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I have the following problem:

Let $f:\Bbb R^2\to \Bbb R$ be $C^1$ and $c>0$. Evaluate $\lim_{t\to 0} {{1}\over {t}}\int_{-ct}^{ct} f(t,u)du$, if this limit exists. If it does not, say why.

Well, the following was how I approached it:

First, let $F(t,u)$ be the function ${d\over du}F(t,u)=f(t,u)$. Then: $$\lim_{t\to 0} {{1}\over {t}}\int_{-ct}^{ct} f(t,u)du=\lim_{t\to 0} {{1}\over {t}} F(t,u)|_{-ct}^{ct}=\lim_{t\to 0} {{1}\over {t}}[F(t,ct)-F(t,-ct)]=2c\lim_{t\to 0} {{1}\over {2ct}}[F(t,ct)-F(t,-ct)]=2cf(0,0).$$

Thus, the conclusion I will get is the limit exists and is equal to $2cf(0,0)$. I wonder whether this approach is right or not. Thanks.

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Each choice of $f$ and $c$ gives a different limit problem, so uniqueness of limit only applies when you compute different values for the limit with the same $f$ and same $c$. –  Hurkyl Feb 24 '13 at 3:34
    
Ah, yes! You are definitely right. I will fix it and ask for suggestions. –  Scorpio19891119 Feb 24 '13 at 3:40
    
@Hurkyl How it looks now? –  Scorpio19891119 Feb 24 '13 at 3:51
    
It will be easier/cleaner to prove the limit exists by directly using the continuity of $f(x,y)$ at $(0,0)$. If $|f(x,y) - f(0,0)| < \epsilon$ over a neighborhood of $(0,0)$. If $t$ is small enough such the the line segment between $(t,-ct)$ to $(t,ct)$ falls inside this neighborhood, you can show $|\frac{1}{t}\int_{-ct}^{ct}f(t,u)du - 2cf(0,0)| \le 2c\epsilon$. –  achille hui Feb 24 '13 at 5:13
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2 Answers

You have the right answer, but I'm not sure I follow your reasoning.

If $\phi(x) = \int_0^{g(x)} f(x,u) du$, then (assuming $f,g$ are sufficiently smooth) we have $\phi'(x) = f(x,{g(x)})g'(x) + \int_0^{g(x)} \frac{\partial f(x, u)} {\partial x} du$.

Let $F(t) = \int_{-ct}^{ct} f(t,u) du = \int_{-ct}^{0} f(t,u) du + \int_{0}^{ct} f(t,u) du = - \int_0^{-ct} f(t,u) du + \int_{0}^{ct} f(t,u) du$. Then using the above we get $F'(t) = -f(t,-ct)(-c)+f(t,ct)(c)+ \int_{-ct}^{ct} \frac{\partial f(t, u)} {\partial x} du$.

The desired value is $F'(0)$ which is $F'(0) = 2 c f(0,0)$.

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I think you have the right idea.

When you define $F(t,u)$, you really mean

$$ \frac{\partial}{\partial u} F(t,u) = f(t,u) $$

I confess I prefer the notation $F_2(t,u)$, where "$F_2$" means "the derivative of $F$ with respect to its second argument" for reasons that will be made clear shortly.

So you computed the integral correctly and got that the limit is

$$ \lim_{t \to 0} \frac{F(t, ct) - F(t, -ct)}{t} $$

and it's the right idea that this looks very much like a limit for computing a derivative.

If we were asked to compute the limit

$$ \lim_{u \to 0} \frac{F(t, cu) - F(t, -cu)}{u} $$

then this is indeed equal to

$$ \lim_{u \to 0} \frac{F(t, cu) - F(t, -cu)}{u} = 2c F_2(t, 0) = 2c \frac{\partial}{\partial u} F(t,u) \mid_{u=0} $$

(Notice how awkward partial derivative notation is becoming!) But that's not actually what we're computing! Your idea of computation is effectively to compute

$$ \lim_{t \to 0} \lim_{u \to 0} \frac{F(t, cu) - F(t, -cu)}{u} $$

but the thing we actually want to compute can be written as

$$ \lim_{t \to 0} \lim_{u \to t} \frac{F(t, cu) - F(t, -cu)}{u} $$

Are we sure these are equal? The answer would be yes if the following limit exists:

$$ \lim_{(t,u) \to (0,0)} \frac{F(t, cu) - F(t, -cu)}{u} $$

Alas I'm rusty on my analysis. There's probably some theorem that says this should definitely be true because $F(t,u)$ is continuously differentiable in a neighborhood of $(0,0)$, but it doesn't spring to mind.

So I'll get my hands dirty and do a differential approximation. Rather than prove the full theorem, I'll do just the limit we're interested in

$$ \lim_{t \to 0} \frac{F(t, ct) - F(t, -ct)}{t} $$ $$ = \lim_{t \to 0} \frac{F(0,0) + t F_1(0,0) + ct F_2(0,0) - F(0,0) - t F_1(0,0) + ct F_2(0,0) + o(t) }{t} $$ $$ = \lim_{t \to 0} 2c F_2(0,0) + o(1) = 2c F_2(0,0) = 2c f(0,0)$$

where I've used little-oh notation to represent the error terms conveniently. It's clear from this calculation why we need $f$ to be differentiable! And you should check that $F(t,u)$ is also continuously differentiable.

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