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Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $g(x) = \int^x_0 f(t) \, dt = \int^x_0 t^{1/3} \, dt = \frac{3}{4}x^{4/3}$, where $f(t) = t^{1/3}$.

Show that the function $h(x) = \int^x_0 g(t) \, dt$ is $C^2$ but not $C^3$ at $x = 0$.

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We need to prove that $\dfrac{9 x^{7/3}}{28}$ is discontinuous at the origin, no? – J. M. Nov 21 '11 at 16:41
Function $f$ is $C^1$ but not $C^0$ at $x = 0$ as $f'(x) = \frac{1}{3}x^{-2/3}$ for $x \neq 0$, and 'undefined' for $x = 0$. – UGPhysics Nov 21 '11 at 16:48
@UGPhysics: You have it the wrong way around: $f$ is $C^0$, but not $C^1$. – Arturo Magidin Nov 21 '11 at 17:33
Yes; now I see. Thank you. – UGPhysics Nov 21 '11 at 17:40
up vote 1 down vote accepted

$f(x)$ is continuous, but not differentiable at $0$; so $f$ is in $\mathcal{C}^0$, but not in $\mathcal{C}^1$, with the problem being at $x=0$.

Since $f(t)$ is continuous by the Second Fundamental theorem of Calculus we have that $g(x)$ is differentiable and: $$\frac{d}{dx} g(x) = \frac{d}{dx}\int_0^x f(t)\,dt = f(x),$$ so $g(x)$ has continuous derivative; but the derivative is not differentiable at $0$. That is, $g\in\mathcal{C}^1$, but not in $\mathcal{C}^2$, with the problem being at $x=0$.

Since $g(x)$ is continuous (it is differentiable), then again by the Second Fundamental Theorem of Calculus we know that $h(x)$ is differentiable, and $$\frac{d}{dx}h(x) = \frac{d}{dx}\int_0^x g(t)\,dt = g(x).$$ Therefore, $\frac{d^2}{dx^2}h(x) = \frac{d}{dx}h'(x) = \frac{d}{dx}g(x) = f(x)$, so $h(x)$ has continuous second derivative; that is, $h\in\mathcal{C}^2$.

However, the second derivative is not differentiable everywhere (not differentiable at $0$), so $h(x)$ does not have a third derivative defined everywhere, and so cannot be $\mathcal{C}^3$, with the problem being at $x=0$.

More generally, if $f(x)$ is in $\mathcal{C}^k$ but not in $\mathcal{C}^{k+1}$, then $\mathcal{F}(x) = \int_0^x f(t)\,dt$ is in $\mathcal{C}^{k+1}$ but not in $\mathcal{C}^{k+2}$, by the Second Fundamental Theorem of Calculus.

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