 Solution

The proposed norm is finite because every continuous function on $[0,1]$ is bounded and integrable. It is nonnegative. If $\Vert f\Vert =0$ and $f(x_0)\ne 0$ at some point, continuity gives a nontrivial interval within $[0,1]$ on which $|f|$ is bounded below by a positive constant, contradicting the zero integral. Thus $\Vert f\Vert =0$ implies $f=0$. The converse is immediate. Absolute homogeneity follows from $|cf(x)|=|c||f(x)|$, and the pointwise inequality $|f(x)+g(x)|\le |f(x)|+|g(x)|$ integrates to the triangle inequality. These establish all the norm axioms, so $V$ is a normed vector space.

To test the Cauchy sequence condition, compare the terms with indices $n$ and $2n$. Put $d_n(x)=\sin (nx)-\sin (2nx)$. The identities $2\sin (nx)\sin (2nx)=\cos (nx)-\cos (3nx)$ and $2{\sin }^2(nx)=1-\cos (2nx)$ giveBecause $|d_n(x)|\le 2$, we have $d_n(x{)}^2\le 2|d_n(x)|$. ThereforeIn particular, for all sufficiently large $n$, these arbitrarily late pairs have norm distance greater than $1/4$. This proves the sine sequence is not Cauchy in the integral norm. Thus the sequence is not Cauchy and does not converge to an element of $V$, since every norm convergent sequence is Cauchy. This direct separation argument does not rely on completeness of $V$.