Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 22I (Analysis of Functions)

The continuous dual space $X^{\prime }$ consists of all bounded linear maps $\Lambda :X\to \mathbb{C}$, withIf $Y=L^p({\mathbb{R}}^n)$ with $1\le p<\infty$ and $q$ is the conjugate exponent, then duality of Lp spaces identifies $Y^{\prime }$ isometrically with $L^q({\mathbb{R}}^n)$, including $q=\infty$ when $p=1$. Every functional has the formand $\Vert {\Lambda }_h\Vert =\Vert h{\Vert }_q$.

Norm convergence means

Weak convergence in $X^{\prime }$ meansNorm convergence implies weak convergence because every $\Phi \in X^{\prime \prime }$ is norm-continuous.

Weak-star convergence means pointwise convergence on the predual:For each $x\in X$, its image $J_{X}x\in X^{\prime \prime }$ under the canonical embedding into the bidual satisfies $(J_{X}x)(\Lambda )=\Lambda (x)$. Thus weak convergence implies weak-star convergence, and altogether

For the requested example, letThe Riemann-Lebesgue lemma gives $f_j\stackrel{\ast }{\rightharpoonup }0$ in $(L^1{)}^{\prime }$. Sincethe same lemma gives $f_j^2\stackrel{\ast }{\rightharpoonup }1/2$. Hence $f=0$ and $g=1/2\ne f^2$, exhibiting the failure of weak-star convergence to commute with squaring.

On $C_c^0({\mathbb{R}}^n)$ equipped with the $L^{\infty }$ norm, point evaluation is linear and satisfieswith equality for a compactly supported function equal to one near zero. Thus $\Vert \Lambda \Vert =1$. The Hahn-Banach theorem extends it, without increasing its norm, to a continuous functional $\tilde{\Lambda }$ on $L^{\infty }({\mathbb{R}}^n)$.

This extension cannot be represented by an $L^1$ function. If $h\in L^1$ satisfied $\tilde{\Lambda }(f)=\int fh$, choose continuous functions $0\le f_k\le 1$ with $f_k(0)=1$ and support in the ball of radius $1/k$. Thenby the absolute continuity of the Lebesgue integral, a contradiction. Therefore $(L^{\infty }{)}^{\prime }$ strictly contains the functionals represented by $L^1$; this is a singular functional on L infinity.

Finally, $L^p({\mathbb{R}}^n)$ is a reflexive Banach space exactly whenThe endpoint spaces $L^1$ and $L^{\infty }$ are not reflexive.