Riesz representation theorem: an application Let $L_n\colon C_c(\mathbb{R})\rightarrow \mathbb{R}$ with
$$ L_nf = \frac{1}{n}\sum_{k=0}^{n}  e^{k/n} f\left(\frac{k}{n}\right).  $$
Prove that $\lim_{n\rightarrow \infty}L_nf=:Lf$ exists and then apply the Riesz representation theorem to $L$.

Until now I proved the existence of the limit. 


  Do you have any idea of how to continue the exercise?
Using the Riemann sum, one readly observes that 
$$ \lim_{n\to\infty}L_nf = \lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n}  e^{k/n} f\left(\frac{k}{n}\right) =\int_0^1e^xf(x)dx $$
for each $f\in C_c(\Bbb R). $ whence set $$Lf= \int_0^1e^xf(x)dx $$

Since $x\mapsto e^x$ is in $L^2(0,1)$ there is not problem if we extend $L$ by taking $f\in L^2(0,1)$

that is 
$$Lf= \int_0^1e^xf(x)dx=\langle f,e^x\rangle_{L^2} ~~~~~ f\in L^2(0,1).$$
this is directly the Riesz representation theorem.
Using the Riemann sum, one readly observes that 
$$ \lim_{n\to\infty}L_nf = \lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n}  e^{k/n} f\left(\frac{k}{n}\right) =\int_0^1e^xf(x)dx $$
for each $f\in C_c(\Bbb R). $ whence set $$Lf= \int_0^1e^xf(x)dx $$

Since $x\mapsto e^x$ is in $L^2(0,1)$ there is not problem if we extend $L$ by taking $f\in L^2(0,1)$

that is 
$$Lf= \int_0^1e^xf(x)dx=\langle f,e^x\rangle_{L^2} ~~~~~ f\in L^2(0,1).$$
this is directly the Riesz representation theorem.