Prove that if $f(x) = \int_{0}^x f(t)\,dt$, then $f(x) = 0$

Question

Prove that if $f(x) = \displaystyle\int_{0}^x f(t)\, dt$ for all $x$, then $f(x) = 0$.

I first differentiated to get $f'(x) = f(x) - f(0)$. Then by the mean value theorem there exists a $c$ in $(0,x)$ such that $f'(c)=\dfrac{f(x)-f(0)}{x}$. Thus, $f'(x) = xf'(c)$. What do I do from here?

A summary of the deliberation in the comments about the necessity of assumptions on $f$:

There was worry that we needed assumptions on $f$ such as continuity in order to exploit FTC.

Thanks to Aloizio and Clark for pointing out that no assumptions need be placed on $f$ as the integral of a Riemann integrable function is continuous. This gives us that $f$ is continuous (since by assumption we have that $f$ is integrable) and thus the Fundamental Theorem of Calculus applies.

Answer 1

Firstly, we know $f$ is continuous, since $\int_0^x f$ is a continuous function when $f$ is Riemann-integrable, or even Lebesgue-integrable (which I'm assuming is the case since otherwise this doesn't make sense). Now, we then have by the FTC that $f$ is differentiable, and it follows that $f'(x)=f(x)$ for all $x$ also due to the FTC. Since $f(0)=0$, we are done.

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"Order of information": We look at the right side, and see that it is a continuous function on $x$ due to the fact that $f$ is integrable. Therefore, the left side is continuous. Looking at the right side now, we have the integration up to $x$ of a continuous function, which is differentiable by the FTC.

Answer 2

Suppose $f$ is continuous. The fundamental theorem of calculus tells us that it is actually differentiable, and moreover by differentiating both sides we get

$$f'(x) = f(x)$$

So $f(x)$ is a function which is its own derivative, and hence is of the form $f(x) = ce^{x}$. What is the constant $c$? Simply compute

$$f(0) = \int_{0}^{0} f(x) = 0$$

to see that $c = 0$, and we are done.

Answer 3

As noted elsewhere, if $f(x)=\int_0^x f(t)\,dt$ for $x\ge 0$, then $f$ is necessarily continuous. Being continuous, there is a point $x_0\in[0,1/2]$ such that $f(x_0)\ge f(x)$ for all $x\in[0,1/2]$. Notice that $f(x_0)\ge f(0)=0$. We then have $$ f(x_0)=\int_0^{x_0}f(t)\,dt\le\int_0^{x_0}f(x_0)\,dt=x_0f(x_0). $$ Thus, either $f(x_0)=0$ or $1\le x_0$, which is absurd. We conclude that $f(x_0)=0$, and so $f(x)\le 0$ for all $x\in[0,1/2]$. In the same way $f(x)\ge 0$ for all $x\in[0,1/2]$. That is, $f$ vanishes on $[0,1/2]$. Repeat this argument on $[1/2,1], [1,3/2],\ldots$ to conclude that $f$ is identically $0$.

Answer 4

Assuming that differentiating $f$ is legal, and using the Fundamental Theorem of Calculus (FTC),

$$f' (x) = f(x)$$

Integrating,

$$f (x) = f_0 \exp (x)$$

We note that $f (0) = 0$. Hence, $f_0 = 0$ and, thus, $f (x) = 0$ for all $x \in \mathbb R$.

Answer 5

We have $f'=f.$

(1). We have $f(0)=\int_0^0f(x)\;dx=0.$

(2). For any $x,$if $f(x)=0$ then $f(y)=0$ for all $y\in [-1+x,1+x]. $ PROOF: For convenience, for $a\ne b$ let $In (a,b)$ denote the open interval between $a$ and $b$.

Now for $0<|y-x|< 1$ we have $$f(y)/(y-x)=(f(y)-f(x))/(y-x)=f'(y')=f(y')$$ for some $y' \in In (y,x),$ by the MVT and by $f'=f$. So there exists $y'\in In (y,x)$ satisfying $$f(y)=(y-x)f(y').$$

So there exists $(y_n)_{n\geq 0}$ with $y_0=y$, and $y_{n+1}\in In (y_n,x), $ where $y_{n+1}$ satisfies $f(y_n)=(y_n-x)f(y_{n+1}).$ Let $M=\max \{|f(z)|: 0\leq |z-x|\leq 1\}.$ Then $$|f(y)|=|f(y_{n+1})\prod_{i=1}^n(y_n-y)|\leq M |y-x|^n,$$ which $\to 0$ as $n\to \infty.$ Therefore $|y-x|<1\implies f(y)=0.$ Then by continuity of $f$ we also have $f(x-1)=f(x+1)=0.$

(3).By (1) and(2) we have $f^{-1}\{0\}\supset [-1,1].$

Now for $n\in N,$ if $f^{-1}\{0\}\supset [-n,n]$ then by (2) we have $f^{-1}\{0\}\supset [n-1,n+1]$ and $f^{-1}\{0\}\supset [-n-1,-n+1],$ giving $$f^{-1}\{0\}\supset [-n-1,-n+1]\cup [-n,n]\cup [n,n+1]=[-n-1,n+1].$$ Hence by induction on $n$ we have $$f^{-1}\{0\}=\mathbb R.$$