of $C^{\infty}$ kernel-functions as defined above, and let $f:\mathbb{R}\longrightarrow \mathbb{R}$ a convex function. Then $f_k(x)):=(f\ast K_k)(x)\equiv \int_{\mathbb{R}} f(x-z)K_k(z)dz$ is a $C^{\infty}$ convex function on $\mathbb{R}$, and $\left\{ f_k \right\}_k$ converges to $f$ uniformly on any compact subset of $\mathbb{R}$
$ \textbf{PROOF}$
It is clear that $f_k$ is a convolution.
I will now present what I have learned from the demonstration.
$ \textbf{-) Convexity in $f_k$.}$
The convexity of $f_k$ is easily deduced from the hypothesis that $f$ is convex, thus:
$f_k((1-\alpha)x+\alpha y)=\int_{\mathbb{R}}f((1-\alpha)x+\alpha y-z)K_k(z)dz=\int_{\mathbb{R}}f((1-\alpha)(x-z)+\alpha(y-z))K_k(z)dz \le (1-\alpha)f_k(x)+\alpha f_k(y)$
$ \textbf{-) $f_k$ to $C^{\infty}$.}$
To see that $f_k$ is of $C^{\infty}$ it is enough to notice that $D(f\ast K_k)=f\ast D(K_k)$ and by hypothesis $K_k$ is of $C^{\infty}$ so you have what you want.
$ \textbf{-) $\left\{ f_k \right\}_k$ converges to $f$ uniformly.}$
As $K$ cancels outside a compact interval then we can assume that
$1=\int _\mathbb{R} K(y)dy=\int_{a}^{b}K(y)dy$ that way then $1=\int_{a}^{b}K(y)dy=\int_{a/k}^{b/k}kK(ky)dy\equiv \int_{a/k}^{b/k}K_k(z)dz\equiv \int _\mathbb{R} K_k(z)dz$
My problem for the moment lies in proving uniform convergence, but I have the following bound for $\left | f(x)-f_k(x) \right |$:
$\left | f(x)-f_n(x) \right |=\left | f(x)-\int_{a/n}^{b/n}nf(x-z)K(nz)dz \right |=\left | f(x)-\int_{a}^{b}f(\frac{nx-y}{n})K(y)dy \right |=\left | \int_{a}^{b}f(x)K(y)dy - \int_{a}^{b}f(\frac{nx-y}{n})K(y)dy \right |=\left | \int_{a}^{b}[f(x)-f(\frac{nx-y}{n})]K(y)dy \right |\leq \int_{a}^{b}\left |[f(x)-f(\frac{nx-y}{n})]K(y) \right |dy \equiv \int_{a}^{b}\left |f(x)-f(\frac{nx-y}{n}) \right |K(y) dy$
At this point, it would be natural to think about trying to limit to $\left |f(x)-f(\frac{nx-y}{n}) \right |$ for any positive epsilon.
If this were to happen, we would have to
$\int_{a}^{b}\left |f(x)-f(\frac{nx-y}{n}) \right |K(y) dy <\varepsilon \int_{a}^{b}K(y)dy=\varepsilon$, since $K$ is a kernel-function
Here I am having problems because to reach this inequality I previously had to fix x and y, and the definition of uniform convergence does not fix these points.
If someone could guide me to be able to conclude this demonstration I would appreciate it and I do not know if I made understand the question that I have but if you have questions I could try to be more specific.