Show that the only functions that satisfy $f''(\xi) = \lambda f(\xi)$, $f'(0) = f(1) = 0$, $\|f\|_2 = 1$ for some $\lambda \in \mathbb{R}$ are the functions:
$$ \phi_n(\xi) = \sqrt{2}\cos\left(\frac{(2n-1)\pi}{2}\xi\right), \quad n =1,2,\dots $$
thus making them an orthonormal basis of $L^2(0,1)$
It's obvious to me why this family of functions satisfy the conditions $f'(0) = f(1) = 0$ and $\|f\|_2 = 1$.
What confuses me is the constant $\lambda$. I fail to see why for every $\lambda$ the solutions are $\phi_n$. Won't positive $\lambda$'s give exponential functions as solutions?
It'd be very helpful if someone could elaborate on how this family of functions is yielded from the original equation $f''(\xi) = \lambda f(\xi)$.
There is something fishy here. if $f(x)= \cos(ax)$ then
$$f^{\prime\prime}(x) = -a^2\cos(ax)$$ so $\lambda$ has to appear in the solution (which is not true for the function you have written down). And yes, this works only for negative $\lambda$.
Just write down what you think is the solution for positive lambda and calculate whehter it satisfies the equation (including initial condition). Then note that there are uniqueness theorems for ordinary differential equations with initial conditions. In other words, if your suggested solution is different from the proposed solution and if it solves the equation and you know it's unique, then the statement from the exercise is false...