set, $I^{\prime },$ of the same type. This set may be described by$$I^{\prime }=I^{\prime }\left( p_1,q_1,p_2,q_2\right) :=\left\{
\left( p^{\prime },q^{\prime }\right) /\;\exists \left( p,q\right)
\in I\;s.t.\;\frac 1p+\frac 1{p^{\prime }}=\frac 1q+\frac
1{q^{\prime }}=1\right\}$$ and it is not difficult to check that $I^{\prime }\left( p_1,q_1,p_2,q_2\right) =I\left( p_1^{\prime
},q_1^{\prime },p_2^{\prime },q_2^{\prime }\right) ,$ where $p_1^{\prime },q_1^{\prime },p_2^{\prime }$ and $q_2^{\prime }$ are defined by $\frac 1{p_1}+\frac 1{p_1^{\prime }}=\frac 1{q_1}+\frac
1{q_1^{\prime }}=\frac 1{p_2}+\frac 1{p_2^{\prime }}=\frac
1{q_2}+\frac 1{q_2^{\prime }}=1.$

Moreover, by Hölder’s inequality, it follows that one has $$\label{dual}\int_0^t\int_{{\cal O}}u\left( s,x\right) v\left(
s,x\right) dxds\le \left\| u\right\| _{I;t}\left\| v\right\|
^{I^{\prime };t},$$ for any $u\in L_{I;t}$ and $v\in L^{I^{\prime };t}.$ This inequality shows that the scalar product of $L^2\left( \left[ 0,t\right] \times {\cal O}\right) $ extends to a duality relation for the spaces $L_{I;t}$ and $L^{I^{\prime };t}.$

Now let us recall that the Sobolev inequality states that$$\label{Sobolev}
\left\| u\right\| _{2^{*}}\le c_S\left\| \nabla u\right\| _2,$$ for each $u\in H_0^1\left( {\cal O}\right) ,$ where $c_S>0$ is a constant that depends on the dimension and $2^{*}=\frac{2d}{d-2}$ if $d>2,$ while $2^{*}$ may be any number in $]2,\infty [$ if $d=2$ and $2^{*}=\infty $ if $d=1.$ Therefore one has$$\left\| u\right\| _{2^{*},2;t}\le c_S\left\| \nabla u\right\|
_{2,2;t},$$ for each $t\ge 0$ and each $u\in 