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If $ f(x) = \int_{-\infty}^x{f(t) d t} $ means $ f(x) = B e^x $

Then $ f(x) = \int_0^x{f(t) d t} $ means $ f(x)=? $

EDIT: The obvious extension: $ f_a(x) = \int_a^x{f_a(t) d t} $ means $ f_a(x)=? $

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2 Answers 2

up vote 7 down vote accepted

Assuming that $f$ is integrable on compact sets, if $f(x) = \int_0^x f(t) dt$, then $f'(x) = f(x)$, and $f(0) = 0$. The (unique) solution is $f(x) = f(0) e^x$, hence $f(x) = 0$ for all $x$.

As an aside, under the same condition on $f$, the first condition means that $f(x) = e^x \int_{-\infty}^0 f(t) dt$. If $c$ is a constant, then $\phi = c f$ satisfies the same functional equation. There is nothing that implies that $\int_{-\infty}^0 f(t) dt = 1$.

The extension also follows suit: Again, $f_a'=f_a$ and this time, $f_a(a) = 0$. The solution is $f_a(x) = e^{x-a} f_a(a)$, hence $f_a(x) = 0$ for all $x$.

Addendum to deal with the latter case when $a$ is infinite:

You need to be a little careful with infinities. Instead of using $f_a(a)$ as the initial condition, note that $f_a(0) = \int_a^0 f(t) dt$, $f_a' = f_a$. hence $f_a(x) = e^x f_a(0)$. This is always valid, and it is always the case that $\lim_{x \to -\infty} f_a(x) = 0$.

If $a$ is finite, then you have $f_a(a) = 0$, which forces $f_a = 0$.

If $a$ is infinite, then you have $\lim_{x\to a} f_a(x) = \lim_{x\to a} \int_a^x f_a(t) dt = 0$.

If $a=+\infty$, then the above expression shows that $f_a(0) =0$, and hence $f_a(x) = 0$.

If $a=-\infty$, then $f_{-\infty}(0)$ is arbitrary, and the solution is $f_\infty(x) = e^x f_{-\infty}(0)$.

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I've corrected my "premise", thanks. –  Mark Hurd Jan 8 '13 at 6:46
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@Mark Even after your edit, $f$ takes value $0$ at $x=a$. So you have a function that equals its derivative, and there is one point where its value is $0$. That's enough to make the function zero everywhere, assuming $f$ was integrable on compact sets to begin with (which is implied by your relation). –  alex.jordan Jan 8 '13 at 8:29
    
@alex.jordan: The function $f(x) = \infty$ satisfies the first, and $f_a = \infty \cdot 1_{(a,\infty)}$ satisfies the latter. If $f(x) \in \mathbb{R}$ for all $x$, does that automatically imply that $f$ is integrable? –  copper.hat Jan 8 '13 at 8:47
    
@copper.hat: Well, at least the second half of your comment makes sense! –  TonyK Jan 8 '13 at 9:08
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@MarkHurd: I have elaborated the solution above. If you want a general formula, it would be $f_a(x) = e^x f_a(0)$ (ie, not $f_a(a)$). –  copper.hat Jan 10 '13 at 1:23

$ f(x) = \int_0^x{f(t)d t} $

So, $ f(x) = F(x)-F(0) $ for some $ F $ where $ F' = f $

I.e. $ F(x) - F'(x) = F(0) $

So $ F(x) = P(x) e^x + c $ ... solve for $ F - F' = 0 $, then adjust the function

Substituting: $ P(x) e^x + c - (P(x) e^x + P'(x)e^x) = P(0) + c $

$ - P'(x)e^x = P(0) $

$ P'(x) = -P(0)e^{-x} $

$ P(x) = e^{-x} $

That makes $ F $ a constant, so $ f(x) = 0 $


Following my edit, although I'd guess copper.hat's path would be simpler, I'm proceeding the same way again:

$ f_a(x) = \int_a^x{f_a(t)d t} $

So, $ f_a(x) = F_a(x)-F_a(a) $ for some $ F_a $ where $ F_a' = f $

I.e. $ F_a(x) - F_a'(x) = F_a(a) $

So $ F_a(x) = P_a(x) e^x + c_a $ ... solve for $ F_a - F_a' = 0 $, then adjust the function

Substituting: $ P_a(x) e^x + c_a - (P_a(x) e^x + P_a'(x)e^x) = P_a(a) + c_a $

$ - P_a'(x)e^x = P_a(a) $

$ P_a'(x) = -P_a(a)e^{-x} $

$ P_a(x) = P_a(a)e^{-x} + d_a $

I.e. $ P_a(a) = P_a(a)e^{-a} + d_a $

$ P_a(a) - P_a(a)e^{-a} = d_a $

So

$ P_a(a) = 0 , d_a = 0 $ or, when $ a\neq 0, P_a(a) =\frac{d_a}{1-e^{-a}} $ for any $ d_a $

That makes $ F_a $ a constant, so $ f_a(x) = 0 $, or,

when $ a\neq 0, F_a(x) = ((\frac{d_a}{1-e^{-a}} )e^{-x} + d_a) e^x + c_a = d_a e^x + g_a$ for any $ d_a $ and $ g_a $.

Thus, $ f_a(x) = d_a e^x $

but $ f_a(a) = d_a e^a = \int_a^a{f_a(t)d t} = 0 $

so $ d_a = 0 $, or $ a = -\infty $ as originally described.

So the only solutions are $ f_a(x) = 0 $ or $ f_{-\infty}(x) = B e^x $.

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