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

[tex]F(y) = \int _a ^b f(x, y)dx[/tex]

is differentiable and that

[tex]F'(y) = \int _a ^b D_2f(x, y)dx[/tex]

Solution:

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

[tex]F(y) = \sup _P \sum _{S \in P}v(S)m_S(f)[/tex]

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

[tex]F'(y) = \lim _{h \to 0}\frac{F(y + h) - F(y)}{h}[/tex]

[tex]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}[/tex]

[tex]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 )[/tex]

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

[tex]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 )[/tex] *

This would give me:

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

[tex]F'(y) \geq \underline{\int _a ^b}D_2f(x, y)dx[/tex]

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

[tex]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[/tex]

where [itex]M_S(f) = \sup \{f(x,y)\ :\ x \in S\}[/itex].

This gives me:

[tex]\underline{\int _a ^b}D_2f(x, y)dx \leq F'(y) \leq \overline{\int _a ^b}D_2f(x, y)dx[/tex]

Now, I would also like to be able to assert * that [itex]D_2f[/itex] 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

[tex]\underline{\int _a ^b}D_2f(x, y)dx = \overline{\int _a ^b}D_2f(x, y)dx[/tex]

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

[tex]\lim _{h \to 0}\frac{F(y + h) - F(y)}{h}[/tex] 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

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# Homework Help: Help with Analysis Problem

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