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The differentiation operator sends a differentiable function to its derivative; its domain and the norms must be specified. On $C^1([-1,1])$ with values in $C([-1,1])$, using the supremum norm on both, it is an unbounded linear operator: $f_n(x)=\sin (nx)$ satisfies $\Vert f_n{\Vert }_{\infty }\le 1$ and $\Vert D{f}_n{\Vert }_{\infty }=n$. Its graph norm is $\Vert f{\Vert }_{\infty }+\Vert f^{\prime }{\Vert }_{\infty }$, which is complete. Indeed, the kernel consists of constants and is complete, while $Sg(x)={\int }_0^{x}g(t)dt$ is a bounded right inverse by the fundamental theorem of calculus; apply completeness of a graph norm from a bounded right inverse.

Let $(x_n)$ be a Cauchy sequence in the graph norm, and set $y_n=T{x}_n$. Since$(y_n)$ is a Cauchy sequence in the Banach space $Y$. Write $y_n\to y\in Y$.

Use the bounded right inverse to remove the part seen by $T$:Thus $w_n\in W$, and the operator norm bound givesThe assumed completeness of $W$ supplies $w\in W$ with $w_n\to w$ in $\Vert \cdot {\Vert }_X$. Put $x=w+Sy\in X$. The linearity of $T$ and the right inverse identity give $Tx=y$. ThereforeWe have proved the completeness of a graph norm from a bounded right inverse: