Examine linear dependence of f_1(x)=x, f_2(x)=x^3…

Examine linear dependence of $f_1(x)=x,\qquad f_2(x)=x^3,\qquad f_3(x)=|x^3|$.

Step 1: Preliminary observation

Since $|x^3|=|x|^3$, \[f_3(x)=|x^3|=\begin{cases}x^3,&x\ge0,\\-x^3,&x<0.\end{cases}\]

$f_3$ is continuous everywhere and, in fact, twice differentiable everywhere ($f_3''(x)=6|x|$ is continuous, even at $x=0$, where $f_3''(0)=0$); but its third derivative fails to exist at $x=0$, so $f_3$ is not real‑analytic there. This is exactly the feature that makes the case interesting. Dependence is examined separately on $(0,+\infty)$, on $(-\infty,0)$, and on all of $\mathbb{R}$.

Step 2: On each half-line

Note the two relations are different: $c_3=-c_2$ is needed on the right, $c_3=+c_2$ on the left. A single triple $(c_1,c_2,c_3)$ cannot satisfy both unless $c_2=c_3=0$ – which is exactly what forces independence globally.

Step 3: On all of $\mathbb{R}$

Suppose $c_1x+c_2x^3+c_3|x^3|\equiv0$ for every $x\in\mathbb{R}$. Splitting into the two half-lines (where $x,x^3$ are themselves independent, by Problem 61, $3^\circ$): \[x>0:\quad c_1x+(c_2+c_3)x^3\equiv0\ \Longrightarrow\ c_1=0,\ \ c_2+c_3=0.\]

\[ x<0:\quad c_1x+(c_2-c_3)x^3\equiv0\ \Longrightarrow\ c_1=0,\ \ c_2-c_3=0. \]

Adding and subtracting $c_2+c_3=0$ and $c_2-c_3=0$ gives $c_2=0$ and $c_3=0$. So $c_1=c_2=c_3=0$ is the only solution.

Equivalent parity argument: $x$ and $x^3$ are odd functions while $|x^3|$ is even. If $c_1x+c_2x^3+c_3|x^3|\equiv0$, the odd part $c_1x+c_2x^3$ and the even part $c_3|x^3|$ must each vanish identically on their own (the zero function is the unique function that is both), giving $c_1=c_2=0$ (odd polynomials independent) and $c_3=0$ (since $|x^3|\not\equiv0$).

The three functions are linearly independent on $\mathbb{R}$.

Step 4: The Wronskian on $\mathbb{R}$

For $x>0$, $f_3=f_2=x^3$, so columns 2 and 3 of $W$ are identical: \[W(x)=\begin{vmatrix}x&x^3&x^3\\1&3x^2&3x^2\\0&6x&6x\end{vmatrix}=0\qquad(x>0).\]

For $x<0$, $f_3=-f_2$, so column 3 is $(-1)$ times column 2, and again $\det=0$. At $x=0$, using the actual values $f_1(0)=0,f_1'(0)=1,f_1''(0)=0$ and $f_2(0)=f_2'(0)=f_2''(0)=0=f_3(0)=f_3'(0)=f_3''(0)$, the first row of $W(0)$ is all zeros, so $W(0)=0$ too.

Thus $W(x)\equiv0$ on all of $\mathbb{R}$.

Step 5: Summary