5
$\begingroup$

Ok, Here's my question:

Let $f(x,y)$ be defined and continuous on a $\le x \le b, c \le y\le d$, and $F(x)$ be defined >by the integral $$\int_c^d f(x,y)dy.$$ Prove that $F(x)$ is continuous on $[a,b]$.

I think I want to show that since $f(x,y)$ is continuous on $[a,b]$, I can use proof by contradiction to get $F'(x)$ continuous on $[a,b]$, which would then imply that $F(x)$ is continuous. But How do I go about setting this up? Any hints would be great. Thank you in advance. Also, this is my first attempt to format everything properly, So I'm sorry if this didn't post properly.

$\endgroup$

3 Answers 3

4
$\begingroup$

This is quite relevant for Rudin 10.1 from Real Analysis on page 246. Function $ f(x,y) $ is a continuous function on a compact set $[a,b]$. Therefore it is uniformly continuous, so the integrated expression is some real value, which can be made arbitrarily small.

In other words, uniform continuity of $f(x,y)$ indeed implies continuity of $F(x)$.

Great question!

$\endgroup$
1
$\begingroup$

Here is a start

$$ F(x+h)-F(x) = \int_c^d f(x+h,y)dy - \int_c^d f(x,y)dy $$

$$ \implies |F(x+h)-F(x)|\leq \int_c^d |f(x+h,y)- f(x,y)|dy. $$

Now, use the assumptions you have been given to finish the proof. Note that, $f$ is continuous on a compact set.

$\endgroup$
2
  • $\begingroup$ Do you mean uniformly continuous? It is stated in the problem that $f$ is continuous. $\endgroup$
    – user10444
    Nov 20, 2013 at 23:03
  • $\begingroup$ @user10444: Yes, It is. $\endgroup$ Nov 21, 2013 at 0:06
-1
$\begingroup$

Consider $f(x,y)=y^{x-1}$ then $F(x)$ defined above is $\frac{y^{x}}{x}$. The definition of continuous requires that the function equal it's limit at the point in question among other things. Clearly $\frac{y^{x}}{x}$ is not defined and thus not equal to it's limit at the point $x = 0$. Yet $f(0,y)=\frac{1}{y}$ has no problem for suitable $y$. Thus the statement you set out to prove is in fact false.

Note If additionally you knew that the $\frac{\partial }{\partial x}F(x,y)$ was continuous, only then could you conclude $F(x)$ continuous on an interval $[a,b]$ as then you would know that both $F(x,y)$'s partials (one of them is $f(x,y)$) where continuous, thus $F$ differentiable and hence continuous.

$\endgroup$
1
  • $\begingroup$ When $x=0$, your antiderivative is $\ln |y|$. $\endgroup$ Dec 19, 2019 at 23:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .