Problem 1.1 (A function that is $C^2$ but not $C^3$)

Question

A function that is $C^2$ but $\operatorname{not} C^3$
Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be the function in Example 1.2(iii). Show that the function $h(x)=\int_0^x g(t) d t$ is $C^2$ but not $C^3$ at $x=0$.

Toğrul Topal · Accepted Answer

\begin{figure} \centering
\begin{tikzpicture}
    \begin{axis}[
        width=5cm,
        xlabel=$x$,
        ylabel=$y$,
        axis lines=middle,
        xmin=0,xmax=2,
        ymin=0,ymax=5,
        clip=false
    ]
        \addplot[domain=0.02:2,samples=100,blue] {1/3*(x^(-2/3))};
        \legend{$f'(x)$}
    \end{axis}
\end{tikzpicture}
\begin{tikzpicture}
    \begin{axis}[
        width=5cm,
        xlabel=$x$,
        ylabel=$y$,
        axis lines=middle,
        xmin=0,xmax=2,
        ymin=0,ymax=5,
        clip=false
    ]
        \addplot[domain=0:2,red] {x^(1/3)};
        \legend{$f(x)$}
    \end{axis}
\end{tikzpicture}
\begin{tikzpicture}
    \begin{axis}[
        width=5cm,
        xlabel=$x$,
        ylabel=$y$,
        axis lines=middle,
        xmin=0,xmax=2,
        ymin=0,ymax=5,
        clip=false
    ]
        \addplot[domain=0:2,green] {0.75*x^(4/3)};
        \legend{$g(x)$}
    \end{axis}
\end{tikzpicture}
\end{figure}

The function $h: \mathbb{R} 	o \mathbb{R}$ given by $h(x)=\int_0^x g$ has its derivative given by $h'=g$ by the fundamental theorem of calculus. Since $g$ is once differentiable, $h$ must be twice differentiable. However, since $g$ is not twice differentiable (at $x=0$), and since $h^{\prime\prime} = g^{\prime}$, $h$ cannot be $C^3$.

Problem 1.1 (A function that is $C^2$ but not $C^3$)

Answers

Comments

Problem 1.1 (A function that is $C^2$ but not $C^3$)

Answers

Comments

Add answer