Let $P([0,1])$ be the space of all the polinomial with complex entries, defined in $[0,1]$ . Show that $||f||_\infty=sup_{t\in [0,1]}|f(t)|$ and $||f||_1= \int_{0}^{1}|f(t)|dt$ are equivalent norms.
Profesor said this would be easy but I've lost half my day thinking on an approach, any hint will be appreciated!, Thanks!
The notion of equivalent norms that I have is the following: Two norms $||\cdot||_1,||\cdot||_2$ over a vector space $V$ are equivalent if there are constants $C_1,C_2>0$ such that for every $x\in V$, $$C_1||x||_1\leq ||x||_2\leq C_2||x||_1.$$
Is it the same definition that you have? If so, under this definition, we can show the norms $||\cdot ||_{\infty}$ and $||f||_1$ are not equivalent.
Suppose they are equivalent, and take $C_1,C_2>0$ such that for every polinomial $f\in P([0,1])$ we have $$C_1||f||_1 \leq ||f||_\infty\leq C_2||f||_1.\hspace{1cm}(*)$$
Take $n$ large enough such that $C_2<n+1$ (that is, $C_2\cdot \frac{1}{n+1}<1$), and consider the polynomial $f=x^n$. Then, we have
$$C_2\cdot ||f||_1=C_2\cdot \int_0^1 |x^n|\,dx=C_2\cdot \dfrac{1}{n+1}<1=||f||_\infty,$$contradicting the inequality $(*)$.
Probably your teacher was thinking about the space of all polynomials with bounded degree, which is a finite-dimensional vector space and it can be proven that for every such space all norms are equivalent. A proof of this can be found here, although I believe any book of functional analysis (and several of linear algebra) will also contain a proof of it.
