Conclusions we can draw about $g(x)$ when $\int_{0}^{L}g(x) \sin \frac{n \pi x}{L}dx = 0$

Question

I have to answer the following question from my PDEs textbook:

Let $g(x)$ denote a function defined on the interval $(0,L)$. What conclusions can you draw about $g(x)$ in the following situations?

1. $\displaystyle \int_{0}^{L} g(x) \sin\frac{n \pi x}{L} dx = 0$, for $n = 1,2,\cdots$

2. $\displaystyle \int_{0}^{L} g(x) \cos \frac{n \pi x}{L} dx = 0$, for $n = 0,1, \cdots$

For Part 1, I said the following:

Since $\displaystyle b_{n} = \frac{2}{L}\int_{0}^{L} g(x) \sin \frac{n \pi x}{L} dx$, for $n = 1,2,\cdots$, we have that $\displaystyle \frac{L}{2}\cdot b_{n} = \int_{0}^{L}g(x) \sin \frac{n \pi x}{L}dx$. Since $L > 0$, if $\displaystyle \int_{0}^{L}g(x) sin \frac{n \pi x}{L}dx = 0$, this means that $b_{n}=0$ for $n = 1, 2, \cdots$. Therefore, $g(x)$ appears to be an even function.

On the other hand, perhaps $g(x)$ is any function (does $g(x)$ necessarily have to be even?) defined only on the "half-interval" $(0, L)$, then we can define $b_{n}=0$ for $n = 1,2,\cdots$, which would mean that $\displaystyle \int_{0}^{L}g(x) \sin \frac{n \pi x}{L}dx = 0$ for $n = 1, 2, \cdots$. Moreover, we can compute the Fourier Coefficients $a_{n}$ for $g(x)$ using $\displaystyle a_{n} = \frac{2}{L}\int_{0}^{L} g(x) \cos \frac{n \pi x}{L}dx$. In this case, the Fourier series for $g(x)$ will have only cosine terms in it and must represent an even function (This is what is known as the half-range Fourier cosine series for $g(x)$).

This is equivalent to extending $g(x)$ to the interval $(-L,0)$ as an even function and then computing the Fourier coefficients in the usual way using $\displaystyle a_{n} = \frac{2}{L}\int_{0}^{L} g(x) \cos \frac{n \pi x}{L}dx$.

I said something analagous for Part 2.

My questions are:

1. Is that all there is to say here about $g(x)$? Are there any other conclusions that can be drawn that I might have missed?

2. Does $g(x)$ necessarily have to be even in Part 1? Could it be, because of what I said in the last paragraph, an odd function or a function that is neither odd nor even that we can extend as an even function on the interval $(-L,0)$?

Answer 1

My first reaction is that both conditions imply that $g(x)=0$.

I'll give a sketch of my reasoning for condition 1:

For ease of typing I'll assume $L=\pi$.

• First of all, it doesn't mean anything to state whether $g(x)$ is even or odd, since it isn't even defined for negative values of $x$. Whatever this function is, we can extend the domain to include negative values however we like, and this won't change any of the properties of $g(x)$ on the original domain.
• Any function* defined on $(0,\pi)$ can be represented as a Fourier Sine series of the form $\Sigma_{n=0}^\infty b_n\sin(nx)$ where $b_n=\frac{2}{\pi}\int_0^\pi g(x)\sin(nx)dx$ by extending $g(x)$ to be an odd function on $(-\pi,\pi)$ and force it to be a series with just sines in it.
• Since all the coefficients end up being zero by assumption, the whole function must be uniformly zero.

$^*$By any function I of course mean any function that meets whatever conditions are necessary for the series to exist