= Solution
For $n\geq1$, $\phi_n$ is <orthogonal> to every <polynomial> in $P_{n-1}$, so $(u,\phi_n)=(u-p_{n-1}^*,\phi_n)$. The <Cauchy-Schwarz inequality> gives
$$
|(u,\phi_n)|\leq E_{n-1}\int_0^1x^2|\phi_n|\,dx
\leq E_{n-1}\left(\int_0^1x^2\,dx\right)^{1/2}\|\phi_n\|_2
=\boxed{E_{n-1}/\sqrt3}.
$$
Using the tail identity from part (b) and shifting the index,
$$
\boxed{\|u-p_n\|_2^2\leq\frac13\sum_{r=n}^\infty E_r^2.}
$$
For the stated power law and $n>1$, the decreasing-function <integral test> gives
$$
\sum_{r=n}^\infty r^{-2a}\leq\int_{n-1}^\infty x^{-2a}\,dx=\frac{(n-1)^{1-2a}}{2a-1}.
$$
Consequently
$$
\boxed{\|u-p_n\|_2\leq\frac K{\sqrt3}\frac{(n-1)^{-a+1/2}}{\sqrt{2a-1}}.}
$$
The assumption $a>1$ is sufficient; this summation estimate itself only needs $a>1/2$.
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