Finding spectral radius if $T^*T$ where $Tf=\int_0^t f(s)ds$

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Let $T:L_2[0,1]\rightarrow L_2[0,1]$ be the Volterra operator (def'n in title) and $T^*f=\int_t^1f(s)ds$ its adjoint. I want to calculate the spectral radius and operator norm. My plan is to find eigenfunctions/eigenvalues then use;

$$(1)r_A=\sup_{\lambda \in \sigma(A)}|\lambda|=||A||$$

but something goes wrong.

Since $T$ is compact, $T^*T$ is compact and it suffices to calculate the point spectrum. We have;

$$-\int_1^t\int_0^sf(s)dxds=\lambda f \iff -f(t)=\lambda f''(t)$$

So $f$ is a simple harmonic oscilator with equation $$f(t)=C_1\sin\left(\frac{t}{\lambda}\right)+C_0\cos\left(\frac{t}{\lambda}\right)$$ We see immediatly that $C_0=0$ since;

$$T^*Tf(0)=\int_t^1 0 ds=0=\lambda f(0)$$

So eigenfunctions of $T^*T$ are $C_1\sin\left(\frac{t}{\sqrt{\lambda}}\right)$ for $|\lambda|>0$.

But this dosn't make any sense since then by (1) $T^*T$ is then unbounded?

EDIT: Solution found here; Eigenfunctions of an integral operator

my mistake was to forget the additional ic $f(1)=0$.