If $\int^{b}_a f > 0$ then there is some interval and $\delta > 0$ on which $f(x) \ge \delta$ (Abbott pp 199 q7.4.4d)

Question

True or False. If $\int^{b}_a f > 0$, then $\exists \; [c,d] \subseteq [a,b]$ and $\delta > 0$ such that $f(x) \ge \delta$ for all $x \in [c,d]$.

1. We need to determine if true or false. How? I tried a figure from Stewart p367:

2. How can we presage to prove the contrapositive? Is a direct proof possible?

3. I don't understand the 'indeed the negation...is a compact set'? We want $\color{red}{\neg}\exists \; █ \; [c,d] \subseteq [a,b] \wedge \delta > 0 \; █ \; \; [ \;f(x) \ge d \; \forall \, x \in [c,d] \; ]$ = $\forall \; \color{red}{\neg} \; █ \; [c,d] \subseteq [a,b] \wedge \delta > 0 \; █ \; [ \;f(x) \ge d \; \forall \, x \in [c,d] \; ]$ = $\forall \; █ \; [c,d] \supset [a,b] \vee \delta < 0 \; █ \; \color{red}{\neg} [ \;f(x) \ge d \; \forall \, x \in [c,d] \; ]$ = $\forall \; █ \; [c,d] \supset [a,b] \vee \delta < 0 \; █ \; \;f(x) < d \; \color{red}{\neg} \;\forall \, x \in [c,d] \; $ = $\forall \; █ \; [c,d] \supset [a,b] \vee \delta < 0 \; █ \; \;f(x) < d \; \; \exists x \in [c,d] \; $

What foundered? Can't have a quantifier at the end ?

Answer 1

I am not completely sure whether your solution is correct. It is correct if $f$ is continuous. This is the reason:

To negate the statement, it should be:

"For all intervals $[c, d]\subset [a, b]$ and $\delta >0$, there is $x\in [c, d]$ such that $f(x) < \delta$."

Now fix $[c, d]$ and let $\delta = 1/n$. Then there is a sequence $x_n \in [c, d]$ such that $f(x_n) < 1/n$. By Bolzano-Weiestrauss theorem, $\{x_n\}$ has a converges subsequence $\{x_{n_k}\}$ which converges to $x\in [c, d]$. (This is where we used the fact that $[c, d]$ is compact).

If $f$ is continuous, then we have $f(x) = \lim_{k\to \infty} f(x_{n_k}) \leq 0$ and the rest follows.

When $f$ is not continuous, one can modify the proof a little bit. Let $\int_a^b f = A >0$. Again assume that the statement is false. Let $\delta = \frac{A}{2(b-a)}$. So all for partition $P$ (as in your answer) we have $c_k \in [x_{k-1}, x_k]$ such that $f(c_k) \leq \delta$. Thus (as in your answer)

$$L(f, P) = \sum_{k=1}^n m_k (x_k - x_{k-1}) \leq \delta (b-a) = A/2\ .$$

Thus $\int_a^b f = \sup_P L(f, P) \leq A/2 <A$, which is a contradiction.

I do not think there is a constructive way to prove this, as there does not seem to be a way to find that interval $[c, d]$.

Answer 2

Negation of the conclusion

The correct formal statement of the conclusion is $$ ∃c,d,δ∀x:([c,d]⊆[a,b]∧ δ>0∧x∈[c,d]) ⟹ f(x)≥δ $$ The negation is $$ ∀c,d,δ∃x:([c,d]⊆[a,b]∧ δ>0∧x∈[c,d])∧f(x)<δ $$ where the defining conditions can again be absorbed into the quantor blocks $$ ∀[c,d]⊆[a,b],δ>0∃x∈[c,d]:f(x)<δ $$

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On the proof of the statement

Now, contrary to the first lines of the proof, this does not mean that there are non-positive values for any interval $[c,d]$ since integrable functions need not be continuous everywhere. But still one can conclude that for any $δ>0$ $$ \inf_{x∈[c,d]} f(x)<δ ⟹ \inf_{x∈[c,d]} f(x)\le 0. $$ So that under the contraposition any lower Darboux sum is non-positive, and since the value of the Riemann integral is equal to the limit of the lower Darboux sums, one would conclude that $\int_a^b f(x)dx\le 0$, in contradiction to the assumption $\int_a^b f(x)dx> 0$.

Answer 3

As usual, the format of the question confuses me: there are many different fonts and it is not clear which of the text is yours, and where the text which is not yours comes from. That is not good in an academic context. But the question is interesting, so I'll make an an exception and answer it.

The given solution seems, in its use of the compactness of $[c,d]$, to be implicitly assuming the continuity of $f$: a continuous function on a compact set which takes arbitrarily small positive values must also take non-positive values. Since $f$ is not assumed to be continuous but only Riemann-integrable, this is not valid...but once we address this point, the solution becomes correct.

Rather, suppose $\int_a^b f > 0$, and seeking a contradiction we suppose that for all nontrivial subintervals $[c,d]$, there is no $\delta >0 $ such that $f(x) \geq \delta$ on for all $x \in [c,d]$. Then the infimum of $f$ on each subinterval is at most $0$. This implies that for any partition $\mathcal{P}$ of $[a,b]$, the lower Darboux sum $L(f,\mathcal{P})$ is at most $0$. But since $f$ is Riemann integrable, $\int_a^b f$ is the supremum of all the lower Darboux sums, so $\int_a^b f \leq 0$, contradiction.