Showing uniform convergence of a 'delta-like' function Suppose we have a sequence of continuous functions $(f_k)_{k\in \mathbb{N}}$ on $[-1,1]$ such that for each $k$ we have $f_k \geq 0$ and for any $\epsilon>0$ we have that $$\lim_{k\rightarrow \infty} \int_{\epsilon \leq |x|\leq 1} f_k(x)dx = 0 \ \ \ \ \ \ (*)$$
and $\int_{-1}^{1} f_k(x)dx = 1$.  Now let $g$ be a continous function on $[-\frac{1}{2},\frac{1}{2}]$ that vanishes at the end points. 
I am trying to show that $$G_k(x) = \int_{-\frac{1}{2}}^{\frac{1}{2}}f_k(x-y)g(y)dy$$
converges uniformly to  $g$ on $[-1/2, 1/2]$.
My attempt:
Since $1 =\int_{-1}^{1} f_k(x)dx = \int_{-1}^{-\epsilon}f_k(x)dx + \int_{-\epsilon}^{\epsilon}f_k(x)dx+\int_{\epsilon}^{1} f_k(x)dx$, sending $k\rightarrow \infty$ gives that $\int_{-\epsilon}^{\epsilon} f_k(x)dx = 1$.
Now fix $x\in[-1/2,1/2]$. 
$$|G_k(x)-g(x)| = \bigg|\int_{-\frac{1}{2}}^{\frac{1}{2}}f_k(x-y)g(y)dy -g(x)\bigg|$$
$$\leq \bigg|\int_{-\frac{1}{2}}^{\frac{1}{2}}f_k(x-y)g(y)dy\bigg|+|g(x)|$$
$$\leq \sup_{x\in[-1/2,1/2]}|g(x)| \bigg(\int_{-1/2}^{1/2}f_k(x-y)dy+1\bigg)$$
$$ = \sup_{x\in[-1/2,1/2]}|g(x)| \bigg(\int_{x-1/2}^{x+1/2}f_k(z)dz+1\bigg)$$
This is where I am stuck and can no longer proceed. I was hoping to send $k\rightarrow \infty$ and then get cancellation of the two terms in the bracket above but there is no minus sign, and moreover the integral bounds are not symmetric so we cannot use $(*)$ yet... 
How should I proceed?
 A: Things become simpler if we extend $g$ and the $f_k$ to all of $\mathbb{R}$ by setting $g(x) = 0$ for $\lvert x\rvert > \frac{1}{2}$ and $f_k(x) = 0$ for $\lvert x\rvert > 1$. Since $g(-1/2) = g(1/2) = 0$ by assumption, the extended $g$ is is continuous. In fact, it is even uniformly continuous, since it vanishes outside a compact set (the interval $\bigl[-\frac{1}{2},\frac{1}{2}\bigr]$). The extended $f_k$ may not be continuous, but they have at worst two points of discontinuity and each is at worst a jump discontinuity, so they are integrable over $\mathbb{R}$. Since $g$ vanishes outside $\bigl[-\frac{1}{2},\frac{1}{2}\bigr]$, we have
$$G_k(x) = \int_{-\frac{1}{2}}^{\frac{1}{2}} f_k(x-y)g(y)\,dy = \int_{-\infty}^{+\infty} f_k(x-y)g(y)\,dy.$$
Now we make a change of variables, $y = x-u$, to write
$$G_k(x) = \int_{-\infty}^{+\infty} f_k(u)g(x-u)\,du.$$
Since the integral of each $f_k$ is $1$, we can write
$$g(x) = \int_{-\infty}^{+\infty} f_k(u)g(x)\,du.$$
Then for an arbitrary $\varepsilon > 0$ we can estimate $\lvert G_k(x) - g(x)\rvert$ by
\begin{align}
\lvert G_k(x) - g(x)\rvert
&= \Biggl\lvert \int_{-\infty}^{+\infty} f_k(u)\bigl(g(x-u) - g(x)\bigr)\,du\Biggr\rvert \\
&\leqslant \int_{-\infty}^{+\infty} f_k(u)\lvert g(x-u) - g(x)\rvert\,du \\
&= \underbrace{\int_{-\varepsilon}^{+\varepsilon} f_k(u)\lvert g(x-u) - g(x)\rvert\,du}_{A_k(\varepsilon)} + \underbrace{\int_{\varepsilon \leqslant \lvert u\rvert} f_k(u)\lvert g(x-u) - g(x)\rvert\,du}_{B_k(\varepsilon)}.
\end{align}
In $A_k(\varepsilon)$, $\lvert u\rvert$ is small, hence $\lvert g(x-u) - g(x)\rvert$ is small, independently of $x$. So $A_k(\varepsilon)$ is small for $\varepsilon$ sufficiently small. In $B_k(\varepsilon)$, $\lvert u\rvert$ is not so small, but $\lvert g(x-u) - g(x)\rvert$ is still bounded - independent of $x$ - and we hypothesis that for large enough $k$, $f_k$ is (on average) small there. So for large $k$, $B_k(\varepsilon)$ is small. So altogether, for small $\varepsilon$ and large $k$, both parts are small, hence $A_k(\varepsilon) + B_k(\varepsilon)$ is also small.
Let's convert the informal observations above into a rigorous argument.
