If $f$ is continuous and $\int_0^1f(xt)dt=0$ for every $x$, then $f\equiv 0$

Let $f$ be continuous and for every $x\in \mathbb{R}$ and $\int_0^1f(xt)\,dt=0$. Prove that $f\equiv 0$

My attempt

Since $f$ is continuous $f(xt)= \displaystyle\lim_{r \to xt}{f(r)}$

$\Rightarrow$ $\int_0^1\displaystyle\lim_{r \to xt}{f(r)dt}=0$

$\Rightarrow$ $\displaystyle\lim_{r \to xt}$ $\int_0^1 {f(r)dt}= 0$

$\Rightarrow \displaystyle\lim_{r \to xt}{f(r)t}|_0^1$

$\Rightarrow \displaystyle\lim_{r \to xt}{(f(r)}=0$

$\Rightarrow f(xt)=0$. Hence $f\equiv 0$

Is my procedure correct? Is there an easier way?

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please fix your latex it is not sure what you mean. Should it be the following? \begin{align} \int_0^1 f(x,t) dt=0 \forall x\in R \end{align} or \begin{align} \int_0^1 f(x\cdot t) dt=0 \forall x\in R \end{align} – cmmndy Dec 13 '12 at 14:07
is f(xt) in the integrand – Miguel Mora Luna Dec 13 '12 at 14:09

Step 1: We have for all $p(t)=a_0\cdot t^0+a_1\cdot t^1+\cdots+a_n\cdot t^n$ that $$\int_{0}^{1} p(xt)\,dx= \int_{0}^{1} a_0\cdot (xt)^0+a_1\cdot (xt)^1+\cdots+a_n\cdot (xt)^n \,dt =0$$ for all $x\in\mathbb{R}$ if, only if, $a_0=0,a_1=0,\dots,a_n=0$. In fact, $$\int_{0}^{1} p(tx) dt = a_0 x +\frac{1}{2}a_1x^2+\frac{1}{3}a_2x^3+\dots+\frac{1}{n+1}a_nx^{n+1}=0$$ for all $x\in\mathbb{R}$. Now use induction on $n\in\mathbb{N}$ and conclud that $a_1=0,a_2=0,\dots a_n=0$.

Step 2: use the classical Weierstrass aproximation theorem

Theorem ( Weierstrass aproximation) For all continuous fuction $f$ defined in a closed interval $[a,b]$ there is a sequence $\{ p_n(\cdot)\}_{n\in\mathbb{N}}$ of polinoms shout that $\lim_{n\to\infty}p_{n}(x)=f(x)$. And the convergence of $\lim_{n\to\infty}p_{n}(x)$ is uniform.

and a integral lema

Lema. Let's $f$ and $f_n$, wthi $n\in\mathbb{N}$, functions in a clused interval $[a,b]$. If the convergence of $\lim_{n\to\infty}f_{n}(x)=f(x)$ is uniform them $$\lim_{n\to\infty}\int_{a}^{b}f_{n}(x)\,dx=\int_{a}^{b}\lim_{n\to\infty} = f_{n}(x)\,dx=\int_{a}^{b} f(x)\, dx$$

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I believe you are correct. Here is another way:

We have for $x\in \mathbb{R}$, $$\int_0^1f(xt)\, dt=0$$ Substituting $u=xt$, $du=xdt$ gives that for $x>0$, $$\frac{1}{x}\int_0^xf(u)\, du=0\Rightarrow \int_0^xf(u)\, du=0$$ Now we invoke the fundamental theorem of calculus, which tells us that $$\frac{d}{dx}\int_0^xf(u)\, du=f(x)$$ But since $$\int_0^xf(u)\, du$$ is constant, it's derivative is zero. Therefore $f(x) = 0$.

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The line

$$\displaystyle\lim_{r \to xt}\int_0^1 f(xt)dt=0$$

is wrong, what is $t$? It needs to be fixed for the outside limit, but a variable inside the integral. Also, since $r \to xt$, $f(r)$ is NOT a constant, so the integral $\int_0^{1} f(r) dt$ is not $f(r)t$.

Here is a simple approach...

$$\int_0^1 f(xt)dt=\frac{1}{x} \int_0^x f(u)du \,.$$

Since $f$ continuous, $f$ has an antiderivative $F$. The above formula implies that for all $x \neq 0$ we have $\frac{F(x)-F(0)}{x}=0$. This means that $F$ is a constant....

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Assuming you mean \begin{align} \int_0^1 f(x\cdot t) dt=0 \forall x\in R \Rightarrow f \equiv 0 \end{align} Proof by contradiction: Assuming $f\not \equiv 0$ then there exist (multiple) intervals on which $f>0$ or $f<0$. Let $[x_0, x_0+\delta]$ be an interval on which (without loss of generality) $f>0$. As we have \begin{align} \int_0^1 f(x\cdot t) dt=0 \forall x\in R \end{align} We can take $x^*=x_0+\delta$ and take $x=x^*$ \begin{align} \int_0^1 f(x^*\cdot t) dt=\int_0^1 f((x_0+\delta)\cdot t) dt = \int_0^{x_0+\delta} f( t) dt \end{align} Now let $x_*=x_0$ we get \begin{align} \int_0^1 f(x_*\cdot t) dt=\int_0^1 f((x_0)\cdot t) dt = \int_0^{x_0} f( t) dt \end{align} If we now take the difference we get \begin{align} \int_0^1 f(x^*\cdot t) dt-\int_0^1 f(x_*\cdot t) dt= \int_{x_0}^{x_0+\delta} f( t) dt>0 \end{align} Which is a contradiction.

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Why does the first interval exists? Think about $f(x)=x \sin(\frac{1}{x})$.... – N. S. Dec 13 '12 at 14:25
You don't need to use the first interval, just use $\int_{x_0}^{x_0+\delta} f( t) dt =\int_0^{x_0+\delta}-\int_0^{x_0}$. – N. S. Dec 13 '12 at 14:27
i see that my assumption is wrong but i don't know how to use your second hint. Feel free to edit – cmmndy Dec 13 '12 at 14:32
Use $x^*=x_0+\delta$ and then again $x^*=x_0$. First one yields $\int_0^{x_0+\delta}=0$ and second one yields $\in_0^{x_0}=0$. – N. S. Dec 13 '12 at 14:34
Looks fine to me. – N. S. Dec 13 '12 at 18:42

Use the First Mean Value Theorem for Integrals to get: $$\int_0^1f(tx) dt = f(t_x\xi_x) ( 1\cdot x- 0.x ) \quad \mbox{ and } t_x\in (0,1),\quad\xi_x\in [0,x]$$ for all $x>0$ and $$f(t_x\cdot\xi_x)=0$$ for all $x>0$. It's well known that if $x$ sweeps across the range $(0, \infty)$ $t_x$ then scans the entire interval $[0,1]$. Then $$f(t_x)=0\quad \forall \;t_x\in(0,1).$$

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