If $f$ has a continuous second derivative, prove that $f(x) = f(0) + f'(0) + \int_0^x (x-t)f''(t)dt$.

Question

This has been asked before, but I am looking for a proof that does not use integration by parts. Instead, I am looking to use what my textbook called the "Identity Criterion", which states that if the derivative of two functions are equal, then they differ at most by a constant.

Here is what I have thus far: By the Second Fundamental Theorem, $$f'(x) = \int_0^x f''(t)dt.$$ Adding something inconspicuous to the right side, $$f'(x) = xf''(x) + \int_0^x f''(t)dt - xf''(x) = \frac{d}{dx} \left[ x\int_0^x f''(t)dt - \int_0^x tf''(t)dt \right] = \dfrac{d}{dx} \left[ \int_0^x (x-t)f''(t)dt \right].$$ By the Identity Criterion, $f(x)$ and $\int_0^x (x-t)f''(t)dt$ must only differ by a constant. But by the First Fundamental Theorem, $$\int_0^x (x-t)f''(t)dt = $$

This is where I start to lose the plot. Any guidance would be useful.

Answer 1

Why not just integration by parts?

\begin{align*} f(x)-f(0)&=\int_{0}^{x}f'(t)dt\\ &=\int_{0}^{x}-\dfrac{d}{dt}(x-t)f'(t)dt\\ &=-(x-t)f'(t)\bigg|_{t=0}^{t=x}+\int_{0}^{x}(x-t)f''(t)dt\\ &=xf'(0)+\int_{0}^{x}(x-t)f''(t)dt. \end{align*}

Edit:

Let \begin{align*} \varphi(x)&=f(x)-\left(f(0)+f'(0)x+\int_{0}^{x}(x-t)f''(t)dt\right)\\ \varphi'(x)&=f'(x)-f'(0)-\int_{0}^{x}f''(t)dt-xf''(x)+xf''(x)\\ \varphi'(x)&=f'(x)-f'(0)-(f'(x)-f'(0))\\ \varphi'(x)&=0, \end{align*} and so $\varphi(x)=c$, but $\varphi(0)=0$ and hence $c=0$.

Answer 2

There is mistake in your formula. The second term is $xf'(0)$ instead of $f'(0)$.

Let $F(x)=\int_0^{x} (x-t)f''(t)dt=x\int_0^{x} f''(t)dt-\int_0^{x} tf''(t)dt$. Then $F'(x)=\int_0^{x} f''(t)dt+xf''(x)-xf''(x)$ by product rule for differentiation. Hence $F'(x)=f'(x)-f'(0)$. Hence $F(x)-f(x)+xf'(0)$ has derivative $0$ which implies it is a constant. Put $x=0$ to see that $c=-f(0)$. This gives $F(x)=f(x)+f(0)-xf'(0)$. This gives the desired identity.