Suppose we are given an arbitrary $\delta > 0$. By the uniform continuity of $g$, there is an $\varepsilon > 0$ such that
$$\lvert x-y\rvert \leqslant \varepsilon \implies \bigl\lvert g(x) - g(y)\rvert \leqslant \frac{\delta}{2}.$$
Then we have
$$A_k(\varepsilon) \leqslant \int_{-\varepsilon}^{+\varepsilon} f_k(u)\cdot \frac{\delta}{2}\,du = \frac{\delta}{2}\int_{-\varepsilon}^{+\varepsilon} f_k(u)\,du \leqslant \frac{\delta}{2} \int_{-\infty}^{+\infty} f_k(u)\,du = \frac{\delta}{2}$$
for all $k$ and all $x\in \mathbb{R}$.
Since $g$ is continuous, and vanishes for $\lvert x\rvert > \frac{1}{2}$, there is an $M\in (0,+\infty)$ with $\lvert g(x)\rvert \leqslant M$ for all $x\in\mathbb{R}$. Then $\lvert g(y) - g(x)\rvert \leqslant \lvert g(y)\rvert + \lvert g(x)\rvert \leqslant 2M$, and hence
$$B_k(\varepsilon) \leqslant 2M \int_{\varepsilon \leqslant \lvert u\rvert} f_k(u)\,du.$$
By $(\ast)$, there is a $k_0$ (depending on $\varepsilon$) such that
$$ \int_{\varepsilon \leqslant \lvert u\rvert} f_k(u)\,du = \int_{\varepsilon \leqslant \lvert u\rvert \leqslant 1} f_k(u)\,du \leqslant \frac{\delta}{4M}$$
for all $k \geqslant k_0$.
Hence
$$\lvert G_k(x) - g(x)\rvert \leqslant A_k(\varepsilon) + B_k(\varepsilon) \leqslant \frac{\delta}{2} + 2M \frac{\delta}{4M} = \delta$$
for all $k \geqslant k_0$ and all $x\in \mathbb{R}$. This is the uniform convergence of the $G_k$ to $g$.
A: Not an answer, but perhaps we can find something useful by showing $G_k(0)$ converges to $g(0)$.
When $x=0$, we have $$\bigg\lvert\int_{-\frac{1}{2}}^{\frac{1}{2}}f_k(-y)g(y)dy -g(0)\bigg\rvert=\bigg\lvert\int_{-\frac{1}{2}}^{\frac{1}{2}}f_k(-y)g(y)dy -\int_{-1}^{1}g(0)f_k(z)dz\bigg\rvert$$
$$=\bigg\lvert\int_{-\frac{1}{2}}^{\frac{1}{2}}f_k(-y)g(y)dy -\int_{-\frac{1}{2}}^{\frac{1}{2}}g(0)f_k(z)dz-\int_{\frac{1}{2}\le\lvert z\rvert\le 1}g(0)f_k(z)dz\bigg\rvert$$
$$\le \bigg\lvert\int_{-\frac{1}{2}}^{\frac{1}{2}}f_k(-y)g(y)dy -\int_{-\frac{1}{2}}^{\frac{1}{2}}g(0)f_k(z)dz\bigg\rvert+\bigg\lvert\int_{\frac{1}{2}\le\lvert z\rvert\le 1}g(0)f_k(z)dz\bigg\rvert.$$
The latter term in the last line can be squeezed as small as possible. Let's not consider it now. By some change of variable, the former term becomes
$$\bigg\lvert\int_{-\frac{1}{2}}^{\frac{1}{2}}f_k(-y)g(y)dy -\int_{-\frac{1}{2}}^{\frac{1}{2}}g(0)f_k(-z)dz\bigg\rvert=\bigg\lvert\int_{-\frac{1}{2}}^{\frac{1}{2}}(g(y)-g(0))f_k(-y)dy\bigg\rvert$$
$$\le\bigg\lvert\int_{-\delta}^{\delta}(g(y)-g(0))f_k(-y)dy\bigg\rvert+\bigg\lvert\int_{\delta\le\lvert y\rvert\le\frac{1}{2}}(g(y)-g(0))f_k(-y)dy\bigg\rvert$$ where $\delta\gt0$ is yet to be chosen.
Now by Hölder inequality, the latter term is bounded by $$\sup_{\delta\le\lvert z\rvert\le\frac{1}{2}}\lvert g(z)-g(0)\rvert\int_{\delta\le\lvert y\rvert\le\frac{1}{2}}\lvert f_k(-y)\rvert dy.$$ The first factor is finite and the second factor goes to $0$ by assumption on the sequence $(f_k)$. So we only have one last term to bound: $$\bigg\lvert\int_{-\delta}^{\delta}(g(y)-g(0))f_k(-y)dy\bigg\rvert.$$ This time we bound $g(y)-g(0)$ by continuity of $g$ at $0$. For any $\varepsilon\gt 0$, choose $\delta\gt0$ so that when $y\in[-\delta,\delta]$, we have $\lvert g(y)-g(0)\rvert\le\varepsilon$. Invoke Hölder inequality again to get a bound for the last term:
$$\bigg\lvert\int_{-\delta}^{\delta}(g(y)-g(0))f_k(-y)dy\bigg\rvert\le\sup_{z\in[-\delta,\delta]}\lvert g(z)-g(0)\rvert\int_{-\delta}^{\delta}\lvert f_k(-y)\rvert dy.$$ The second factor is finite (less than or equal to $1$ by assumption on $(f_k)$) and the first factor is bounded by $\varepsilon$.
