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Let $g$ be a continuous periodic function on $mathbb{R}$ with $g(x + 1) = g(x)$. Assume that $int_{0}^{1}g(x)dx = 0$.

(a) Let $f$ be continuous on all $mathbb{R}$

Show that $lim_{ntoinfty} int_{0}^{1}f(x)g(nx)dx = 0$

(b) Let $f$ be Lebesgue integrable on all $mathbb{R}$

Show that $lim_{ntoinfty} int_{0}^{1}f(x)g(nx)dx = 0$

My attempt:

(a)

For part (a) I dont know how to proceed.
I have a solution that uses the following:

$$lim_{ntoinfty} int_{0}^{1}f(x)g(nx)dx = int_{0}^{1}f(x)int_{0}^{1}g(x)dx$$ and by hypothesis the results follow. There is another approach ?

(b) (This needs verification)

$g(x)$ is continuous and periodic on $mathbb{R}$, then $exists B > 0$ such that $|g(x)|< B$.

$f$ Lebesgue integrable on $[0,1]$ $rightarrow forall epsilon > 0, exists h$ step function such that $int_{0}^{1}|f - h| < frac{epsilon}{B}, forall x in [0,1]$

then

$$| int_{0}^{1}f(x)g(nx)dx | leq int_{0}^{1}|f(x) - h(x)| |g(nx)|dx + |int_{0}^{1} h(x)g(nx)dx| leq epsilon + |int_{0}^{1} h(x)g(nx)dx|$$

We must show that $lim_{nto infty}int_{0}^{1} h(x)g(nx)dx|$

Since $h$ is a step function,

$$h(x) = sum_{i=1}^{k}lambda_ichi_{O_i}(x), mbox{ where } bigcup_{i=1}^{k}O_i = [0,1]$$

Hence, for $M = max_{i = 1,...,k}|lambda_i|$

$$int_{0}^{1} h(x)g(nx)dx = sum_{i=1}^{k}lambda_iint_{O_i}g(nx)dx$$
$$leq Msum_{i=1}^{k}int_{O_i}g(nx)dx$$
$$= Mint_{0}^{1}g(nx)dx| = frac{M}{n}int_{0}^{n}g(x)dx = Mint_{0}^{n}g(x) = 0$$

Then

$$|int_{0}^{1}f(x)g(nx)dx| < epsilon$$

For part (a):
Let $I=int_{0}^{1}|g(x)|dx$. Let $epsilon>0$ be given. Note that
$f$ is uniformly continuous on $[0,1]$, so there exists $delta>0$
such that $|f(x)-f(y)|<frac{epsilon}{I+1}$ whenever $x,yin[0,1]$
with $|x-y|<delta$. Choose $Ninmathbb{N}$ such that $frac{1}{N}<delta$.
By considering the substitution $y=nx$, we may rewrite
begin{eqnarray*}
int_{0}^{1}f(x)g(nx)dx & = & frac{1}{n}int_{0}^{n}f(frac{y}{n})g(y)dy\
& = & frac{1}{n}sum_{k=1}^{n}int_{k-1}^{k}f(frac{y}{n})g(y)dy.
end{eqnarray*}
Let $ngeq N$ be arbitrary. Denote $f_{k}=f(frac{k}{n})$, for $k=1,2,ldots n$.
By periodicity of $g$, we have $int_{k-1}^{k}g(y)dy=int_{0}^{1}g(y)dy=0$.
Therefore
begin{eqnarray*}
& & |int_{k-1}^{k}f(frac{y}{n})g(y)dy|\
& = & |int_{k-1}^{k}f(frac{y}{n})-f_{k}g(y)dy|\
& leq & int_{k-1}^{k}|f(frac{y}{n})-f_{k}||g(y)|dy\
& leq & int_{k-1}^{k}frac{epsilon}{I+1}|g(y)|dy\
& = & frac{epsilon}{I+1}int_{0}^{1}|g(y)|dy\
& = & epsiloncdotfrac{I}{I+1}\
& leq & epsilon.
end{eqnarray*}
Now it is clear that
begin{eqnarray*}
& & |int_{0}^{1}f(x)g(nx)dx|\
& leq & frac{1}{n}sum_{k=1}^{n}|int_{k-1}^{k}f(frac{y}{n})g(y)dy|\
& leq & frac{1}{n}sum_{k=1}^{n}epsilon\
& = & epsilon.
end{eqnarray*}

For (b): Since $g$ is continuous, there exists $M>0$ such that $|g(x)|leq M$
for all $xin[0,1]$. By periodicity of $g$, it follows that $|g(x)|leq M$
for all $xinmathbb{R}$. Recall that Lebesgue integrable function
can be approximated by integrable continuous function in $||cdot||_{1}$-norm.
More precisely, given $epsilon>0$, there exists a continuous function
$h:mathbb{R}rightarrowmathbb{R}$ such that $int_{0}^{1}|f-h|<epsilon$.

Let $epsilon>0$ be given. Choose continuous function $h:mathbb{R}rightarrowmathbb{R}$
such that $int_{0}^{1}|f-h|<frac{epsilon}{2M}.$ By part (a), there
exists $N$ such that $|int_{0}^{1}h(x)g(nx)dx|<frac{epsilon}{2}$
whenever $ngeq N$. Now let $ngeq N$ be aribitrary, then
begin{eqnarray*}
& & |int_{0}^{1}f(x)g(nx)dx|\
& leq & |int_{0}^{1}[f(x)-h(x)]g(nx)dx|+|int_{0}^{1}h(x)g(nx)dx|\
& leq & Mint_{0}^{1}|f(x)-h(x)|dx+frac{epsilon}{2}\
& < & epsilon.
end{eqnarray*}