Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function. Write down the negation of the statement $$\exists \alpha>0 \quad \forall s>0 \quad \exists x>1 \quad\left|\int_1^x \frac{f(t)}{(t-1)^\alpha} d t\right|<s.\tag1$$ Prove whether statement $(1)$ or its negation is true.
Symbolically, the negation of statement $(1)$ is
$$
\forall \alpha>0 \quad \exists s>0 \quad \forall x>1 \quad\left|\int_1^x \frac{f(t)}{(t-1)^\alpha} d t\right|\geqslant s .
$$
I feel like the negation is easy to disprove as you can just have $f(t)=0$. However, I believe that disproving the negation is not enough to prove the original statement in this case, as it is only showing the original statement to be true for the specific case $f=0,$ not for all $f.$ Our lecturer has confirmed this to be true.
I am unsure of how to proceed with the general function for a proof of the original statement.
Best Answer
Please understand that $(1)$ and its negation $(1\text n)$ are not statements, as they contain the free variable $f;$ here, neither happens to have a definite truth value. Let $C$ denote the set of all continuous functions from $\mathbb R$ to $\mathbb R.$ Now, these are statements: $$\forall f\in C\quad \exists \alpha>0 \quad \forall s>0 \quad \exists x>1 \quad\left|\int_1^x \frac{f(t)}{(t-1)^\alpha}\,\mathrm d t\right|<s.\tag2$$ $$\forall f\in C\quad\forall \alpha>0 \quad \exists s>0 \quad \forall x>1 \quad\left|\int_1^x \frac{f(t)}{(t-1)^\alpha}\,\mathrm d t\right|\geqslant s.\tag3$$ $$\exists f\in C\quad\forall \alpha>0 \quad \exists s>0 \quad \forall x>1 \quad\left|\int_1^x \frac{f(t)}{(t-1)^\alpha}\,\mathrm d t\right|\geqslant s.\tag{2n}$$
Exhibiting $f\equiv0$ disproves statement $(3);$ however, it is $(2\text n)$ rather than $(3)$ that negates $(2).$
No: proving statement P is in fact logically equivalent to disproving its negation.
Does your lecturer actually want you to choose between proving $(2)$ and $(2\text n) ?$ This is not actually how they have written their exercise.
In any case, we can prove statement $(2)$ like this. Put $\alpha=1$ and observe that the integrand is continuous on $(1,\infty);$ so, by the FTC, for each $b$ and $x$ such that $x>b>1,$ for some antiderivative $F,$ $\left|\int_b^x \frac{f(t)}{(t-1)^\alpha}\,\mathrm d t\right|$ equals $|F(x)-F(b)|;$ we want to show that $\displaystyle\lim_{x\to b^+}\lim_{b\to1^+}|F(x)-F(b)|=0$ that is, $\displaystyle\lim_{x\to b^+}\lim_{b\to1^+}F(x)-F(b)=0;$ this is indeed true, by definition, since the antiderivative $F$ is continuous on $(1,\infty).$