I'm doing exercise 7.4 of Berezanksky's Functional Analysis Vol I. I need to compute the weak limit, $l$, of a sequence of linear functionals $(l_n)_{n=1}^\infty$ defined on $C[0,1]$ with the sup-norm where
$$ l_n(f) = n\int_0^\frac{1}{n} f(x)\ \text{d}x $$
I have little intuition for what the weak limit of this is. I first thought that $l(f)=0$ for all $f$, then $l(f)=\|f\|_\infty$. I couldn't prove either of these. I've been told that $l(f)=f(0)$. This makes sense but I'm having trouble showing it. Here's my attempt:
\begin{align*} |l_n(f)-l(f)| &= \bigg|n\int_0^\frac{1}{n} f(x)\ \text{d}x-f(0)\bigg| \\ &= \bigg|n\int_0^\frac{1}{n} f(x)\ \text{d}x - n\int_0^\frac{1}{n} f(0)\ \text{d}x\bigg| \\ &= \bigg|n\int_0^\frac{1}{n} f(x)-f(0)\ \text{d}x\bigg| \\ &\leq n\int_0^\frac{1}{n}|f(x)-f(0)|\ \text{d}x \end{align*}
Now it feels like I need to make some observation about the final line to see that $|l_n(f)-l(f)|$ can be made arbitrarily small. I've drawn some diagrams and cannot see this. Is my direction even correct?
Hint: What does the continuity of $f$ mean at $x=0$?
Another way: From the MVT, we have that $$\int_0^{1/n} f=\frac{1}{n}f(c)$$ for some $c \in (0, 1/n)$. This means that $$n\int_0^{1/n} f=f(c)$$ Which gives us that $$n\int_0^{1/n} f \to f(0)$$ as desired.