If $f$ is bdd, integrable, $F(x) = \int_a ^x f$ for $x \in [a,b]$, prove $F$ unif continuous, differentiable & continuous at any $c$, $F'(c) = f(c)$

Question

Prove if $f$ is bounded and integrable on $[a,b]$ and if we define $F(x) = \int_a ^x f$ for $x \in [a,b]$, then

1.$\ F$ is uniformly continuous on $[a,b]$

2.$\ F$ is differentiable at any point $c \in [a,b]$ s.t. $f $ is continuous at $c$, and $F'(c) = f(c)$

My professor is calling this the 'Fundamental Theorem of Calculus Version 2'. He gave us the hint to use the following:

If $f$ is integrable on rectangle $A$, we have

$\quad\bullet$ if $ m\le f \le M$ on $A$, then $mv(A) \le \int_A f \le Mv(A)$

$\quad\bullet |f|$ is integrable on $A$ and $|\int_A f| \le \int_A|f|$

Where $v(A)$ is the volume of rectangle $ A$

Honestly, I'm unsure why I would need these hints? They don't seem to relate (at least to me) to the FTC at all. How do I use them in a proof of this??

Answer 1

Hint for (1). For $a\leq x\leq y\leq b$, $$|F(x)-F(y)|=\left|\int_x^y f(t)dt\right|\leq \int_x^y |f(t)|dt.$$ Hint for (2). Consider the difference $$\frac{F(c+h)-F(c)}{h}-f(c)=\frac{1}{h}\int_c^{c+h} (f(t)-f(c)) dt.$$

Answer 2

You will need the second for the uniform continuity. $f $ is bounded means there exist $M\ge 0$ such that $$(\forall x\in [a,b]) \;\; |f (x)|\le M $$

thus for $x,y \in [a,b],$

$$|F (x)-F (y)|=|\int_x^yf|\le M|x-y|$$ this proves the uniform continuity of $F $ at $[a,b] $.