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?
Thanks for your help
 A: 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 $f(x) = 0$.
A: 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....
A: 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 
  $$
\lim_{n\to\infty}\int_{a}^{b}f_{n}(x)\,dx=\int_{a}^{b}\lim_{n\to\infty}
=
f_{n}(x)\,dx=\int_{a}^{b} f(x)\, dx
$$

A: 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.
A: 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
$$
f(t_x)=0\quad  \forall \;t_x\in(0,1).
$$
