# $E$ is separable Hilbert space. Suppose for all measurable $A$, $\big|\int_Af_e\,d\mu\big|\leq b\mu(A)$. Then $|f|\leq b$ a.e?

$$E$$ is separable Hilbert space with norm $$|\cdot|$$ defined by hermitian form. $$f\in L^1(E)$$ with $$L^1(E)$$ integrable functions with target in $$E$$. For each unit vector $$e\in E$$,$$f_e=\langle f,e\rangle$$. Suppose for all measurable $$A$$ and all $$e$$, $$\big|\int_Af_e\,d\mu\big|\leq b\mu(A)$$. Then $$|f|\leq b$$ a.e.

It is clear that $$|f_e(x)|\leq b$$ a.e.

$$\textbf{Q:}$$ How do I deduce $$|f|\leq b$$ a.e? Since $$E$$ is separable Hilbert space, I have $$f=\sum_e f_e e$$. I need $$\big|\int_Af\big|\leq b\mu(A)$$ for all measurable sets. WLOG, $$e$$ is a set of orthonormal basis. Then $$\int_Af=\int_A f_ee$$ this indicates all coordinates $$\big|\int_Af_e\big|\leq b$$. Now $$\big|\int_Af\big|=\sqrt{\sum_e\big|\int_Af_e\big|^2}$$ but I might not be able to bound this by $$b$$.

Ref. Lang, Real and Function Analysis Chpt VI, Sec 6, Cor 5.20

• $A$ isn't fixed, it should give you a hint that we need to take a proper set $A$. – Jakobian Aug 13 at 16:12

## 1 Answer

Let $$D \subset E$$ be countable dense and not containing $$0$$. Thus, for every $$v \in E$$, $$\|v\|_E=\sup_{a \in D}\,\frac{\langle v,\,a\rangle}{\|a\|_E}$$.

There is thus a full measure set $$A_0$$ such that for all $$d \in D$$, $$x \in A_0$$, $$\langle f(x),\,d\rangle \leq b\|d\|_E$$.

By taking suprema, you conclude that for all $$x \in A_0$$, $$\|f(x)\|_E \leq b$$.