I was asked this question-
Let $f$ be a real continuous function satisfying $f(0)=0$ and for each natural number $n$
$$n^2\int_{x}^{x+\frac 1 n} f(t)\;\text{d}t=nf(x)+0.5$$
Then find the value of $f(1729)$
I couldn't make much of a progress. But I tried using
$$F(x)=\int_0^x f(x)\;\text{d}x$$
so that
$$\int_{x}^{x+\frac 1 n} f(t)\;\text{d}t=F\left(x+\frac 1 n\right)-F(x)$$
and then maybe letting $n$ go to infinity such that we can think about the behaviour of
$$\lim_{n\to \infty} n^2\cdot \left(F\left(x+\frac 1 n\right)-F(x)\right)$$
But, soon I realised that we don't even know whether this limit exists, since we don't have any idea of whether the limit in the RHS exists, i.e., we don't know whether
$$\lim_{n\to \infty} nf(x)$$
exists. However, if this would somehow exist, we could have argued that the limit of the LHS also exists, and thus,
$$\left(F\left(x+\frac 1 n\right)-F(x)\right)=\frac {h(x)}{n^2}$$
But still, there's no progress.
Another approach that I tried was to sum up the equation from $1$ to $k$ to get
$$\sum_{n=1}^k n^2\int_{x}^{x+\frac 1 n} f(t)\;\text{d}t = \frac {k(k+1)}2 f(x) + \frac k 2$$
But how to calculate the sum on the LHS? I tried writing it as
$$\sum_{n=1}^k n^2\cdot \left(F\left(x+\frac 1 n\right)-F(x)\right)=\sum_{n=1}^k n^2\cdot F\left(x+\frac 1 n\right) – \sum_{n=1}^k n^2\cdot F(x)$$
where the second term maybe easy to calculate, but what about the first term?
Any help would be appreciated. Also, I'm not really sure about the tags- feel free to edit them.
Best Answer
Note that the given condition can be written as $$\frac{F(x+1/n) - F(x)}{1/n} = f(x) + \frac{1}{2n}$$ Now differentiate both sides to get $$\frac{f(x+1/n) - f(x)}{(\color{red}{x+}1/n) \color{red}{-x} } = f'(x)$$ by mean value theorem, there exists $\xi_{n,x} \in (x, x+1/n)$ such that $$f'(\xi_{n,x}) = f'(x)$$ Hence, by Rolle's theorem, there exists $\eta_{n,x} \in (\xi_{n,x}, x)$ such that $$f''(\eta_{n,x}) = 0$$ Since $\xi_{n,x}$ squeezes $\eta_{n,x}$ to $x$ as $n\to\infty$, we have $\eta_{n,x} \xrightarrow{n\to\infty} x$. It follows that $$0 = \lim_{n\to\infty}f''(\eta_{n,x}) = f''(x)$$ for all $x$. Hence, $f(x)$ is a linear function.
Now, let $f(x) = ax + b$ and finish it.