# Showing uniform continuity

I've been attempting to show that a function $f$ on the actual period $[a,b]$ which pleases

$$f(x)=f(a)+\int_a^xf'(s)\,ds\qquad\text{(f' defined almost everywhere)}$$

has to be evenly continual on $[a,b]$.

Given that the problem over amounts outright connection I recognize that I can show what I require using the evidence that outright connection - from its basic definition - indicates consistent connection: I have actually seen an evidence of that. Nonetheless, I would love to show the above without entailing an additional kind of connection while doing so.

I recognize that - given that what I have actually mentioned is additionally equal to there existing any kind of integrable function instead of $f'$ - the evidence needs to not entail the buildings of the by-product. Nonetheless, in developing a bound I get just regarding

$$\left|f(y)-f(x)\right|=\left|\int_x^yf'(s)ds\right|\leq\int_x^y\left|f'(s)\right|\,ds$$

Is it feasible to show that an integrable first by-product (or without a doubt any kind of integrable function) is bounded in sup standard?

Give thanks to you.ยข Marko

5
2022-06-07 14:38:27
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We think that $\|f'\|_{L^1} = \int_{a}^{b} |f'|\,dt \lt \infty$. Allow $A_{n} = \{x \,:\,|f'(x)| \leq n\}$. Place $g_{n} = [A_{n}] f'$, where $[A_n]$ represents the particular function of $A_n$. After that we have actually $g_{n} \to f'$ virtually anywhere, and also, as Nate mentions in his comment, controlled merging indicates that $\int_{a}^{b} |g_n - f'|\,dt \to 0$ as $n \to \infty$ (the integrand is bounded by the integrable function $2|f'|$).

Currently, offered $\varepsilon \gt 0$, pick $n$ so huge that $\int_{a}^{b} |g_n - f'|\,dt \lt \varepsilon /2$. As $|g_{n}|$ is bounded by $n$, we have that $|\int_{x}^{y} g_{n}(t)\,dt| \leq n|y-x|$. Hence, $$\left\vert \int_{x}^{y} f'(t)\,dt\right\vert \leq \left\vert\int_{x}^{y} |f'-g_n|\,dt\right\vert + \left\vert \int_{y}^{x} |g_n(t)|\,dt\right\vert \leq \varepsilon/2 + n \cdot |y-x|$$ and also for $\delta = \frac{\varepsilon}{2n}$ we get for all $x,y$ with $|y-x| \lt \delta$ that $$|f(y) - f(x)| = \left\vert \int_{x}^{y} f'(t)\,dt \right\vert \leq \varepsilon/2 + \delta n \lt \varepsilon$$ which is the really definition of consistent connection of $f$.

Actually, we get the a lot more basic price quote that for $\mu(E) \lt \delta$ we have $\int_{E} |f'|\,dt \lt \varepsilon$. Yet that is specifically outright connection of $f$.

It is certainly not real that an integrable first by-product is bounded in the sup - standard. As an example, for $f(x) = \sqrt{x}$ we have $f'(x) = \frac{1}{2\sqrt{x}}$ which is not bounded yet integrable on $[0,1]$.

4
2022-06-07 15:03:31
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As you stated, $f$ is definitely continual, and also revealing this isn't actually tougher than revealing consistent connection straight.

If $g$ is integrable and also $\varepsilon>0$ is offered, there is a $\delta>0$ such that $m(A)<\delta$ indicates $\int_A|g|<\varepsilon$. To see this, you can as an example first take $h$ bounded by $M>0$ such that $\int_a^b|g-h|<\frac{\varepsilon}{2}$, and afterwards take $\delta = \frac{\varepsilon}{2M}$.

As soon as you have this, you have $|\int_x^y g|<\varepsilon$ whenever $|x-y|<\delta$. And also as stated, this includes revealing outright connection. Boundedness of $f'$ would indicate the more powerful problem of Lipschitz connection.

3
2022-06-07 14:55:25
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