Are Linear Transformations of Derivatives and Integrations Linearly Independent?

Problem 463

Let $W=C^{\infty}(\R)$ be the vector space of all $C^{\infty}$ real-valued functions (smooth function, differentiable for all degrees of differentiation).
Let $V$ be the vector space of all linear transformations from $W$ to $W$.
The addition and the scalar multiplication of $V$ are given by those of linear transformations.

Let $T_1, T_2, T_3$ be the elements in $V$ defined by
\begin{align*}
T_1\left(\, f(x) \,\right)&=\frac{\mathrm{d}}{\mathrm{d}x}f(x)\\[6pt]
T_2\left(\, f(x) \,\right)&=\frac{\mathrm{d}^2}{\mathrm{d}x^2}f(x)\\[6pt]
T_3\left(\, f(x) \,\right)&=\int_{0}^x \! f(t)\,\mathrm{d}t.
\end{align*}
Then determine whether the set $\{T_1, T_2, T_3\}$ are linearly independent or linearly dependent.

We prove that the set $\{T_1, T_2, T_3\}$ are linearly independent.
Suppose that we have a linear combination
\[c_1T_1+c_2T_2+c_3 T_3=0,\tag{*}\]
for some real numbers $c_1, c_2, c_3$.
Note that $0$ is the zero linear transformation $0(f(x))=0$ for all $x$, and it is the zero vector in the vector space $V$.
The equality (*) is an equality as linear transformations.

We need to show that $c_1=c_2=c_3=0$.

Let us consider the function $f(x)=1$ in $C^{\infty}(\R)$.
Then we have
\begin{align*}
T_1\left( 1 \right)&=0\\
T_2\left( 1 \right)&=0\\
T_3\left( 1 \right)&=\int_{0}^x \! 1\,\mathrm{d}t=x.
\end{align*}
These yield that
\begin{align*}
0=c_1T_1(1)+c_2T_2(1)+c_3 T_3(1)=c_3x,
\end{align*}
and hence $c_3=0$.

Next, consider the function $f(x)=x$.
We have
\begin{align*}
T_1\left( x \right)=1\\
T_2\left( x \right)=0
\end{align*}
and
\begin{align*}
0=c_1T_1(x)+c_2T_2(x)=c_1.
\end{align*}
Thus $c_1=0$.

Finally, consider the function $f(x)=x^2$.
We compute
\begin{align*}
T_2\left( x^2\right)=2,
\end{align*}
and the equality (*) together with $c_1=c_3=0$ gives
\begin{align*}
0=c_2T_2\left( x^2 \right)=2c_2.
\end{align*}

So $c_2=0$, and we have proved that $c_1=c_2=c_3=0$.
Thus we conclude that the set $\{T_1, T_2, T_3\}$ is linearly independent.

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For each specific $f(x)$, we get some restrictions on $c_1, c_2, c_3$. The choice $f(x) = e^x$ gives the restriction $c_1+c_2+c_3 = 0$ but this is not strong enough to show that $c_1=c_2=c_3=0$. Also this does not mean that $T_1, T_2, T_3$ are dependent. This just say that for your choice of $f(x)$, the coefficient must be like this. You could use other functions $f(x)$ to get other restrictions and solve them to get $c_1+c_2+c_3 = 0$. In my case, I used three functions $f(x)= 1, x, x^2$ and get three equations which yields $c_1=c_2=c_3=0$.

What if we use f(x)=e^x then we have c1*e^x+c2*e^x+c3*e^x=0 => c1=-1, c2=1/2, c3=1/2, and the are linearly dependent.

For each specific $f(x)$, we get some restrictions on $c_1, c_2, c_3$. The choice $f(x) = e^x$ gives the restriction $c_1+c_2+c_3 = 0$ but this is not strong enough to show that $c_1=c_2=c_3=0$. Also this does not mean that $T_1, T_2, T_3$ are dependent. This just say that for your choice of $f(x)$, the coefficient must be like this. You could use other functions $f(x)$ to get other restrictions and solve them to get $c_1+c_2+c_3 = 0$. In my case, I used three functions $f(x)= 1, x, x^2$ and get three equations which yields $c_1=c_2=c_3=0$.