Prove $\sup _{t \in [0,1]} |P(t)|$ is a norm Let $X$ be the vector space containing all polynomials with real coefficients. For every $P \in X$, define $N_1 (P)=\sup _{t \in [0,1]} |P(t)|$
I am having trouble proving the part when you show $||P||=0 \iff P=0$.
So $\|P\|= N_1(P)=  \sup _{t \in [0,1]} |P(t)|$
($\implies$) Let $\|P\|=0$, then $\sup |P(t)|=0$, which forces $P(t)=0$ for all $t$ in our region. Hence $P=0$
($\impliedby$) Let $P=0$, then $P(t)=0$ for all $t$. So $\|P\|=N_1 (0 ) = \sup |0| = 0$ as required.
My lecturer said that this is not enough. She also said that in the forward direction, I cant really conclude that $P=0$...
Can someone give a full proof of this please.
 A: Probably because she is considering that all your polynomials are polynomial functions defined in all of $\Bbb R$. You have shown that $P$ is zero on all of $[0,1]$ but not on all of $\Bbb R$. But a degree $n$ polynomial is determined by its value on $n+1$ distinct points. Certainly $[0,1]$ has more than $n+1$ points, for all $n \geq 0$, so it follows that $\sup_{t \in [0,1]}|P(t)| = 0$ implies $P(t) = 0$ for all $t \in \Bbb R$ (which is stronger than the $P(t) = 0$ for all $t \in [0,1]$).

($1$): About $P(t)$ being determined by its values on $n+1$ points: write $P(t) = a_0+a_1t+\cdots +a_nt^n$. If you know $y_0 = P(t_0), y_0 = P(t_1),\cdots, y_n = P(t_n)$, then you can solve the system $$\begin{cases} P(t_0) = y_0 \\ \vdots \\ P(t_n) = y_n\end{cases}$$for $a_0,\cdots, a_n$, and then you know $P(t)$ for all values of $t$.
($2$): About the analysis far from $[0,1]$ being important: you can define non polynomial functions ($C^\infty$, etc) that are non-zero outside $[0,1]$, but zero on $[0,1]$ - your proof would fail if we were working with arbitrary bounded functions instead of polynomials.
The point is that $P\big|_{[0,1]} \equiv 0 \implies P\equiv 0$ uses strongly the fact that $P$ is a polynomial, so we can apply fact $1$ above.
