If i assume the sentence:
$$ \text{If} \ f \in L^{1}(0,1) \ \text{then} \ \displaystyle\lim_{n \rightarrow \infty} \displaystyle\int_{0}^{1} f(x) \sin (nx) \ dx = 0 \ (n \in N)$$
is obvious to conclude that
$$\text{If} \ f \in L^{1}(0,1) \ \text{then} \ \displaystyle\lim_{\lambda \rightarrow \infty} \displaystyle\int_{0}^{1} f(x) \sin (\lambda x) \ dx = 0 \ (\lambda \in R)$$
I am not seeing how to show this...
Someone can give me a hint ?
Thanks in advance
Consider the function $$F(\lambda):=\int_0^1 f(x) \sin (\lambda x) \mathrm{d}x. $$ The sentence you have written merely proves that $F(\lambda) \to 0$ along the sequence $\lambda_n=n$. If you want to prove that the limit exists as $\lambda \to \infty$ continuously, you must show that for any sequence $(\lambda_n)_{n=1}^\infty$ tending to $\infty$, $F(\lambda_n) \to 0$ (This is the Heine definition of continuity). In short, the answer is no, it is not that obvious to generalize your statement.
However, there is a more general statement of the Riemann-Lebesgue lemma, from which the conclusion is immediate.
