Verify that a function is a norm and show that there is a positive constant C such that the inequality satisfies Denote $P_n[0,1]$ to be the set of polynomials of degree less than or equal to $n$ on the interval $[0,1]$. For any $u\in C[0,1]$, define a projection $\mathbb{P}u\in P_n[0,1]$ as
\begin{equation}
\int_0^1(u-\mathbb{P}u)v\ dx=0, \forall v\in P_{n-1}[0,1], \text{ and  } u(1)=\mathbb{P}u(1).
\end{equation}


*

*Verify that 
\begin{equation}
\|u\|=\max_{1\leq k\leq n-1}\left\{\left|\int_0^1ux^kdx\right|,|u(1)|\right\}
\end{equation}
is a norm on $P_n[0,1]$.


I kind of understand how to do this one. One simple question, if I would like to prove that this function is a norm on $P_n[0,1]$, should I take $\|\mathbb{P}u\|$ show that $\|\mathbb{P}u\|\geq0$, $\|\alpha\mathbb{P}u\|=|\alpha|\|\mathbb{P}u\|$ and $\|\mathbb{P}u+\mathbb{P}v\|\leq\|\mathbb{P}u\|+\|\mathbb{P}v\|$ or use $u\in C[0,1]$ instead of $\mathbb{P}u$? How to show case $\|\cdot\|=0$?


*Show that for any $u\in C[0,1]$, there is a positive constant $C$ such that
\begin{equation}
\|\mathbb{P}u\|_{L^\infty}\leq C \|u\|_{L^\infty}
\end{equation}


I'm totally confused about this one.
 A: For 1 the function defined is a pseudo norm, it is not a norm. Let
me use $\rho(u)$ to denote the right hand side above.
In particular, there is a $p \neq 0$, $p \in P_n[0,1]$ such that 
$\rho(p) = 0$.
Let $p(x) = \sum_{k=0}^n \alpha_k x^k$. Then the following system of $n$ linear equations
in $n+1$ unknowns has a solution $\alpha \neq 0$:
$\int_0^1 p(x) x^m dx = \sum_{k=0}^n \alpha_k {1 \over m+k+1} = 0$, for $m=1,...,n-1$ and $p(1) = \sum_{k=0}^n \alpha_k = 0$.
For 2, it is straightforward to compute an explicit expression for ${\mathbb P}$ using the shifted Legendre polynomials $\tilde{P}_k$ (https://en.wikipedia.org/wiki/Legendre_polynomials#Shifted_Legendre_polynomials) as a basis for the functions $x \mapsto x^k$. In particular, the $\tilde{P}_k$
are orthogonal (with the inner product $\langle f, g \rangle =\int_0^1 f(x)g(x) dx$)
and $\tilde{P}_k(1) = 1$.
Since ${\mathbb P}u \in P_n[0,1]$, we can write ${\mathbb P} u = \sum_{k=0}^n \alpha_k(u)\tilde{P}_k$
where the $\alpha_k$ are linear functionals on $C[0,1]$.
We have
$\langle \tilde{P}_m, {\mathbb P}u \rangle = \alpha_m(u) \langle \tilde{P}_m, \tilde{P}_m \rangle$, and for $m=0,...,n-1$ we are given that $\langle \tilde{P}_m, {\mathbb P}u \rangle = \langle \tilde{P}_m, u \rangle$, hence
$\alpha_m(u) = { \langle \tilde{P}_m, u \rangle \over \langle \tilde{P}_m, \tilde{P}_m \rangle}$ for $m=0,...,n-1$.
For $m=n$ we are given that
$({\mathbb P} u)(1) = u(1) = \sum_{k=0}^n \alpha_k(u)\tilde{P}_k(1) = \sum_{k=0}^n \alpha_k(u)$, and so
$\alpha_n(u) = u(1) - \sum_{k=0}^{n-1} \alpha_k(u)$.
Since $|\alpha_k(u)| \le { 1\over \langle \tilde{P}_k, \tilde{P}_k \rangle } \|\tilde{P}_k \|_2 \|u\|_2 \le { 1\over \langle \tilde{P}_k, \tilde{P}_k \rangle } \|\tilde{P}_k \|_2 \|u\|_\infty $
for $k=0,...,n-1$ we see that these $\alpha_k$ are bounded linear
functionals with respect to $\|\cdot\|_\infty$, and hence so is $\alpha_n$. It follows that ${\mathbb P}$ is a bounded linear
functional and hence there is some $C$ such that
$\| \mathbb{P} u \|_\infty \le C \|u\|_\infty$.
Aside:
The shifted Legendre polynomials can be generated by applying Gram
Schmidt to the functions $x \mapsto x^k$ using the above inner product and scaling the results so that $\tilde{P}_k(1) = 1$.
