Prove $\lim_{n \to \infty} \int_a^b f(x+y_n)g(x)dx = \int_a^bf(x)g(x)dx$

Question

Prove $\displaystyle \lim_{n \to \infty} \int_a^b f(x+y_n)g(x)dx = \int_a^bf(x)g(x)dx$ where:

$\displaystyle f:[a,b+1] \to \mathbb{R}, g:[a,b] \to \mathbb{R} $ are continuous. $\displaystyle 0 \le y_n \le 1, \forall n \in \mathbb{N}, \lim_{n\to\infty}y_n = 0$.

We haven't covered uniform convergence yet but it seems that if I could prove $f(x + y_n)$ uniformly converges to $f(x)$ then I could give the following:

EDIT: $f$ is continuous over $[a,b+1]$ which is compact so it is uniformly continuous. Additionally $\forall n \in \mathbb{N}, 0 \le y_n \le 1 $. Thus $x + y_n \in [a.b+1], \forall x \in [a,b], \forall n \in \mathbb{N}$. Let $\varepsilon > 0,$ Let $y = x + y_n$, $\exists \delta > 0, \forall x ,y \in [a,b+1], |x - y| < \delta \implies |f(x) - f(y)| < \varepsilon$. Needs some work.

$\displaystyle \lim_{n \to\infty} \int_a^b f(x+y_n)g(x) dx = \int_a^b \lim_{n \to\infty}f(x+y_n)g(x) dx = \int_a^b f(x)g(x) dx$.

Without Uniform Convergence: I am not sure what to use without uniform convergence.

Answer 1

I don't think you need uniform convergence here.

Let $$A_n = \int_a^bf(x+y_n)g(x)dx\\A = \int_a^bf(x)g(x)dx.$$

Then since both functions are nice so that both integrals exist, we have$$ |A_n-A| = \left|\int_a^b(f(x+y_n)-f(x))g(x)dx\right|\le \int_a^b \left|f(x+y_n)-f(x)\right||g(x)|dx.$$ Since $f(x)$ is uniformly continuous, we can pick an $N$ such that $|f(x+y_n)-f(x)| < \epsilon$ for all $n>N$ and $x\in [a,b]$ so $$|A_n-A| < \epsilon \int_a^b |g(x)|dx.$$

(Since $g(x)$ is nice, the integral is just a number so we're done.)

Answer 2

As you've said, since $f$ and $g$ are continuous over compact intervals, they are uniformly continuous over those intervals. We want to show that the product $fg$ is uniformly continuous so that we can push the limit into the integral.

For $\epsilon > 0$, there exists $\delta > 0$ such that $|x-y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon$ and $|g(x) - g(y)| < \epsilon$. We can write \begin{align*} |f(x)g(x) - f(y) g(y)| &\leq |f(x)g(x) - f(x)g(y)| + |f(x)g(y) - f(y) g(y)| \\ &= |f(x)|\cdot |g(x) - g(y)| + |g(y)| \cdot |f(x) - f(y)| \\ &< \max(f) \cdot \epsilon + \max(g) \cdot \epsilon \\ &< M \epsilon \to 0 \end{align*} for some $M < \infty$. The maxima of $f$ and $g$ are attained since the functions are continuous on a closed interval. Thus $fg$ is uniformly continuous, and we conclude that \begin{align*} \lim_n \int f(x + y_n) g(x) dx = \int \lim_n f(x+ y_n) g(x) dx \stackrel{(*)}{=} \int f(x)g(x)dx \end{align*} where $(*)$ follows from the continuity of $f$.