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Let $f:[a,b] \mapsto \mathbf{R}$ be a continuous function such that $f(x) \geq 0$ for all x $\in [a, b].$ Suppose $\int_a^b f = 0.$ Show that $f(x) = 0$ for all x $\in [a,b]$

Joe Mama
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2 Answers2

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Suppose there exists $c \in [a,b]$ so that $f(c) > 0$. Since $f$ is continuous, there exists $r > 0$ so that $f(x) > 0$ for $x \in [c - r, c + r]$. Can you see how this, combined with the fact that $f \geq 0$, contradicts the fact that $\int_a^b f dx = 0$?

Ian Dulchinos
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    I think the "Can you see" part becomes easier if you first find an interval on which $f$ is not only positive but above some fixed positive bound, like $f(c)/2$. – Andreas Blass May 10 '20 at 23:22
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Suppose there exists some $ x_{0}\in\left[a,b\right] $ such that $ f\left(x_{0}\right)>0 \cdot $

Let's use the fact that $ f $ is continuous :

There exists some $ \eta \in\left(0,\min\left(x_{0}-a,b-x_{0}\right)\right) $ such that : $$ \left(\forall x\in\mathcal{B}\left(x_{0},\eta\right)\right),\ \left|f\left(x\right)-f\left(x_{0}\right)\right|<\frac{f\left(x_{0}\right)}{2} $$

Thus : \begin{aligned}\int_{a}^{b}{f\left(x\right)\mathrm{d}x}&=\int_{a}^{x_{0}-\eta}{f\left(x\right)\mathrm{d}x}+\int_{x_{0}-\eta}^{x_{0}+\eta}{f\left(x\right)\mathrm{d}x}+\int_{x_{0}+\eta}^{b}{f\left(x\right)\mathrm{d}x}\\ &\geq\int_{x_{0}-\eta}^{x_{0}+\eta}{f\left(x\right)\mathrm{d}x}\geq\int_{x_{0}-\eta}^{x_{0}+\eta}{\frac{f\left(x_{0}\right)}{2}\,\mathrm{d}x}=\eta f\left(x_{0}\right)>0\end{aligned}

Which is not possible. That is why : $$ \left(\forall x\in\left[a,b\right]\right),\ f\left(x\right)=0 $$

CHAMSI
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