Let
f be a bounded integrable non-negative function continous on [a, b].
Prove:
integral from a to b f(x) dx = 0 implies f(x) = 0 for every x
I believe I understand why it is true, I'm just not sure how to write the formal proof.
Let
f be a bounded integrable non-negative function continous on [a, b].
Prove:
integral from a to b f(x) dx = 0 implies f(x) = 0 for every x
I believe I understand why it is true, I'm just not sure how to write the formal proof.
What have you tried? Here is a weird approach.
Suppose that $\displaystyle f\ne 0$ then there exists some $\displaystyle x\in[a,b]$ such $\displaystyle f(x)\in(0,\infty)$ and since this set is open there exists some open ball $\displaystyle B_{\varepsilon}(f(x))\subseteq(0,\infty)$ and by continuity there exists some $\displaystyle B_{\delta}(x)$ such that $\displaystyle f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x))\subseteq(0,\infty)$. In other words, there exists some open ball around the point where $\displaystyle f(x)\ne0$ which is entirely positive. Take the concentric closed ball $\displaystyle B_{\frac{\delta}{2}}$. This is a closed and bounded interval and thus, by the Heine-Borel theorem, compact. Thus, since $\displaystyle f:B_{\frac{\delta}{2}}[x]\mapsto\mathbb{R}$ is continuous we know that $\displaystyle f$ assumes a minimum on $\displaystyle B_{\frac{\delta}{2}}[x]$. In particular, $\displaystyle \inf_{x\in B_{\frac{\delta}{2}}[x]}f(x)=\varepsilon>0$. Thus, $\displaystyle \int_a^b\text{ }f=\int_a^{x-\frac{\delta}{2}}\text{ }f+\int_{x-\frac{\delta}{2}}^{x+\frac{\delta}{2}}\text{ }f+\int_{x+\frac{\delta}{2}}^b \text{ }f\geqslant 0+\frac{\varepsilon \delta}{2}+0>0$. This of course is a contradiction.
Hello,
Why can't we just say that (without HB theorem) :
if there exists $\displaystyle m\in[a,b]$ such that $\displaystyle f(m)>0$, then by definition of the continuity, $\displaystyle \forall \epsilon>0, \exists \delta>0,\forall x\in[a,b], (|x-m|<\delta)\Rightarrow (|f(x)-f(m)|<\epsilon)$
In particular, if we let $\displaystyle \epsilon=\frac{f(m)}{2}$, $\displaystyle |f(x)-f(m)|<\epsilon \Leftrightarrow -\tfrac{f(m)}{2}<f(x)-f(m)<\tfrac{f(m)}{2} \Leftrightarrow \tfrac{f(m)}{2}<f(x)<f(m)$
So there exists $\displaystyle \delta>0$ such that $\displaystyle \forall x\in]m-\delta,m+\delta[$, $\displaystyle f(x)>\tfrac{f(m)}{2}>0$
Hence $\displaystyle \int_a^b f(x) ~dx=\int_a^{m-\delta} \underbrace{f(x)}_{\geqslant 0} ~dx+\int_{m+\delta}^b \underbrace{f(x)}_{\geqslant 0} ~dx+\int_{m-\delta}^{m+\delta} \underbrace{f(x)}_{> \frac{f(m)}{2}} ~dx$
So $\displaystyle \int_a^b f(x) ~dx>0+0+2\delta \cdot\tfrac{f(m)}{2}=\delta f(m)>0$
which is a contradiction.
So m can't exist, and hence $\displaystyle f(x)=0 ~,~ \forall x\in[a,b]$