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.