= Solution
The <L1 norm> properties of absolute homogeneity and the <triangle inequality> follow from the corresponding properties of the absolute value and the <integral>. If $\|f\|_1=0$, the nonnegative <continuous function> $|f|$ has zero integral. Were $|f(x_0)|>0$, continuity would make it bounded below by a positive number on a nontrivial interval, contradicting the zero integral. Thus $f=0$, proving positive definiteness.
Put $M=\|f\|_\infty$ and choose the continuous <cutoff function>
$$
\eta_n(x)=\min\{1,nx,n(1-x)\},\qquad g_n=\eta_n f.
$$
Then $g_n(0)=g_n(1)=0$, $\|g_n\|_\infty\leq M$, and $f-g_n$ is supported in the two endpoint intervals of total length $2/n$. Therefore
$$
\|f-g_n\|_1\leq\frac{2M}{n}\longrightarrow0.
$$
Now suppose $\int_0^1fg=0$ for every $g\in\mathcal S$. Apply this to the above $g_n$. By the <uniform norm> bound,
$$
0\leq\int_0^1f^2
=\int_0^1f(f-g_n)
\leq\|f\|_\infty\|f-g_n\|_1\longrightarrow0.
$$
Hence $\int_0^1f^2=0$, and continuity again gives \b[$f=0$.]
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