Prove that the function $f$ is identically zero by the given condition. Let $f$ be a continuous function on $\mathbb R$ such that $f(x)=0$ for $|x|>1$. Suppose $\int_{-1}^{1} f(x-t)\ dx =0$ for all $t \in \mathbb R$. Show that $f$ is identically equal to zero.
I have tried to solve it but I fail.Would anyone please give some hint to proceed.
Thank you in advance.
 A: Just a hint
Put $u=x-t $.
the integral becomes
$$\int_{-1-t}^{1-t}f (u)du=0$$
if $0 \le t\le1$ then
$$\int_{-1}^{1-t}f (u)du=0$$
thus
$$\forall v\in [0,1] \int_{-1}^vf (u)du=0$$
and by differentiation
$$f (v)=0$$
by the same for $[-1,0] $.
A: Clearly $f$ is uniformly continuous.
We have $$\int_{-1}^1f(x-t) \ dx = 0\hspace{2cm} (1)$$ for all $t \in \mathbb{R}$. Replacing $t$ by $-t$ we get, $$\int_{-1}^1f(x+t) \ dx = 0$$ for all $t \in \mathbb{R}$.
This gives $$\int_{-1}^1 \{f(x+t) -f(t) \} \ dx +2f(t) =0$$ for all $t \in \mathbb{R}$. Let $\epsilon >0$ be given. Then there exist $\delta >0$ such that $|x|<\delta \implies |f(x-t)-f(x)|<\epsilon$
Choose $\delta <1$ if necessary, we get
$$0<2\epsilon + 2f(t)$$
 for all $t \in \mathbb{R}$. As $\epsilon >0 $ was arbitrary we get $f(t) > 0$ for all $t \in \mathbb{R}$
Now putting $t=0$ in $(1)$ we have $\int_{-1}^1 f(x) \ dx =0$. This gives $$\int_{\mathbb{R}} f(x) \ dx =0$$ As $f \geq 0$, it follows that $f \equiv 0$.
A: I cannot restrict myself at last by answering my own question.I have an answer analogous to one of my comments which is as follows $:$
First of all let us state a well known theorem $:$
Theorem $:$


Let $I=[a,b] \subset \mathbb R$ and let $f : I \longrightarrow \mathbb R$ be a continuous on $I$. Let $J=[c,d] \subset \mathbb R$ and let $u : J \longrightarrow \mathbb R$ be differentiable on $J$ and $u(J) \subset I$; $v : J \longrightarrow \mathbb R$ be differentiable on $J$ and $v(J) \subset I$. If $g : J \longrightarrow \mathbb R$ be defined by $g(t)=\int_{u(t)}^{v(t)} f(x) dx$ for $t \in J$, then $g'(t) = (f \circ v)(t).v'(t) - (f \circ u)(t).u'(t)$ for all $t \in J$.


Now lets proceed to answer my own question with the help of the above theorem.
It is very easily seen that the integral $\int_{-1}^{1} f(x-t)dx$ can be transformed to the integral $\int_{-1-t}^{1-t} f(x)dx$.Now take $I = [-3,3] \subset \mathbb R$ and also take $u(t) -1-t$ , $t \in [-2,2]$, $v(t)=1-t$ , $t \in [-2,2]$ and $J=[-2,2]$.Then clearly by the given hypothesis $f$ is continuous on $I$; $u(J) \subset I$ and $v(J) \subset I$.Then by the above theorem the function $g : J \longrightarrow \mathbb R$ defined by $g(t)=\int_{u(t)}^{v(t)} f(x) dx$ , $t \in J$  is differentiable on $J$ and $g'(t) = (f \circ v)(t).v'(t) - (f \circ u)(t).u'(t)$ for all $t \in J$.
Now after some calculation we have $g'(t) = f(-1-t) - f(1-t)$ for all $t \in J$.By the given hypothesis $g(t) = 0$ for all $t \in J$.Hence so is $g'(t)$ and consequently $f(-1-t) = f(1-t)$ for all $ t \in J$   $-$$-$$-$$-$$-$ $(1)$.
First put $t=-2 \in J$ then we have $f(1)=f(3)=0$ since it is given that $f(x)=0$ for $|x| >1$.Now put $t=0$ then we have $f(-1)=f(1)=0$.Again for any $c \in (-1,1)$ we have $1-c \in (0,2) \subset J$ so taking $t=1-c$ we have from $(1)$  $f(c)=f(-2+c)=0$ since $|-2+c|>1$ because $-2+c \in (-3,-1)$ for $c \in (-1,1)$.This shows that $f(x)=0$ for all $x \in [-1,1]$.
Therefore $f(x)=0$ for all $x \in \mathbb R$.Which proves our claim.
