I have a linear functional $\phi_n:(C[0,1],\Vert{\cdot}\Vert_\infty) \to \mathbb{R}, n\in\mathbb{N}$ defined by $$\phi_n(x)=\int_0^1t^nx(t)dt$$
I have to calculate an upper bound for $\Vert\phi_n\Vert$ which is independent from $n$.
I thought about using the Cauchy–Schwarz inequality by saying that $$\vert\phi_n\vert=\vert\langle{t^n,x(t)\rangle}\vert\leq\Vert{t^n}\Vert\cdot\Vert{x(t)\Vert}$$ but I don't know how that would be independet from $n$. I'm new to this area of mathematics so I don't really have a clue which other approach I should use to tackle this problem.
I would be really cool if someone could give me some advice or push me in the right direction. Thanks!
I'd say you're almost there: just realize that your domain is $[0, 1]$, a compact set, and that $x$ is continuous.
That means that there are no asymptotes in the output space/codomain for $x$ (which is $\mathbb{R}$) for any value your input set/domain. This means that the value $||x(t)||_\infty$, the supremum of $x(t)$ calculted over $t \in [0, 1]$, does exist and indeed is a finite value.
Finally, $\forall t \in [0, 1], \forall n \in \mathbb{N}, 0 \le t^n \le t \le 1$. So $\forall n \in \mathbb{N}, ||t^n||_\infty = 1$ (for the given domain).
Conclusion: By the previous, and the Cauchy-Schwartz inequality, $||\phi_n||_\infty \le ||x(t)||_\infty$
Ie, the trick was to bound the $t^n$ term to remove the $n$ term from the desired bound.