Prove the limit of parametric integral:

Question

Suppose that:
1). $\varphi_{n}(x) \geq 0$, $\forall n \in \mathbb{N}$ on $[-1,1]$.
2). $\varphi_{n}(x) \rightrightarrows 0$ as $n \to \infty$ on $(0< \varepsilon \leq |x|\leq 1)$.
3). $\int_{-1}^{1} \varphi_{n}(x)dx \to 1$ as $n \to \infty$.
Prove that if $f \in C[-1,1]$, then $$\lim_{n \to \infty} \int_{-1}^{1} f(x)\varphi_{n}(x)dx=f(0)$$

Answer 1

Here's an answer, modulo a few details, under the assumption that $\rightrightarrows$ means 'converges to'.

Let's split this up into two parts: where $|x|<\varepsilon$ and $|x|\geq \varepsilon$: $$ \int _{-1}^1 f(x)\varphi_n(x)\,dx = \int _{|x|<\varepsilon} f(x)\varphi_n(x)\,dx +\int _{|x|\geq \varepsilon} f(x)\varphi_n(x)\,dx $$ By the First Mean Value Theorem for Integrals, since $\varphi_n$ is non-negative, we have $$ \int _{|x|<\varepsilon} f(x)\varphi_n(x)\,dx +\int _{|x|\geq \varepsilon} f(x)\varphi_n(x)\,dx $$ $$ = f(\xi_1)\int _{|x|<\varepsilon} \varphi_n(x)\,dx + (f(\xi_2)+f(\xi_3))\int _{|x|\geq \varepsilon}\varphi_n(x)\,dx, $$for some $\xi_1,\xi_2,\xi_3$. Now by the second assumption, we can choose $n$ such that the second integral is arbitrarily small, using the continuity and hence boundedness of $f$ on $[-1,1]$. The first integral will integrate to $1$ by assumption; however, $\xi_1\in(-\varepsilon,\varepsilon)$. In the limit the first term then approaches $f(0)\cdot 1$, as we'd hope.

Answer 2

We compute

\begin{align*} \left| \int_{-1}^1 f(x) \varphi_n(x)dx - f(0) \right| &\le \left| \int_{-1}^1 f(x) \varphi_n(x)dx - \int_{-1}^1 f(0)\varphi_n(x)dx \right| + \left| \int_{-1}^1 f(0) \varphi_n(x)dx - f(0)\right| \\ &= \left| \int_{-1}^1 (f(x) -f(0))\varphi_n(x)dx\right| + |f(0)| \left| \int_{-1}^1 \varphi_n(x)dx -1\right| \end{align*}

and the second term goes to $0$ from assumption $3)$. For the first term, we have

\begin{align*} \left| \int_{-1}^1 (f(x) -f(0))\varphi_n(x)dx\right| &\le \left| \int_{-\varepsilon}^\varepsilon (f(x) -f(0))\varphi_n(x)dx\right| + \left| \int_{-1}^{-\varepsilon}(f(x) -f(0))\varphi_n(x)dx\right| + \left| \int_\varepsilon^1 (f(x) -f(0))\varphi_n(x)dx\right| \end{align*}

and by the dominated convergence theorem and assumption $2)$ the second and third terms go to $0$. Finally, since $f$ is continuous on $[-1,1]$ it is also uniformly continuous on $[-1,1]$ so given any $\delta > 0$ there exists $\varepsilon > 0$ such that $\sup_{|x| < \varepsilon} |f(x)-f(0)| < \delta$ so

\begin{align*} \left| \int_{-\varepsilon}^\varepsilon (f(x) -f(0))\varphi_n(x)dx\right| &\le \int_{-\varepsilon}^\varepsilon |f(x) -f(0)|\varphi_n(x)dx \\ &\le \delta \int_{-\varepsilon}^\varepsilon \varphi_n(x)dx \\ &\le \delta \int_{-1}^1 \varphi_n(x)dx. \end{align*}

Since $\delta$ was arbitrary and $\sup_{n}\int_{-1}^1 \varphi_n(x)dx < \infty$, we conclude

\begin{align*} \lim_{n \rightarrow \infty} \left| \int_{-1}^1 f(x) \varphi_n(x)dx - f(0) \right| &=0 \end{align*}

as desired.