Consider the following transport equation: $$y_t(t,x)=y_x(t,x), \ \ (t,x)\in(0,\infty)\times(-\infty,\infty) $$ with initial state $y(0,x)=y_0(x)\in L^2(-\infty,\infty)$.
By the method of characteristics, we can prove that the solution $y$ is given explicitly by $$y(t,x)=y_0(x+t).$$ Clearly, the solution have the same regularity of the initial state.i.e $y\in L^2((0,\infty)\times(\infty,\infty)).$ Now, by using the semigroup approach, we can prove that the derevative operator is skew-adjoint, and hence, it generates a group of linear operators. Furthermore, from the semigroup theory, we have the regularity: $$y\in C(0,\infty;L^2(-\infty,\infty)).$$ My question: How $y$ can be continuous knowing that from the exact formula it doesn't?.
The assertion from semigroup theory is that it is continuous in time, namely with domain $(0,\infty)$, with values in $L^2((-\infty,\infty))$. In other words, for each $t \in (0,\infty)$ and $\epsilon >0$ there exists $\delta >0$ such that $$ |t-s| < \delta \Rightarrow \Vert y(t,\cdot) - y(s,\cdot) \Vert_{L^2} < \epsilon. $$ This is not the same thing as asserting that the solution $y$ is continuous in space and time, or even that $y(t,\cdot)$ is continuous in space for some $t$.
Now, from the solution generated by characteristics we actually have the same conclusion since $$ \Vert y(t,\cdot) - y(s,\cdot) \Vert_{L^2}^2 = \int_{-\infty}^\infty |y(t,x) - y(s,x)|^2 dx = \int_{-\infty}^\infty |y_0(x + t) - y_0(x+s)|^2 dx \\ = \int_{-\infty}^\infty |y_0(x+t-s) - y_0(x)|^2 dx, $$ and so $$ \Vert y(t,\cdot) - y(s,\cdot) \Vert_{L^2} = \Vert y_0(\cdot + t-s) - y_0 \Vert_{L^2}. $$ Continuity in time then follows from continuity of translations in $L^2$, namely $$ \lim_{h \to 0 } \Vert f(\cdot +h) - f \Vert_{L^2} =0 $$ for $f \in L^2$.