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#1 h May 3, 2025, 6:02 AM

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Find all continuous function $f, g\colon\mathbb{R}\to\mathbb{R}$ satisfied
$(x - k)f(x) = \int_k^x g(y)\mathrm{d}y$ for all $x\in\mathbb{R}$ and all $k\in\mathbb{Z}$ .

This post has been edited 1 time. Last edited by bakkune, May 3, 2025, 6:04 AM

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#2 h May 3, 2025, 7:03 AM

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$(x-k)f(x)+f'(x)=g(x)$ assuming that $f$ is differentiable, which is a linear first-order differential equation. The integrating factor is $\mu(x)=e^{\int (x-k)\,\text dx}=e^{x^2/2-kx},$ and the general solution is $f(x)=\frac{\int g(x)\mu(x)\,\text dx+C}{\mu(x)}=\left(\int e^{x^2/2-2k}g(x)\,\text dx+C\right)e^{xk-x^2/2}.$ All I learned to solve were differential equations, so I'm an idiot who doesn't know about general functional equations :fool:

:fool:

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#3 h May 3, 2025, 7:22 AM

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HacheB2031 wrote:

$(x-k)f(x)+f'(x)=g(x)$ assuming that $f$ is differentiable, which is a linear first-order differential equation. The integrating factor is $\mu(x)=e^{\int (x-k)\,\text dx}=e^{x^2/2-kx},$ and the general solution is $f(x)=\frac{\int g(x)\mu(x)\,\text dx+C}{\mu(x)}=\left(\int e^{x^2/2-2k}g(x)\,\text dx+C\right)e^{xk-x^2/2}.$ All I learned to solve were differential equations, so I'm an idiot who doesn't know about general functional equations :fool:

:fool:

Note that $k$ is not a fixed integer, while your solution of $f(x)$ contains $k$ , which seems wrong to me

Moreover, $f(x)$ might not be differentiable at $x = k\in\mathbb{Z}$ (however, it does differentiable at $x\notin\mathbb{Z}$ ).

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#4 h May 4, 2025, 3:58 AM

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It turn out that the problem is very simple, and that the integral is redundant. The problem could be generalized to: find all functions (not necessary continuous) $f, F\colon\mathbb{R}\to\mathbb{R}$ satisfied
$(x - k) f(x) = F(x) - F(k)$ for all $x\in\mathbb{R}$ and all $k\in\mathbb{Z}$ .

Solution. Denote the given expression by $E(x, k)$ . For a fixed integer $k$ , we have the following assertions:
$\begin{align*} E(k+1, k)\colon\quad & f(k+1) = F(k+1) - F(k), \\ E(k+2, k)\colon\quad & 2f(k+2) = F(k+2) - F(k), \\ E(k+2, k+1)\colon\quad & f(k+2) = F(k+2) - F(k+1). \end{align*}$ Thus, $2f(k+2) = f(k+2) + f(k+1)$ , or equivalently, $f(k+2) = f(k+1)$ . Since this holds for all $k\in\mathbb{Z}$ , we can deduce that $f(k) = f(0)$ for all $k\in\mathbb{Z}$ . Now, for a fixed real number $\alpha\in (0, 1)$ , we have the following assertions:
$\begin{align*} E(k+\alpha, k)\colon\quad & \alpha f(k+\alpha) = F(k+\alpha) - F(k), \\ E(k+\alpha, k+1)\colon\quad & (\alpha - 1)f(k+\alpha) = F(k+\alpha) - F(k+1). \end{align*}$ We then have $f(k + \alpha) = F(k + 1) - F(k) = f(k + 1) = f(0)$ . Since this holds for all $k\in\mathbb{Z}$ and $\alpha\in (0, 1)$ , we can conclude that $f(x) = f(0)$ for all $x\in\mathbb{R}$ , in other words, $f(x) = c$ for an arbitrary fixed constant $c\in\mathbb{R}$ . Finally, we have $F(x) - cx = F(k) - ck$ . Is it also clear that $F$ must be of the form $cx + b$ for an arbitrary fixed constant $b\in\mathbb{R}$ .

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