Past exam of the functional analysis I course of the University of Oxford / 2023 / 1

Both terms in the graph norm are finite and nonnegative. Moreover, $\Vert x{\Vert }_T=0$ forces $\Vert x{\Vert }_X=0$, hence $x=0$; conversely $T0=0$ gives $\Vert 0{\Vert }_T=0$. This proves definiteness of a norm.

The linearity of $T$ and absolute homogeneity of a norm give, for each real $a$,Applying the triangle inequality in both normed spaces givesThus

Since $W$ is proper, $T$ is a nonzero linear functional. Choose $e\in X$ with $Te=1$, by rescaling any vector outside its kernel. For every $x\in X$,If $ae\in W$, applying $T$ gives $a=0$. Consequently this is a direct sum, and the quotient vector space $X/W$ has dimension one. This computes the codimension of a vector subspace:

To prove boundedness without assuming completeness, use the linear functional with closed kernel criterion directly. Since $W$ is a closed set and $e\notin W$, the distanceis positive. If $Tx\ne 0$, then $e-x/(Tx)\in W$, soFor $Tx=0$ the resulting estimate is immediate. ThereforeHere $X^{\ast }$ denotes the continuous dual space.

The bounded linear functional from part (i) satisfies $|Tx|\le \Vert T\Vert \Vert x{\Vert }_X$, where $\Vert T\Vert$ is its operator norm. HenceThese are equivalent norms. Applied to $x_n-x_m$, the inequalities show that a sequence is a Cauchy sequence for either norm exactly when it is one for the other. Applied to $x_n-x$, they show that the two norms give the same limits in $X$. Thus every Cauchy sequence has a limit in one norm precisely when it does in the other:

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:

By the mean value theorem, a continuously differentiable function with zero derivative on $[-1,1]$ is constant. Thus the kernel of the differentiation operator isIt is a closed set by the finite-dimensional subspace is closed theorem, applied in the normed space $(C^1([-1,1]),\Vert \cdot {\Vert }_{\infty })$. It is proper because the function $x\mapsto x$ has nonzero derivative.

For $n\ge 1$, take $u_n(x)=\sin (nx)$. Its supremum norm is at most one, whereas $u_n^{\prime }(0)=n$ and $|u_n^{\prime }(x)|\le n$. Hence

Choose the continuously differentiable functionsThe estimate $\sqrt{a^2+b^2}\le a+b$ for $a,b\ge 0$ givesThus $f_n\to |\cdot |$ by uniform convergence, and the triangle inequality gives $\Vert f_n-f_m{\Vert }_{\infty }\le n^{-1}+m^{-1}$. The sequence is a Cauchy sequence in $X$, but its unique possible supremum norm limit is $x\mapsto |x|$, which is not differentiable at zero. That limit is outside $X$, so

The kernel $W=⟨1⟩$ is complete: the map $a\mapsto a1$ is an isometry from $\mathbb{R}$ with its absolute-value norm onto $W$ with the supremum norm.

Define a linear map $S:Y\to X$ byThe fundamental theorem of calculus gives $Sg\in C^1([-1,1])$ and $TSg=g$. Also,Thus $S$ is a bounded right inverse, and $Y=C([-1,1])$ is a Banach space. Applying part (c),

Take the normed space $X=C^1([-1,1])$ with the supremum norm, and let $T:X\to C([-1,1])$ be the inclusion $Tf=f$. This is an injective linear map, so its kernel is the complete zero space. Its graph norm isThe Cauchy sequence $f_n(x)=\sqrt{x^2+n^{-2}}$ from part (d)(ii) still has no limit in $X$ for this norm, because its only possible limit is $|x|$. Consequently