Paper 3, Section I, 2F2 F

Analysis II | Part IB, 2018

For a continuous function f=(f1,f2,…,fm):[0,1]→Rmf=\left(f_{1}, f_{2}, \ldots, f_{m}\right):[0,1] \rightarrow \mathbb{R}^{m}, define

∫01f(t)dt=(∫01f1(t)dt,∫01f2(t)dt,…,∫01fm(t)dt)\int_{0}^{1} f(t) d t=\left(\int_{0}^{1} f_{1}(t) d t, \int_{0}^{1} f_{2}(t) d t, \ldots, \int_{0}^{1} f_{m}(t) d t\right)

Show that

∥∫01f(t)dt∥2⩽∫01∥f(t)∥2dt\left\|\int_{0}^{1} f(t) d t\right\|_{2} \leqslant \int_{0}^{1}\|f(t)\|_{2} d t

for every continuous function f:[0,1]→Rmf:[0,1] \rightarrow \mathbb{R}^{m}, where ∥⋅∥2\|\cdot\|_{2} denotes the Euclidean norm on Rm\mathbb{R}^{m}.

Find all continuous functions f:[0,1]→Rmf:[0,1] \rightarrow \mathbb{R}^{m} with the property that

∥∫01f(t)dt∥=∫01∥f(t)∥dt\left\|\int_{0}^{1} f(t) d t\right\|=\int_{0}^{1}\|f(t)\| d t

regardless of the norm ∥⋅∥\|\cdot\| on Rm\mathbb{R}^{m}.

[Hint: start by analysing the case when ∥⋅∥\|\cdot\| is the Euclidean norm ∥⋅∥2\|\cdot\|_{2}.]

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