Past exam of the functional analysis I course of the University of Oxford

Source: Oxford's released examination PDF, pages 280–283, paper dated 9 June 2023. Questions 1, 2 and 3 occupy printed pages 2, 3 and 4, respectively.

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

For a normed space $X$, let $X^{\ast }$ be its continuous dual space and $X^{\ast \ast }=(X^{\ast }{)}^{\ast }$ its bidual space. Its canonical embedding into the bidual isThe definition of a reflexive space in this setting isThe Hahn-Banach theorem makes $J_X$ a linear map that is an isometry. For any normed space $Z$, its continuous dual space is a Banach space: a Cauchy sequence $({\ell }_n)$ in the operator norm has scalar limits $\ell (z)={\lim }_n{\ell }_n(z)$, and the uniform Cauchy bound proves both boundedness of $\ell$ and $\Vert {\ell }_n-\ell \Vert \to 0$. Apply this with $Z=X^{\ast }$ to see that $X^{\ast \ast }$ is complete. If $J_X$ is onto, the isometry transfers this completeness to $X$. Therefore

The answer isIndeed, Lebesgue measure on $E\subset {\mathbb{R}}^n$ is a sigma-finite measure, and duality of Lp spaces gives $(L^1(E){)}^{\ast }=L^{\infty }(E)$. The restricted Lebesgue measure is a non-atomic measure and $E$ has positive measure, so the Dual of L infinity is larger than L1: the canonical integral representation does not exhaust $(L^{\infty }(E){)}^{\ast }$.

For $1<p<\infty$, the Holder conjugate exponent also lies strictly between one and infinity. Duality of Lp spaces then holds in both directions, in agreement with Reflexivity of Lp spaces. For $p=\infty$, the first required identification already fails.

Let $p^{\prime }$ be the Holder conjugate exponent of $p$, with $1/\infty =0$. The indicator function $1_{(-h,h)}$ has Lp norm $(2h{)}^{1/p^{\prime }}$ in $L^{p^{\prime }}$, including the value one when $p^{\prime }=\infty$. The Holder inequality therefore givesThe Lebesgue integral is unchanged by equality almost everywhere, so $T_h$ is well-defined on the Lp space; its linearity and the bound put it in the continuous dual space $X^{\ast }$.

For $p<\infty$, the function $f_h=(2h{)}^{-1/p}{1}_{(-h,h)}$ has Lp norm one and satisfiesFor $p=\infty$, use $f_h=1_{(-h,h)}$, whose L-infinity norm is one and whose image is $2$. Thus all endpoints are included in

Put $M=2^{1/p^{\prime }}$, the common operator norm from part (b). Given $\epsilon >0$, choose $N$ with $\Vert f-f_N{\Vert }_p<\epsilon /(3M)$. Since $f_N\in Y$, the scalar limit of $T_h{f}_N$ exists. The Cauchy criterion for a function limit gives $\delta >0$ such thatUsing the triangle inequality and the uniform operator norm bound,This is the requested estimate. The completeness of $\mathbb{R}$ makes the scalar Cauchy criterion for a function limit sufficient, hence ${\lim }_{h\to 0^{+}}{T}_{h}f$ exists and $f\in Y$.

For $u,v\in Y$ and real $a,b$, the linearity of $T_h$ gives $T_h(au+bv)=a{T}_{h}u+b{T}_{h}v$, which also has a limit. Thus $Y$ is a linear subspace. The preceding argument shows it contains every norm convergence limit of its sequences, so the sequential criterion for closed sets in a metric space gives

Use $D=C([-1,1])$, restricted to $(-1,1)$ and regarded as a linear subspace of $L^p((-1,1))$. The density of continuous functions in Lp on an interval says that $D$ is a dense subspace whenever $1\le p<\infty$.

For $g\in D$, the supremum norm givesIn particular $D\subseteq Y$, with zero limit. Part (c), together with $D$ being a dense subspace, implies $Y=X$. To identify the limit throughout $X$, for $f\in X$ and $g\in D$ useFirst choose $g$ in the dense subspace to make the first term arbitrarily small, then let $h\to 0^{+}$. This is convergence on a dense subspace under a uniform operator bound, and provesThe zero linear functional belongs to the continuous dual space $X^{\ast }$.

When $p=\infty$, $p^{\prime }=1$ and $T_{h}f=h^{-1}{\int }_{-h}^{h}f(x)dx$. Take the bounded measurable functionIt represents an element of the Lp space $L^{\infty }((-1,1))$, with L-infinity norm at most one. Sincewhere the lower endpoint contributes zero, its symmetric Lebesgue integral givesFor $h_n=e^{-2\pi n}$ and ${\tilde{h}}_n=e^{-(2\pi n+\pi )}$, both sequences tend to zero through $(0,1)$, but $T_{h_n}f=-1$ and $T_{{\tilde{h}}_n}f=1$. Therefore

Writing $\mathcal{B}(X)$ for the Banach space of bounded linear operators on $X$, the spectrum isHere $I$ is the identity operator. Since $X$ is a Banach space, the bounded inverse theorem makes this equivalent to failure of bijectivity of $\lambda I-T$.

The spectral radius isFor a nonzero complex Banach space, the spectrum of a bounded linear operator is nonempty and compact, so the supremum is a maximum. If the zero space is allowed, its empty spectrum is assigned spectral radius zero.

