Consider the operator $T:L^1([0,1]) \longrightarrow L^1([0,1])$ given by $$Tu(x) = x^2u(x).$$ $T$ is linear and bounded. I'm asked the point spectrum of this operator and the eigenspaces corresponding to the eigenvalues of $T$.
The point spectrum of $T$ is $$\sigma_p(T) = \{\lambda: \text{ker} (\lambda I-T)\neq \{0\}\}. $$ Let $\lambda \in \sigma_p(T) $. Then $$\lambda u(x) - x^2u(x)=u(x)(\lambda-x^2)=0 ~~\forall x \in [0,1]~~~(*) $$ for some $u \neq 0$ in $L^1$, which is impossible because it seems to me that $(*) $ is true only when $u$ is zero in $L^1$. Then the point spectrum should be empty and there are no eigenvalues. So where am I wrong?
You're correct. First note that every spectral value lies inside $[0,1].$ If $(\lambda-x^2)u=0,$ then $u=0$ on $\{\sqrt{\lambda}\}^c$ and so $u=0$ a.e.