How Does Continuity Ensure Differentiability in Integral Functions?

[AKG]

Problem:

Let $f\ :\ [a,b]\times [c,d] \to \mathbb{R}$ be continuous, and suppose that $D_2f$ exists and is continuous. Show that the function $F\ :\ (c,d)\to \mathbb{R}$ defined by:

$$F(y) = \int _a ^b f(x, y)dx$$

is differentiable and that

$$F'(y) = \int _a ^b D_2f(x, y)dx$$

Solution:

Let P vary over all partitions of [a, b]. Then we want to show that $\forall y \in (c,d)$, F'(y) exists and is given by the equation above.

$$F(y) = \sup _P \sum _{S \in P}v(S)m_S(f)$$

where v(S) is the volume of S, and $m_S(f) = \inf \{f(x,y)\ :\ x \in S\}$

$$F'(y) = \lim _{h \to 0}\frac{F(y + h) - F(y)}{h}$$

$$F'(y) = \lim _{h \to 0}\frac{\sup _P\sum _{S \in P}v(S)m_S(f(x, y+h)) - \sup _P\sum _{S \in P}v(S)m_S(f(x,y))}{h}$$

$$F'(y) = \lim _{h \to 0}\left (\sup _P\sum _{S \in P}v(S)m_S\left (\frac{f(x, y+h)}{h}\right ) - \sup _P\sum _{S \in P}v(S)m_S\left (\frac{f(x,y)}{h}\right )\right )$$

From here, I'd like to be able to say:

$$F'(y) \geq \sup _P \sum _{S \in P} v(S)m_S \left (\lim _{h \to 0}\frac{f(x, y+h) - f(x, y)}{h}\right )$$ *

This would give me:

$$F'(y) \geq \sup _P \sum _{S \in P} v(S)m_S(D_2f(x, y))$$

$$F'(y) \geq \underline{\int _a ^b}D_2f(x, y)dx$$

Now, if * follows validly from the previous line, then I should also be able to say:

$$F'(y) \leq \inf _P \sum _{S \in P}v(S)M_S\left (\lim _{h \to 0}\frac{f(x, y+h) - f(x, y)}{h}\right ) = \overline{\int _a ^b}D_2f(x, y)dx$$

where $M_S(f) = \sup \{f(x,y)\ :\ x \in S\}$.

This gives me:

$$\underline{\int _a ^b}D_2f(x, y)dx \leq F'(y) \leq \overline{\int _a ^b}D_2f(x, y)dx$$

Now, I would also like to be able to assert * that $D_2f$ is integrable. Is the fact that it is continuous, and that [a, b] is bounded enough to show that? If so, then I have that

$$\underline{\int _a ^b}D_2f(x, y)dx = \overline{\int _a ^b}D_2f(x, y)dx$$

This proves that F is differentiable, because to be differentiable, the limit:

$$\lim _{h \to 0}\frac{F(y + h) - F(y)}{h}$$ must exist, and by the above, it does exist and its value is exactly what we were expected to show it was.

So, can someone help me see the justification for * and *? Or am I wrong entirely, and are those two things false? If so, I could really use a hint as to how to go about solving the problem. Thanks,

AKG

 

[wais]

The justification for * and * comes from the fundamental theorem of calculus, which states that if f is continuous on [a, b] and F is the function defined by F(x) = \int_a^x f(t)dt, then F is differentiable on [a, b] and F'(x) = f(x). In this case, we have a similar situation where F(y) is defined as an integral with respect to x, and we can use the same theorem to show that F'(y) = \int_a^b D_2f(x, y)dx.

To show that D_2f is integrable, we can use the fact that f is continuous on [a, b] and [c, d] to argue that it is bounded, and therefore Riemann integrable. This allows us to use the same argument as before to show that F'(y) exists and is equal to the integral of D_2f over [a, b].

Overall, your proof is correct and well-written. Just make sure to mention the use of the fundamental theorem of calculus and the integrability of D_2f in your explanation. Good job!