Let $C[0, 1]$ denotes the complex vector space of all continuous functions $f : [0, 1] \to \mathbb C$, and that for any such function, $ ||f||_{\infty} := sup|f(x)| x \in [0,1]$ Prove that $(C[0, 1], ||·||_{\infty})$ is a normed space.
Also i want to show that $<·,·> : C[0, 1] \times C[0, 1] \to \mathbb C$ defined by $<f, g> := \int_{0}^{1} f(x) g(x) dx$ is an inner product.
COMMENT.-(1) You need $$\begin{cases}\|f\|_{\infty}=0\iff f=0\\\|\lambda f\|_{\infty}=|\lambda|\|f\|_{\infty}\\\|f+g\|\le \|f\|_{\infty}+\|g\|_{\infty}\end{cases}$$ This is deduced from the continuity with the basic fact that $f$ is bounded since $[0,1]$ is compact. Respect to the triangular inequality, you have for all $x\in[0,1$] the inequality with reals $|f(x)+g(x)|\le |f(x)|+|g(x)|$ from which you can finish.
(2) $<f, g> = \int_{0}^{1} f(x) g(x) dx$ is easily bilinear and because it is associated to the quadratic form $\int_{0}^{1} [f(x)]^2dx$ then it is definite positive since it is not canceled but for $f = 0$ (i.e. $f(x)=0$ for all $x\in[0,1]$). Besides you have $\|f\|=\sqrt{\int_{0}^{1} [f(x)]^2dx}$. Furthermore you can have the cosinus formed by two functions (the main motivation for inner products is some euclidean geometrization of the space) $$\cos(f,g)=\frac{\int_{0}^{1} f(x) g(x) dx}{\sqrt{\int_{0}^{1} [f(x)]^2dx\int_{0}^{1} [g(x)]^2dx}}$$