Let $V = C(0, 1)$ on which we have defined the inner product $\langle f, g\rangle =\int_0^1 f (x)g(x)dx$, and let $W = \mathbb{R}\{e^x\} \subset V$ . Find an infinite set of elements of $W^⊥$.
The set $A=\{ae^{kx}|k\in \mathbb{R}\}$ is Linearly independent, so is the set $B=\{e^x,e^{2x}\}$, Using the Gram-Schmidt we find the orthogonal set $B'=\{e^x, e^{2x}-\frac{2(e^3-1)}{3(e^2-1)}e^x\}$. Is it enough to say the set $\{k(e^{2x}-\frac{2(e^3-1)}{3(e^2-1)}e^x)|k\in\mathbb{R}\}$ is a set with infinite elements of $W^⊥$?
If we only need an infinite sets of elements of $W^{\perp}$, we can try with $g(x)=ax+b.$ Then, $$\langle e^x, ax+b\rangle =\int_0^1 e^x(ax+b)\;dx=\ldots=a+(e-1)b.$$ Choosing $a=1-e$, $b=1$ we get the inifinite set of vectors: $$\{\lambda[(1-e)x+1]:\lambda\in\mathbb{R}\}\subset W^{\perp}.$$