Finding constant for orthonormality

Question

Assume the problem

\begin{align} f''(x) &= \lambda f(x) \\ f'(0) &= f(1) = 0 \\ \end{align}

with solution:

$$ \phi_n(x) = c \cdot\cos\left( \frac{(2n-1)\pi}{2}x\right), \quad n = 1,2,\dots $$

Now, if we want the solution to satisfy $\|f\|_2 = 1$, then:

$$ |c|\left(\int_0^1 \cos^2\left( \frac{(2n-1)\pi}{2}x\right) \right)^{\frac12} =1 \iff c = \pm \sqrt{2} $$

But I don't think $c = -\sqrt{2}$ $\,$ is an acceptable value if we want $\phi_n(x)$ to be the orthonormal basis of $L^2(0,1)$. Any suggestions why we reject the negative value?

Answer 1

I think you are confused. The constant $c$ could have all the values you want, but the functions belonging to $L^2(0,1)$ are functions in which the $x$ are in $(0,1)$ it is not the same.

Edit: In you could also have more than one ortonormal basis for a $L^p$ space. A nice exercise for you would be to see if both constant define the same orthomoral system due to the $cos$ properties.

Answer 2

There is nothing wrong with taking $c=-\sqrt{2}$. Indeed, if you take an orthonormal basis $(e_n)$ and replace some of the $e_n$'s with their negatives, they remain an orthonormal basis (it does not change $\langle e_n,e_m\rangle$ for $m\neq n$ since $(-1)\cdot 0=0$, and it does not change $\langle e_n,e_n\rangle$ since you get two factors of $-1$ which cancel out).