= Solution
{c}
First verify that $V_0$ is a <vector space>. Sums and scalar multiples preserve <continuity> and the condition at zero. For every <partition of an interval> $D$, the <triangle inequality> and <absolute homogeneity of a norm> of the <absolute value> imply
$$
V(f+g,D)\leq V(f,D)+V(g,D),\qquad V(\lambda f,D)=|\lambda|V(f,D).
$$
Taking <suprema> yields
$$
V(f+g)\leq V(f)+V(g),\qquad V(\lambda f)=|\lambda|V(f).
$$
For $\lambda=0$ the second equality follows directly from $V(0)=0$. In particular, <finite variation> is preserved, so $V_0$ is closed under these operations.
Nonnegativity is immediate. For any $t\in[0,1]$, use the <partition of an interval> $(0,t,1)$, omitting repeated endpoints if necessary, to get
$$
|f(t)|=|f(t)-f(0)|\leq V(f).
$$
Thus $V(f)=0$ forces $f=0$, and conversely $f=0$ has zero <total variation of a function>. These observations prove that $\|f\|=V(f)$ is a <norm> on $V_0$.
Applying the same bound to $f_n-f$ gives
$$
\boxed{\|f_n-f\|_\infty\leq V(f_n-f)=\|f_n-f\|.}
$$
Therefore convergence in the <anchored variation norm> implies <uniform convergence> on $[0,1]$. The anchoring condition $f(0)=0$ removes the nonzero constant <functions> which otherwise have zero <total variation of a function>.
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