Find eigenvalues for $T(f) = \int_{-\infty}^x tf(t)dt$ Let $V$ be the linear space of all functions continuous on $(-\infty, \infty)$ and such that that the integral $\int_{-\infty}^x tf(t)\,dt$ exists. If $f \in V$, let $g=T(f)$ be defined as $g(x) = \int_{-\infty}^x tf(t)\,dt$. Prove that every $\lambda < 0$ is an eigenvector and determine the eigenfunctions corresponding to $\lambda$.
We know that $T(f) = \int_{-\infty}^x tf(t)\,dt = \lambda f(x)$. (Not sure if this is right so far.) So know what do I do? 

You wrote "We know that $T(f) = \int_{-\infty}^x tf(t)\,dt = \lambda f(x)$."
You could say we know that $T(f) = \int_{-\infty}^x tf(t)\,dt$ for every $f$ in the domain, and we know that if $f$ is an eigenfunction with eigenvalue $\lambda$ then the second equality holds.  That doesn't indicate that any such function exists.
Differentiating both sides of the equation
$$
\int_{-\infty}^x tf(t)\,dt=\lambda f(x)
$$
we get
$$
xf(x) = \lambda f'(x)
$$
so that
$$
\frac x \lambda\,dx = \frac{df} f.
$$
Hence
$$
\frac{x^2}{2\lambda} + C = \log |f(x)|.
$$
$$
Ae^{x^2/(2\lambda)} = f(x). \tag 1
$$
This holds if $f$ is an eigenfunction with eigenvalue $\lambda$, so we've shown that if any such eigenfunction exists, then it looks like $(1)$.
Now ask yourself: For which values of $\lambda$ is it the case that $\displaystyle\int_{-\infty}^x t e^{t^2/(2\lambda)} \, dt$ exists for every $x$?
