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Let $Tx(t)=x'(t)$ in $L_{p}(0,1)$, $1\leq p<\infty$. $\textrm{Dom}(T)=\left\{ x\in L_{p}(0,1):x\in AC[0,1],x'\in L_{p}(0,1)\right\}$.

Since the Lebesgue measures of $\{0\}$ and $\{1\}$ are zero, $x\in\textrm{Dom}(T)\Rightarrow x\in L_{p}[0,1]$ and $x'\in L_{p}[0,1]$.

$\textrm{ker}T=\left\{ x\in\textrm{Dom}(T):Tx=0\right\} =\left\{ x\in\textrm{Dom}(T):x'(t)=0\textrm{ for a.e. }t\in[0,1]\right\} $

Suppose $[x]\in\textrm{Dom}(T)/\textrm{ker}T\Rightarrow[x]=\left\{ x+y:y\in\textrm{ker}T\right\} $.

Let $\tilde{T}:\textrm{Dom}(T)/\textrm{ker}T\rightarrow L_{p}(0,1)$ such that $\tilde{T}[x]=x'(t),\: x\in[x]$.

$\textrm{ker}\tilde{T}=[0]=\textrm{ker}T\Rightarrow\tilde{T}$ is one-to-one.

$\Rightarrow\exists\tilde{T}^{-1}:L_{p}(0,1)\rightarrow \textrm{Dom}(T)/\textrm{ker}T$ such that $\tilde{T}^{-1}y=[x]$ where $x'(t)=y,\forall t\in[0,1],\ x\in\textrm{Dom}(T)$.

Suppose $x'(t)=y(t),\forall t\in[0,1]$.

$x\in AC[0,1]\Rightarrow x(t)=x(0)+\int_{0}^{t}y(s)ds$

Pick $y\in L_{p}[0,1]$ such that $y=Tx,\ x\in\textrm{Dom}(T)$.

$\left\Vert \tilde{T}^{-1}y\right\Vert =\left\Vert [x]\right\Vert =\inf_{z\in\textrm{ker}T}\left\Vert x+z\right\Vert \leq\left\Vert x+x(0)\right\Vert $

Let $z(t)=x(0),\forall t\in[0,1]\Rightarrow z\in\textrm{ker}T$.

Thus, \begin{eqnarray*} \left\Vert \tilde{T}^{-1}y\right\Vert & \leq & \left(\int_{0}^{1}\left|x(t)-x(0)\right|^{p}dt\right)^{\frac{1}{p}}\\ & = & \left(\int_{0}^{1}\left|\int_{0}^{t}y(s)ds\right|^{p}dt\right)^{\frac{1}{p}}\\ & \leq & \left(\int_{0}^{1}\left(c_{p}\cdot\int_{0}^{t}|y(s)|^{p}ds\right)dt\right)^{\frac{1}{p}}\\ & \leq & \left(\int_{0}^{1}\left(c_{p}\cdot\int_{0}^{1}|y(s)|^{p}ds\right)dt\right)^{\frac{1}{p}}\\ & = & c_{p}^{\frac{1}{p}}\left\Vert y\right\Vert _{L_{p}}\\ & < & +\infty \end{eqnarray*}

My questions:

(1) Why do we need $c_{p}$? I guess $\left|\int_{0}^{t}y(s)ds\right|^{p}\leq\int_{0}^{t}|y(s)|^{p}ds$.

(2) Why $\left(\int_{0}^{1}\left(c_{p}\cdot\int_{0}^{1}|y(s)|^{p}ds\right)dt\right)^{\frac{1}{p}}=c_{p}^{\frac{1}{p}}\left\Vert y\right\Vert _{L_{p}}$? By definition, $\left\Vert f\right\Vert _{L_{p}}=\left[\int|f|^{p}d\mu\right]^{\frac{1}{p}}.$

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up vote 1 down vote accepted

1) We do need $c_p$ because of the following estiamtion which can be obtained by means of Minkowskii's inequality applied to functions $y(s)$ and $1$: $$ \left|\int\limits_0^t y(s)ds\right|\leq \int\limits_0^t |y(s)|\cdot 1 ds\leq \left(\int\limits_0^t |y(s)|^p ds\right)^{1/p} \left(\int\limits_0^t 1^q ds\right)^{1/q}= $$ $$ \left(\int\limits_0^t |y(s)|^p ds\right)^{1/p} t^{1/q}\leq \left(\int\limits_0^t |y(s)|^p ds\right)^{1/p} $$ Hence $$ \left|\int\limits_0^t y(s)ds\right|^p\leq \int\limits_0^t |y(s)|^p ds $$ This result coincides with your claim but it can be different if you have considered intervals with measure not equal to $1$.

2) The answer to the second question is much more easier. You just had to note that $\int\limits_0^1 |y(s)|^p ds$ is a number which, of course, doesn't depends on $t$. So $$ \left(\int\limits_0^1\left(c_p\cdot\int\limits|y(s)|^p ds\right)dt\right)^{1/p}= \left(\left(c_p\cdot\int\limits_0^1|y(s)|^p ds\right)\int\limits_0^1dt\right)^{1/p}= $$ $$ \left(c_p\cdot\int\limits_0^1|y(s)|^p ds\right)^{1/p}= c_p^{1/p}\left(\int\limits_0^1|y(s)|^p ds\right)^{1/p}= c_p^{1/p}\Vert y\Vert_{L_p} $$

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