Prove all eigenvalues on operator over continuous functions are strictly negative. $G:C_0^0[0,1] \to C_0^0[0,1]$ is the operator defined as $$G(f)(x) = \int_0^1 \chi_{[0, x]}(t)(x-t)f(t) dt - x \int_0^1 (1-t)f(t)dt$$where $\chi_{[0,x]}$ is the characteristic function of $[0,x]$ and $C_0^0[0,1]$ are continuous functions over $[0,1]$ which are $0$ at the endpoints. Prove that any eigenvalues of $G$ are negative. We are using the inner product $f \cdot g = \int_0^1 f(x)g(x) dx$ and the $L_2$ norm that comes from that inner product.
There is a hint that says to try integration by parts. I have try that on anything I could think to try it with, but never got anywhere.
 A: It's just a matter of transformations.
Let $F(x)=\int_0^xf(t)dt$. We have $F(0) = 0$ and $F' = f$.
Via integration by parts, we have $$\int_0^x (x - t)f(t)dt = (x - t)F(t)\vert_0^x + \int_0^xF(t)dt = \int_0^x F(t)dt.$$ Therefore:
$$G(f)(x) = \int_0^x F(t) dt - x\int_0^1F(t)dt.$$
We then calculate the inner product:
$$\langle f, G(f)\rangle = \int_0^1 f(x)G(f)(x)dx = \int_0^1f(x)\left(\int_0^xF(t) dt\right)dx - \int_0^1f(x)\left(x\int_0^1F(t)dt\right)dx.$$
The first term is:
$$\left(F(x)\int_0^xF(t)dt\right)\vert_0^1 - \int_0^1F(x)^2dx = F(1)\int_0^1F(t)dt - \int_0^1F(x)^2dx.$$
For the second term, we have:
$$\int_0^1xf(x)dx = xF(x)\vert_0^1 - \int_0^1F(x)dx = F(1) - \int_0^1F(x)dx.$$
Putting together, we obtain:
$$\langle f, G(f)\rangle = -\int_0^1F(x)^2dx - \left(\int_0^1F(x)dx\right)^2 \leq 0,$$
with equality if and only if $F(x) = 0$ for all $x \in [0, 1]$, i.e. $f = 0$.
This shows that any eigenvalue is strictly negative. Namely, if $f\neq 0$ is an eigenfunction with eigenvalue $\lambda$, then we have $$\lambda \langle f, f\rangle = \langle f, \lambda f\rangle = \langle f, G(f)\rangle < 0.$$

... So this account of yours is used only for asking questions about eigenvalues?
