# $\lim _{n \to \infty} \int_{0}^1 f \left(\frac{x}{n}\right) dx =0$ find $f(0) ?$

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 .

• 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. Commented Nov 14, 2018 at 10:57
• Do you intuitively understand what the answer is? Have you tried some simple functions for $f$, and some small numbers for $n$? Commented Nov 14, 2018 at 11:02
• @EeveeTrainer: no, that would be with an argument of the form $k/n$ and an ordinary summation.
– user65203
Commented Nov 14, 2018 at 11:25

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 $$-\varepsilon for all $$-\delta.

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

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

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

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, $$\lim _{n \to \infty} \int_{0}^1 f \left(\frac{x}{n}\right) \, dx \ge a > 0$$

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 $$f(0)=0$$.

• Can you explain how does the fundamental theorem of calculus imply $\lim_{h\to 0}\frac {1}{h}\int_{0}^{h}f(t)\,dt=f(0)$? Commented Nov 15, 2018 at 8:18
• @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 $F'(0)=\lim_{h\to 0} \dfrac{1}{h}\int_{0}^{h}f(t)\,dt$. Commented Nov 15, 2018 at 11:58