Question Number 4716 by 123456 last updated on 28/Feb/16
![lets f:[0,T]→R does? (1/T)∫_0 ^T f(t)dt≤(√((1/T)∫_0 ^T [f(t)]^2 dt))≤(1/T)∫_0 ^T ∣f(t)∣dt](Q4716.png)
$$\mathrm{lets}\:{f}:\left[\mathrm{0},\mathrm{T}\right]\rightarrow\mathbb{R} \\ $$$$\mathrm{does}? \\ $$$$\frac{\mathrm{1}}{\mathrm{T}}\underset{\mathrm{0}} {\overset{\mathrm{T}} {\int}}{f}\left({t}\right){dt}\leqslant\sqrt{\frac{\mathrm{1}}{\mathrm{T}}\underset{\mathrm{0}} {\overset{\mathrm{T}} {\int}}\left[{f}\left({t}\right)\right]^{\mathrm{2}} {dt}}\leqslant\frac{\mathrm{1}}{\mathrm{T}}\underset{\mathrm{0}} {\overset{\mathrm{T}} {\int}}\mid{f}\left({t}\right)\mid{dt} \\ $$
Commented by prakash jain last updated on 29/Feb/16
![Let us say T=1 f(t)= { (2,(0<t≤0.5)),(4,(0.5<t≤1)) :} (1/T)∫_0 ^T f(t)dt=1+2=3 (√((1/T)∫_0 ^T [f(t)]^2 dt))=(√(4×.5+16×.5))=(√(10))>3 (1/T)∫_0 ^T ∣f(t)∣dt=3<(√(10)) The first and third part on inequality evalutes to same value if f(t)≥0. So the given condition is not true.](Q4717.png)
$$\mathrm{Let}\:\mathrm{us}\:\mathrm{say}\:\mathrm{T}=\mathrm{1} \\ $$$${f}\left({t}\right)=\begin{cases}{\mathrm{2}}&{\mathrm{0}<{t}\leqslant\mathrm{0}.\mathrm{5}}\\{\mathrm{4}}&{\mathrm{0}.\mathrm{5}<{t}\leqslant\mathrm{1}}\end{cases} \\ $$$$\frac{\mathrm{1}}{\mathrm{T}}\underset{\mathrm{0}} {\overset{\mathrm{T}} {\int}}{f}\left({t}\right){dt}=\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$$$\sqrt{\frac{\mathrm{1}}{\mathrm{T}}\underset{\mathrm{0}} {\overset{\mathrm{T}} {\int}}\left[{f}\left({t}\right)\right]^{\mathrm{2}} {dt}}=\sqrt{\mathrm{4}×.\mathrm{5}+\mathrm{16}×.\mathrm{5}}=\sqrt{\mathrm{10}}>\mathrm{3} \\ $$$$\frac{\mathrm{1}}{\mathrm{T}}\underset{\mathrm{0}} {\overset{\mathrm{T}} {\int}}\mid{f}\left({t}\right)\mid{dt}=\mathrm{3}<\sqrt{\mathrm{10}} \\ $$$$\mathrm{The}\:\mathrm{first}\:\mathrm{and}\:\mathrm{third}\:\mathrm{part}\:\mathrm{on}\:\mathrm{inequality}\:\mathrm{evalutes} \\ $$$$\mathrm{to}\:\mathrm{same}\:\mathrm{value}\:\mathrm{if}\:{f}\left({t}\right)\geqslant\mathrm{0}. \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{given}\:\mathrm{condition}\:\mathrm{is}\:\mathrm{not}\:\mathrm{true}. \\ $$