Simple Sequence of Functions I've been trying to find a sequence of continuous functions that converge to the function $$h = \begin{cases} 1 \quad x \in [0,\frac{1}{2}] \\
-1 \quad x \in (\frac{1}{2},1] \end{cases}$$  
such that for each $n$, $$\int_{0}^{\frac{1}{2}} f_n \, dx - \int_{\frac{1}{2}}^{1} f_n \, dx = 1.$$ 
This has lead me to consider the sequence of functions whose "connecting strip" gets steeper and steeper. Though, the complication arises when trying to make sure each $f_n$ satisfies the above. To simplify the matter, I've defined $f_n$ to be $1$ on $[0,\frac{1}{2}]$ for all $n$. Thus, does it suffice to solve, given some $a \in \mathbb{N}$, $$\int_{\frac{1}{2}}^{x} 1-a(t-\frac{1}{2}) \, dt - \frac{(\frac{1}{2} -x)(1-a(x-\frac{1}{2}))}{2} = -\frac{1}{2}$$
for $x$? The equation above comes from the fact that my "connecting line" has value $1$ at $\frac{1}{2}$, and I need to see precisely how far "down" the connecting line need go to satisfy the first equation. Clearly this will have two solutions as one could consider the solution where the "connecting" line does not cross the $x$-axis, but this won't converge to $h$.  
As some context, this is part of a question showing that closed and convex does not imply existence of a unique element with minimal norm in an arbitrary Banach Space, in this case $C([0,1])$ with the sup norm. (Is it correct to say that $h$ is the function with smallest norm in the space of all functions with at most $1$ discontinuity?)   
I'm trying to show that no element of $C$ satisfying the first equation has smallest norm, since $h$ has the smallest norm in a space containing $C$, and indeed $f_n$ can converge to $h$ point-wise and so the norms may get progressively smaller.
 A: The thing you want is a sequence of continuous functions that converge pointwise to zero but which have constant non-zero integral over some set $[a,b]$.
This will give you the result, because if you look at
$$h_n(x)=\begin{cases}1 \qquad x\in [0,\frac{1}{2}]\\ -1 \quad x \in (\frac{1}{2}+\frac{1}{n},1]\\\text{else linear interpolation between lines}\end{cases}$$
the expression $\int_0^{\frac{1}{2}} h_n - \int_\frac{1}{2}^1 h_n$ is bounded from below by 0 and from above by 1. So if we add this function that converges pointwise to zero but has constant integral to this expression with the right prefactors we can make the integral stay at any value at all while retaining pointwise convergence to the function we want.
To construct the desired thing lets look at what triangles do, the integral over
$$f(x)=\begin{cases}1-\vert x\vert \quad& x\in[-1,1] \\0 & \text{else}\end{cases}$$
is $1$ and the function is continuous. By squeezing the triangle together via $f_n(x) = 2^{n}f(2^{n} x)$ the integral is unchanged and the set where $f$ is nonzero becomes $[-2^{-n},2^{-n}]$. So $f_n$ converges pointwise to zero everywhere except for the point $0$. To get pointwise convergence everywhere we'll shift the point of the triangle around in a nice way.
By doing a translation $g_n(x) = f_n(x-\sum_{k=0}^{n-1}2^{-k})$ the peak will always be at $\sum_{k=0}^{n-1}2^{-k}=2-2^{-n+1}$ whereas the function has width $2^{-n+1}$. The support is then
$$\text{supp}(g_n)= [2-2^{-n+1}-2^{-n},2-2^{-n+1}+2^{-n}]=[2-3\cdot2^{-n},2-2^{-n}] \subset [0,2]$$
The intersection of the supports of $g_n$ and $g_{n+2}$ ist: $\{x \ \vert \ 2-3\cdot2^{-n-2}≤ x ≤ 2-2^{-n}\}$. But since $3\cdot2^{-n-2} < 2^{-n}$ this set is empty. Since the centre of the peak is moving ever to the right this means that every point will evaluate to something non-zero for at most 2 values of $n$ and the functions converge to 0 pointwise.
With the sequence in our pocket its easy to do a rescaling to get a new sequence on $[0,\frac{1}{2}]$ with integral $1$, add this sequence to $h_n$ with the prefactator $1-\int_0^{\frac{1}{2}}h_n + \int_\frac{1}{2}^1 h_n$ so that the desired integral expression is always $1$ (since these prefactors are bounded the new sequence will still converge to $h$).
