Equivalence of sup and Manhattan norm

Question

I'd like to show the following statement to complete my proof: for every $n \in \mathbb{N}$ there exists a constant $C_{n}$ such that for every polynomial $f(t)=a_{0}+a_{1} t+\ldots+a_{n} t^{n}$ of degree $\leq n$ with real coefficients holds $$ |a_{0}|+\ldots+|a_{n}| \leq C_{n} \sup _{t \in[0,1]}|f(t)| . $$ I then want to use the equivalence of norms in finite-dimensional vector spaces because I already got the inequality in the backward direction, which is obvious: $$ \sup_{t \in[0,1]}|f(t)| \leq|a_{0}|+\ldots+|a_{n}| . $$

Answer 1

The space $\mathcal{P}_{n}$ of polynomials of degree $\leq n$ is a vector space of dimension $n+1$. The two expressions $$ \left|a_{0}\right|+\ldots+\left|a_{n}\right|=\|f\|_{1} $$ and $$ \sup _{t \in[0,1]}|f(t)|=\|f\|_{C[0,1]} $$ are the norms in $\mathcal{P}_{n+1}$. These two norms are equivalent as the vector space is finite, from which the assertion follows.