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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ continuous and $$\int_0^1 f(tx) \,dx=0, \forall t\in \mathbb{R}$$ Show that $f \equiv 0$. $$$$ $\int_0^1 f(tx) \, dx=0$

$u=tx \Rightarrow du=t \, dx$ $x=0 \rightarrow u=0, x=1 \rightarrow u=t$

So $\frac{1}{t}\int_0^t f(u) \, du=0 \Rightarrow \int_0^t f(u) \, du=0$ $$$$ Could I show that $f \equiv 0$ using the above relation?

Michael Hardy
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Mary Star
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    You can show $\int_a^b f(x),dx=0$ for all $a$, $b$. This will easily imply your result. – David Mitra May 09 '14 at 23:04
  • @DavidMitra Using that $$\int_0^t{f(u)}du=0, \forall t \in \mathbb{R}$$?? – Mary Star May 09 '14 at 23:06
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    I edited in order to change f(x)dx to f(x),dx$ (in particular, $t,dx$ definitely looks better than $tdx$), and I noticed that you're enclosing every function being integrated in {curly braces}, thus: \int_0^1{f(tx)}dx. Is there a reason for doing that? – Michael Hardy May 09 '14 at 23:07
  • @MichaelHardy No, there is no particular reason...Is it wrong to write it in this way?? – Mary Star May 09 '14 at 23:09
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    Yes. $\int_0^b f-\int_0^a f=\int_a^b f$. – David Mitra May 09 '14 at 23:11
  • Just wondering about $f(x)=(x-0.5)$, the integral results in zero in the specified interval but f is not identical to zero. – NoChance May 09 '14 at 23:29
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    @EmmadKareem not just ∫f(x)dx=0, but ∫f(tx)dx=0,for every real number t – Seth May 10 '14 at 13:11
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    @MaryStar : We frequently see absurd over-use of curly braces here --- things like { {{{\int}}} { {{\cos}}{{x}^{{2}}} } }, that makes editing harder. And I wonder where it's coming from. Is it all just use of absurd software found on the web somewhere, or are there also people who feel a need to do it that way? – Michael Hardy May 10 '14 at 15:26
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    @Seth, thanks for your explanation, but $f(x)= t(x-0.5)$ would still give the same zero result. What am I missing? – NoChance May 10 '14 at 17:05
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    @EmmadKareem $f(tx)\neq tf(x)$ – Seth May 10 '14 at 17:22
  • @MichaelHardy I thought that I had to enclose the integrated function in curly braces...I didn't know that it is better without them... Thanks for noticing it! – Mary Star May 10 '14 at 17:31
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    @Seth, thanks again, however, in this case it still won't make a difference, never mind, I was just curious. – NoChance May 10 '14 at 19:58

2 Answers2

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Note that you have shown $F(t)=\int_0^t f(u)\,du=0$ for all $t$. So, because of the fundamental theorem of calculus $0=F(t)=\int_0^t f(u)\,du$ is differentiable and $F'(t)=f(t)=0$ $\forall t.$ That is, $f\equiv 0.$

mfl
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Using the transformation $x\mapsto t^{-1}x$, we get for $t>0$ $$\int_{0}^{1}f(tx)\;dx=\frac{1}{t}\int_{0}^{t}f(x)\;dx=0.$$

Since $t$ is arbitrary and $t^{-1}\neq0$, we find $\int_{0}^{a}f(x)\;dx=0$ for every $a>0$. This implies that $f\equiv0$ on $[0,\infty)$, as $f$ is assumed continuous (I assume you know how to prove this part).

To get $f\equiv0$ on $(-\infty,0)$ just take $t<0$ and use the fact that $\int_{0}^{-a}f(x)\;dx=-\int_{-a}^{0}f(x)\;dx$ for all $a>0.$

Sargera
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