= Solution
The <range of the Volterra integration operator> is $\{g\in H^1(0,1):g(0)=0\}$, using the representative of a <Sobolev space> element that is an <absolutely continuous function>. Indeed $Ku$ has <weak derivative> $u$ and zero initial trace; conversely the <fundamental theorem of calculus> reconstructs such a $g$ from its derivative. This range contains smooth compactly supported functions and is dense in $L^2(0,1)$.
The given step is in $L^2$, and its value at the single midpoint is immaterial. It cannot be the image of an $L^2$ function: such an image is continuous, whereas no continuous representative agrees almost everywhere with zero on the left half and one on the right half. Its <distributional derivative> is a <Dirac delta distribution>, not an $L^2$ function. Thus
$$
\boxed{f\in\overline{\mathcal R(K)}\setminus\mathcal R(K).}
$$
For a direct approximation, replace the jump by a linear ramp of width $1/n$ centered at $1/2$. Each ramp starts at zero and has an $L^2$ derivative, so lies in the range, while its squared $L^2$ error is $1/(12n)$. Its derivative <norm> is $\sqrt n$, illustrating unstable differentiation.
The supplied <SVD> gives $K(\sigma_j u_j)=\sigma_j^2v_j$, or more simply $K^\dagger(\sigma_jv_j)=u_j$. Since $\sigma_jv_j\to0$ but $\|u_j\|=1$, \b[the Moore–Penrose inverse is discontinuous]. Its domain is the dense, nonclosed range above, and there it is the <weak derivative>.
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