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Suppose $f = f(x,t)$. Suppose $f$ is continuous $\forall x, t \in \mathbb{R}$. Then suppose $$ \int_{x_1}^{x_2} f(x, t) dx = 0 \hspace{1cm} \forall x_1, x_2 \in \mathbb{R}. $$ Why is it that $ f(x, t) = 0$ for arbitrary $x$?

alex
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    Isn't some kind of continuity for $f(x,t)$ needed for this? – coffeemath Jul 11 '18 at 04:23
  • I believed so; but from where I read it, there was no restraints put on the function $f$. It's an abstraction from a pde's book; maybe it is safe to assume that $f$ is continuous over all $x$ and $t$ in $\mathbb{R}$? I will edit my question to include this assumption. – alex Jul 11 '18 at 04:36
  • Which PDE book? Where does this come from? – Wraith1995 Jul 11 '18 at 04:39
  • Currently unreleased, being used as lecture notes at my university. – alex Jul 11 '18 at 04:40
  • User coffeemath incorrectly suggested continuity is needed to make the statement, and indeed the proof significantly simplifies when $f$ is continuous, but I voted to close the question because OP's original hypothesis (without continuity) suffices. – user217285 Jul 11 '18 at 04:55
  • @Nitin: that question deals with measurable $f$; this with continuous $f$. – Robert Lewis Jul 11 '18 at 04:55
  • @RobertLewis I was looking at OP's original question – user217285 Jul 11 '18 at 04:57
  • @Nitin: yeah, I know, he changed it up because of the comments . . . – Robert Lewis Jul 11 '18 at 04:57

2 Answers2

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Since $f(x, t)$ is continuous in $x$ and $t$ and the integral is independent of $t$, we can assume we are dealing with a function of $x$ alone. Therefore we may suppress the variable $t$ in what follows.

Suppose

$\exists y \in \Bbb R,\; f(y) > 0; \tag 1$

then by the continuity of $f$, there exist $x_1, x_2 \in \Bbb R$ such that

$x_1 < y < x_2 \tag 2$

and

$f(x) > 0, \; \forall x \in [x_1, x_2];\tag 2$

since $[x_1, x_2]$ is compact, $f(x)$ attains a minimum $m > 0$ on this interval, that is

$f(x) \ge m, \; x \in [x_1, x_2]; \tag 3$

then

$\displaystyle \int_{x_1}^{x_2} f(x) dx \ge \int_{x_1}^{x_2} m dx = m(x_2 - x_1) > 0; \tag 4$

this contradicts our assumption that

$\displaystyle \int_{x_1}^{x_2} f(x) dx = 0; \tag 5$

thus we cannot have $f(y) > 0$; a similar argument shows we may also rule out $f(y) < 0$; thus

$f(x) = 0, \; \forall x \in \Bbb R. \tag 6$

Robert Lewis
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Let $t \in \mathbb R$ be fixed. Define $g(s):=f(s,t)$ for $s \in \mathbb R$ and

$G(x):= \int_0^x g(s) ds$

for $x \in \mathbb R$.

Then we have $G(x)=0$ for all $x \in \mathbb R$. Since $g$ is continuous, why see that $G$ is an anti-derivative of $g$. Hence $g=0$ on $\mathbb R$.

Fred
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