Vtyjkyuk

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.

Just a hint

Put $u=x-t $.
the integral becomes

$$int_-1-t^1-tf (u)du=0$$
if $0 le tle1$ then

$$int_-1^1-tf (u)du=0$$

thus
$$forall vin [0,1] int_-1^vf (u)du=0$$
and by differentiation
$$f (v)=0$$

by the same for $[-1,0] $.

Clearly $f$ is uniformly continuous.

We have $$int_-1^1f(x-t) dx = 0hspace2cm (1)$$ for all $t in mathbbR$. Replacing $t$ by $-t$ we get, $$int_-1^1f(x+t) dx = 0$$ for all $t in mathbbR$.

This gives $$int_-1^1 f(x+t) -f(t) dx +2f(t) =0$$ for all $t in mathbbR$. 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

$$0<2epsilon + 2f(t)$$
for all $t in mathbbR$. As $epsilon >0 $ was arbitrary we get $f(t) > 0$ for all $t in mathbbR$

Now putting $t=0$ in $(1)$ we have $int_-1^1 f(x) dx =0$. This gives $$int_mathbbR f(x) dx =0$$ As $f geq 0$, it follows that $f equiv 0$.

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.