The Neumann series criterion states that $\Vert A\Vert <1$ for a bounded linear operator on a Banach space implieswith convergence in the operator norm. In particular, $\Vert T\Vert <1$ suffices for invertibility of $I-T$.

If $|\lambda |>\Vert T\Vert$, apply this criterion to $A=T/\lambda$. It givesThus no such $\lambda$ belongs to the spectrum, and

For each integer $j\ge 1$, if $|\lambda {|}^j>\Vert T^j\Vert$, then ${\lambda }^{j}I-T^j$ is invertible by the same argument. The commuting factorizationthen gives a bounded two-sided inverse for $\lambda I-T$, namely the displayed sum multiplied by $({\lambda }^{j}I-T^j{)}^{-1}$. Hence every $\lambda \in \sigma (T)$ satisfies $|\lambda {|}^j\le \Vert T^j\Vert$. Taking the supremum over the spectrum and then the infimum over $j$ proves

For a continuous function $g$ on $[0,1]$, the L-infinity norm of its restriction to $(0,1)$ equals its supremum norm on $[0,1]$. Indeed, values arbitrarily close to a maximum occur on an interval of positive Lebesgue measure, even if the maximum lies at an endpoint. Also, continuous functions equal almost everywhere on $(0,1)$ are equal everywhere on $[0,1]$. Thus restriction embeds $C([0,1])$ isometrically as a linear subspace of $X$.

The point evaluation functional $E_{x_0}(g)=g(x_0)$ is complex-linear and satisfies $|E_{x_0}(g)|\le \Vert g{\Vert }_{\infty }$. The constant function one attains equality, so $\Vert E_{x_0}\Vert =1$. The complex Hahn-Banach theorem extends this bounded linear functional from that linear subspace to $X$, preserving its operator norm. Therefore

Write $\psi ={\psi }_{x_0}$; its value at $x_0$, if that point is in $(0,1)$, may be assigned arbitrarily because Lp spaces identify functions equal almost everywhere. Its L-infinity norm is one.

The supplied estimate shows that $\psi$ cannot agree almost everywhere with any continuous function $q$: choosing $g=-q$ would give $0\ge |q(x_0)|+1$. Henceis a direct sum, so every element has a unique expression $g+c\psi$.

For any complex $a$ with $|a|\le 1$, define the linear functionalIf $c\ne 0$, the supplied estimate, applied to $g/c$, and absolute homogeneity of a norm giveFor $c=0$, the same bound follows from point evaluation. Thus $\Vert L_a\Vert \le 1$, and $L_a(1)=1$ makes its operator norm exactly one. The complex Hahn-Banach theorem extends $L_a$ to a norm-one linear functional on all of $X$, with value $a$ at $\psi$.

Conversely every norm-one extension satisfies $|{\ell }_{x_0}(\psi )|\le \Vert \psi {\Vert }_{\infty }=1$. We have therefore determined all possible values:In particular, taking $a=0$ and $a=1$ proves nonuniqueness of norm-preserving point evaluation extensions.

Because the linear functionals ${\ell }_0,{\ell }_1$ agree with endpoint point evaluations on continuous functions,The endpoint values also make $e_0,e_1$ linearly independent. By its definition, the image of a linear map $T$ lies in $E=⟨e_0,e_1⟩$. Conversely $T{e}_j=e_j$, so $T$ acts as the identity operator on $E$. ThusFor every $f$, $Tf\in E$, whence $T^{2}f=Tf$. It is a linear projection, and an annihilating polynomial is

Let $M=\Vert |e_0|+|e_1|{\Vert }_{\infty }$. Since both linear functionals have operator norm one, for every $f\in X$ the triangle inequality givesTaking the essential supremum proves $\Vert T\Vert \le M$.

For the reverse bound, continuity and the extreme value theorem give $x_{\ast }\in [0,1]$ with $|e_0(x_{\ast })|+|e_1(x_{\ast })|=M$. Choose complex $a_0,a_1$ of modulus one so that $a_j{e}_j(x_{\ast })=|e_j(x_{\ast })|$, taking $a_j=1$ if $e_j(x_{\ast })=0$. The continuous functionhas supremum norm one by the triangle inequality and its endpoint values. Since ${\ell }_j(f_{\ast })=a_j$,The output is continuous, so its L-infinity norm equals its supremum norm, including when $x_{\ast }$ is an endpoint. Therefore $\Vert T\Vert \ge M$, and the exact operator norm isThis is the norm of a two-endpoint interpolation projection.

From $T{e}_0=e_0\ne 0$, one is an eigenvalue. Zero is also an eigenvalue: the nonzero continuous function $q(x)=x(1-x)$ has both endpoint values zero, so $Tq=0$.

For $\lambda \notin \{0,1\}$, the linear projection identity $T^2=T$ gives an explicit bounded inverse:Multiplying on either side by $\lambda I-T$ yields $(I-T)+T=I$. There are no other points in the spectrum, and thus the spectrum of a nontrivial bounded projection here is

Let $M=\Vert T\Vert$, as calculated in part (ii). Its endpoint values ensure $1\le M<\infty$. Since $T^j=T$ for all $j\ge 1$,This verifies the spectral radius formula explicitly: