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Let $f\in \mathcal{C}^0\big([0,1],\mathbb{R}\big)$ and, for every $h\in(0,1]$, $$I(h)=\int_0^1\dfrac{f(hx)}{x^2+1}dx$$

For $\varepsilon >0$, show there exists $\eta>0$ such that for every $h\in(0,\eta)$,

$$\left|I(h)-f(0)\frac{\pi}{4}\right|\leq \varepsilon$$

Since $\dfrac{\pi}{4} = \displaystyle\int_0^1\dfrac{1}{x^2+1}dx$

$$\left |\int_0^1\dfrac{f(hx)}{x^2+1}dx -\int_0^1\dfrac{f(0)}{x^2+1}dx \right|\leq\varepsilon$$

And now, I have got no idea how to solve this problem. I think, I should show that :

$$|h|\leq \eta \implies \left|I(h)-f(0)\frac{\pi}{4}\right|\leq \varepsilon$$

but I'm not sure.

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Note that we can write

$$\begin{align} \left|\int_0^1 \frac{f(hx)}{x^2+1}\,dx-f(0)\frac\pi4\right|&=\left|\int_0^1 \frac{f(hx)-f(0)}{x^2+1}\,dx\right|\\\\ &\le \int_0^1 \frac{|f(hx)-f(0)|}{x^2+1}\,dx\\\\ & \le\frac\pi4 \sup_{x\in [0,1]}|f(hx)-f(0)| \end{align}$$

For any $\epsilon>0$, there exists a number $\eta>0$ such that $|f(hx)-f(0)|<4\epsilon/\pi$ whenever $|hx|<\eta$.

Since $x\in [0,1]$, then whenever $|hx|<|h|<\eta$, we have $|f(hx)-f(0)|<4\epsilon/\pi$.

Hence, $\sup_{x\in [0,1]}|f(hx)-f(0)|<4\epsilon/\pi$ whenever $|h|<\eta$.

Putting it all together, we have for $|h|<\eta$,

$$\left|\int_0^1 \frac{f(hx)}{x^2+1}\,dx-f(0)\frac\pi4\right|<\epsilon$$

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  • $\begingroup$ Tanks for this answer but how to show, please $$\int_0^1 \frac{|f(hx)-f(0)|}{x^2+1}\,dx \le \frac\pi4 \sup_{x\in [0,1]}|f(hx)-f(0)|$$ $\endgroup$ – Stu Aug 17 '17 at 19:41
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    $\begingroup$ You're welcome. My pleasure. Well, first $\left|\int_1^1 g(x)\,dx\right|\le \int_0^1 |g(x)|\,dx$. Second, $\int_0^1 \frac{1}{x^2+1}\,dx=\frac\pi 4$. Third, $\int_0^1 |g(x)h(x)|\,dx\le \sup_{x\in [0,1]}|g(x)|\int_0^1 |h(x)|\,dx$. Is that sufficient for you to continue? $\endgroup$ – Mark Viola Aug 17 '17 at 19:43
  • $\begingroup$ Understood and clever!! $\endgroup$ – Stu Aug 17 '17 at 19:48
  • $\begingroup$ Over a scale from 1 to 5, may you evaluate the difficulty of this exercise, please? $\endgroup$ – Stu Aug 17 '17 at 19:53
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    $\begingroup$ @Stu Difficulty is relative. This was relatively easy for me now. But this might be difficult for those less experienced. $\endgroup$ – Mark Viola Aug 17 '17 at 19:55
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One way to knock this out is the dominated convergence theorem. Take any sequence $(h_n)$ of non-zero numbers such that $h_n \to 0$ and define $$g_n(x) = \frac{f(h_nx)}{1+x^2}, \,\,\,\, x \in [0,1].$$ Then $$g_n(x) \to g(x) = \frac{f(0)}{1+x^2}$$ pointwise on $[0,1]$ and since $f$ is continuous, it is bounded so $$\lvert g_n(x) \rvert \le \frac{M}{1+x^2} \le M$$ for some constant $M$. Since the range of integration is compact, we can apply the dominated convergence theorem to show that $$\lim_{n\to\infty} \int^1_0 g_n(x) dx = \int^1_0 \lim_{n\to \infty} g_n(x) dx = f(0) \frac \pi 4$$ as desired.

Alternatively, if you need to do it directly, notice that \begin{align*}\left \lvert \int^1_0 \frac{f(hx)}{1+x^2}dx - \int^1_0\frac{f(0)}{1+x^2}dx \right \rvert &= \left \lvert\int^1_0 \frac{f(hx) - f(0)}{1+x^2} dx \right \rvert \\ &\le \int^1_0 \frac{\lvert f(hx) - f(0) \rvert}{1+x^2}dx. \end{align*} Since $f$ is continuous on a compact set, it is uniformly continuous. Hence for any $\epsilon > 0$, there is $\delta > 0$ such that for all $x,y$ with $\lvert x - y \rvert < \delta$, we have $\lvert f(x) - f(y) \rvert < \epsilon$. Then for $0 < h < \delta$, we will also have $0 < hx < \delta$ (for any $x \in (0,1)$) and so $\lvert f(hx) - f(0) \lvert < \epsilon.$ For such $h$, we see \begin{align*}\left \lvert \int^1_0 \frac{f(hx)}{1+x^2}dx - \int^1_0\frac{f(0)}{1+x^2}dx \right \rvert &\le \int^1_0 \frac{\epsilon}{1+x^2}dx = \epsilon\frac \pi 4. \end{align*} which is the same thing you wanted up to rescaling $\epsilon$.

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  • $\begingroup$ Uniform continuity is not needed to bound $|f(hx)-f(0)|$ uniformly on $x$ in $(0,1)$, only the (simple) continuity of $f$ at $0$. $\endgroup$ – Did Aug 17 '17 at 19:21
  • $\begingroup$ Thanks for the note! I noticed this after I posted and actually began to edit in order to change that but it didn't feel necessary since it wasn't really an error; I just used a stronger property to subsume a weaker one. $\endgroup$ – User8128 Aug 17 '17 at 19:32
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    $\begingroup$ You need not invoke a theorem as powerful as the DCT here. $\endgroup$ – Mark Viola Aug 17 '17 at 19:38
  • $\begingroup$ Sure, that's why I gave the "direct proof" as well. $\endgroup$ – User8128 Aug 17 '17 at 19:47

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