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let $f:\mathbb{R} \to \mathbb{R}$ be a continuous function such that $$\lim _{n \to \infty} \int_{0}^1 f \left(\frac{x}{n}\right) dx =0$$

Then what we can say about $f(0) \ $ ??

how to approach this problem . Any hint .

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  • $\begingroup$ My gut instinct on seeing that $x/n$ would be to express the integral as some sort of Riemann sum. I don't know if it'll help, but it's something. $\endgroup$ Commented Nov 14, 2018 at 10:57
  • $\begingroup$ Do you intuitively understand what the answer is? Have you tried some simple functions for $f$, and some small numbers for $n$? $\endgroup$ Commented Nov 14, 2018 at 11:02
  • $\begingroup$ @EeveeTrainer: no, that would be with an argument of the form $k/n$ and an ordinary summation. $\endgroup$
    – user65203
    Commented Nov 14, 2018 at 11:25

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You have to couple the continuity of the function $f$ with the convergence of the series:

Let $\varepsilon>0$. Thus, there exists a $\delta>0$ such that $-\varepsilon<f(y)-f(0)<\varepsilon$ for all $-\delta<y<\delta$.

Since in the integral, $0<x<1$ and therefore $0<x/n<1/n$ there exists an $N$ such that $x/n<\delta$ for all $n\geq N$. This implies that:

$f(0)-\varepsilon<\int_{0}^1 f \left(\frac{x}{n}\right) dx<f(0)+\varepsilon$ for all $n\geq N$.

Which finally yields $\lim_{n\to\infty} \int_{0}^1 f \left(\frac{x}{n}\right) dx = f(0)$.

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We must have $f(0)=0$.

Suppose wlog that $f(0)>0$. Fix an interval $[a,b]$ with $a>0$ and $f(0)\in [a,b]$.

Since $f$ is continuous, there is $\delta>0$ such that $f(x) \in [a,b]$ whenever $|x| < \delta$.

But then for all $n$ such that $1/n < \delta$ we have $$ \int_{0}^1 f \left(\frac{x}{n}\right) \, dx \ge \int_{0}^1 a \, dx = a $$ Therefore, $$ \lim _{n \to \infty} \int_{0}^1 f \left(\frac{x}{n}\right) \, dx \ge a > 0 $$

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The key here is the Fundamental Theorem of Calculus.

Consider the substitution $x=nt$ and then the given limit condition is equivalent to $$\lim_{n\to\infty} n\int_{0}^{1/n}f(t)\,dt=0$$ On the other hand since $f$ is continuous at $0$ we have via Fundamental Theorem of Calculus $$\lim_{h\to 0}\frac {1}{h}\int_{0}^{h}f(t)\,dt=f(0)$$ Putting $h=1/n,n\in\mathbb {N} $ we get $$\lim_{n\to\infty} n\int_{0}^{1/n}f(t)\,dt=f(0)$$ It should now be clear that $f(0)=0$.

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  • $\begingroup$ Can you explain how does the fundamental theorem of calculus imply $\lim_{h\to 0}\frac {1}{h}\int_{0}^{h}f(t)\,dt=f(0)$? $\endgroup$
    – Lemon
    Commented Nov 15, 2018 at 8:18
  • $\begingroup$ @Hawk: consider $F(x) =\int_{0}^{x}f(t)\,dt$. Then FTC says that $F'(x) =f(x) $ and in particular $F'(0)=f(0)$. By definition $F'(0)=\lim_{h\to 0} \dfrac{1}{h}\int_{0}^{h}f(t)\,dt$. $\endgroup$
    – Paramanand Singh
    Commented Nov 15, 2018 at 11:58

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