Analysis Problem of the Day 56

Today’s problem appeared as Problem 8 on the Stanford Fall 2022 Analysis Qual:

Problem 8. Let $V$ be the vector space of real-valued Borel measurable functions on $[0,1].$ Show that there is no seminorm $p$ on $V$ such that $f_n \to f$ in measure if and only if $p(f_n-f) \to 0.$

Solution: We first claim that any such seminorm is in fact a norm, i.e. $f=g$ a.e. if and only if $p(f)=p(g).$ If $f \not = g$ and $p(f)-p(g)=p(f-g)=0,$ then the constant sequence $f-g$ is such that $p(f-g) \to 0,$ which implies $f - g \to 0$ in measure, i.e. $f=g$ a.e. Now, without loss of generality, suppose $p(\chi_{[0,1]})=1.$ Consider the decomposition $\chi_{[0,1]}(x) = \sum_{i=0}^{n-1} \chi_{[\frac{i}{n},\frac{i+1}{n}]}(x).$ By linearity of seminorms, the triangle inequality, and the pigeonhole principle, it follows that $p(\chi_{[\frac{j}{n},\frac{j+1}{n}]}) \geq \frac{1}{n}$ for at least one $j.$ Then, defining $f_n = n \chi_{[\frac{j_n}{n},\frac{j_n+1}{n}]}$ for $j_n$ as above, it follows that $p(f_n) \geq 1$ for all $n$ but $f_n \to 0$ in measure since the support of $f_n$ has measure at most $\frac{1}{n}.$ Thus, such a seminorm $p$ cannot exist.

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