How can we describe $\ker (T)$ and $\ker( T^*)$ in this case? Suppose we have a Hilbert space $H=L^2([0,\infty))$. Also, assume $Tf(x)=f(x+1)$ ($f$ is in $H$). Then how should we determine $T^*$ based on functional analysis knowledge? Also, how can we describe $\ker(T)$ and $\ker( T^*)$? 
 A: Recall the inner product property defining the adjoint operator : 
$$
\forall f,g \in L^2([0,\infty[), \quad \langle Tf, g \rangle = \langle f, T^*g \rangle. 
$$
Let $f,g \in L^2([0,\infty[)$. Then 
$$
\langle Tf,g \rangle = \int_0^{\infty} f(x+1) g(x) \, dx. 
$$
You want to find $h = h(g) \in L^2([0,\infty[)$ which satisfies the following for all $f \in L^2([0,\infty[)$ : 
$$
 \langle f, h \rangle = \int_0^{\infty} f(x) h(x)\, dx = \int_0^{\infty} f(x+1) g(x) \, dx. 
$$
Let
$$
h(x) \overset{def}=
\begin{cases}
g(x-1) & \text{ if } x \ge 1  \\
0      & \text{ if } 0 \le x \le 1. 
\end{cases}
$$
It is clear that $h \in L^2([0,\infty[)$. By a simple change of variables,
$$
\int_0^{\infty} f(x+1) g(x) \, dx = \int_0^{\infty} f(x+1) h(x+1) \, dx = \int_0^{\infty} f(x) h(x) \, dx = \langle f,h \rangle.
$$
You should think about what it means for $T$ and $T^*$ to send an element of $L^2([0,\infty[)$ to zero, the answer should become obvious. 
Hope that helps,
A: As for ${\rm ker }(T)$: observe that for the purpose of determining $Tf$ nobody is paying any attention to the values of $f$ in the interval $(0,1)$. So, it is an easy guess that ${\rm ker }(T)=\{f\in L^2(0,\infty):f_{(1,\infty)}\equiv 0\}\simeq L^2(0,1)$; this implies in particular that ${\rm ran}(T^*)$ is the orthogonal subspace $\{f\in L^2(0,\infty):f_{(0,1)}\equiv 0\}\simeq L^2(1,\infty)$. On the other hand, each $g\in L^2(0,\infty)$ can be seen as the image under $T$ of some $f$ - more precisely, of $f$ defined by $f(x+1)=g(x)$: the value of $f$ in $(0,1)$ is irrelevant, so you can take e.g. $f(x)=0$ for any $x\in (0,1)$. Accordingly, the orthogonal subspace ${\rm ker }(T^*)$ to ${\rm ran }(T)$ is trivial.
Finally, follow Daniel Fischer's suggestion in the comments and a simple variable substitution to find $T^*$. If you are unsure about your own solution, use the knowledge about ${\rm ker}(T^*)$ and ${\rm ran}(T)$ to double-check your answer.
