Finding null space and image Let $T(f(x)) = g(x) = \int_{-1}^{1}f(t)(t-x)^2dt$ be a linear transformation from $V$ to $V$ where $V$ is the vector space of continuous functions on $[-1 , 1]$ . Find $\text{Nul}(T)$ and $\text{Im}(T)$ . I really have no idea about the solution . Can $x$ be considered constant in the integral ? And then how to evaluate integral ? 
 A: Here are some ideas, too long for a comment. I would be interested to see how the kernel and image can be characterized more precisely. 
If $f\in \text{ker}\ T$, then
$\tag 1 \int_{-1}^{1}f(t)(t-x)^2dt=\int_{-1}^{1}t^2f(t)dt-2x\int_{-1}^{1}tf(t)dt+x^2\int_{-1}^{1}f(t)dt=0$ 
and linear independence of $\{1,x,x^2\}$ implies that each of these integrals vanishes individually.
So, $f\in \text{ker} \ T\Rightarrow f\in K:=\{f\in C([-1,1]):\int_{-1}^{1}t^nf(t)dt=0;\ n=0,1,2\}$. On the other hand, if $f\in K,$ then $(1)$ holds so in fact $\text{ker} T=K$.
If $Tf\in \text{Im}\ T,$ then 
$\tag2 Tf(x)=\int_{-1}^{1}t^2f(t)dt-2x\int_{-1}^{1}tf(t)dt+x^2\int_{-1}^{1}f(t)dt$
so $\text{Im}\ T$ is a subspace of $\text{span}\{1,x,x^2\}$ and Moreover, $\text{ker}\ T \perp \text{Im}\ T.$ 
We need to find useful relations among the three integrals that appear in $(2)$ and simplify if possible the conditions on the $f$ in the kernel. 
A: $T$ is a convolution and one should better write $T[f](x)$, because $T$ acts on the function (argument in square brackets) and is then evaluated at $x$ (round brackets). The kernel (or null space ) is defined as $\rm{ker}(T): = \{f \in V: \forall x \in [-1,1]: T[f](x) = 0\}$. Rearranging the given definition of $T[f](x)$, we find that $f \in \rm{ker}(T)$ iff
\begin{equation}
\int_{-1}^{1}f(t) t^2\, \rm{d}t- 2x \int_{-1}^{1}f(t) t \, \rm{d}t + x^2 \int_{-1}^{1}f(t) \rm{d}t = 0, \quad \forall x \in [-1,1],\tag{1}\label{1}
\end{equation}
which implies that all three integrals must vanish because as stated in (\ref{1}) the expression must vanish for all $x \in [-1,1]$. That is, $f \in \rm{ker}(T)$, iff
\begin{align}
&\int_{-1}^{1}f(t) t^2\, \rm{d}t  = \int_{-1}^{1}f(t) \left(\frac{2}{3}P_{2}(t)+\frac{1}{3}P_{0}(t)\right)\, \rm{d}t= 0,\tag{2}\label{2}\\
&\int_{-1}^{1}f(t) t\, \rm{d}t = \int_{-1}^{1}f(t) P_{1}(t)\, \rm{d}t = 0,\tag{3}\label{3}\\ 
&\int_{-1}^{1}f(t)\, \rm{d}t  = \int_{-1}^{1}f(t)P_{0}(x)\, \rm{d}t= 0,\tag{4}\label{4}
\end{align}
where $P_n(x)$ in the middle expressions of each line are Legendre polynomials with $n$ the degree of the polynomial. Since Legendre polynomials are orthogonal polynomials on $[-1,1]$, all three conditions are fulfilled for any linear superposition of Legendre polynomials of degree larger than 2.
Hence, 
\begin{equation}
\rm{ker}(T)  = \rm{span}\left(\{P_{2n+1}(x)\}_{n=1}^{\infty}\right).\tag{5}\label{5}
\end{equation}
Since the Legendre polynomials are a complete orthogonal basis for 

any piecewise continuous function $f(x)$ with finitely many discontinuities in the interval $[−1,1]$

[wikipedia], we found the kernel of $T$ on the entire domain $V$, since $V$ is restricted to continuous functions on $[-1,1]$.
The image of $T$ is given by
\begin{equation}
\rm{im}(T) = \rm{ker}(T)^{\perp} = V\setminus \rm{ker}(T) = \rm{span}\left(\{P_{0}(x), P_{1}(x), P_{2}(x)\}\right).\tag{6}\label{6}
\end{equation}
A: As other have pointed out the image of $T$ is a subspace of the $2^{nd}$ degree polynomials
To find the image let us only look at how $T$ operates on polynomials
$$ f(t) = \sum_{k=1}^n a_kt^k$$
\begin{align*}
\int_{-1}^1 f(t) (t-x)^2 dt &= \int_{-1}^1f(t)(t^2 + x^2 -2tx) \;dt \\
&= \int_{-1}^1 \sum_{k = 1}^n a_k t^{k+2} \;dt + x^2 \int_{-1}^1 \sum_{k = 1}^n a_k t^k \; dt -2x \int_{-1}^1 \sum_{k = 1}^n  a_k t^{k+1} \; dt \\
&=  \sum_{k =1}^n \frac{a_k}{k+3} (1 - (-1)^{k+3}) + x^2 \sum_{k = 1}^n \frac{a_k}{k+1} (1 - (-1)^{k+1}) -2x \sum_{k = 1}^n \frac{a_k}{k+2} (1-(-1)^{k+2}) \\
&= \sum_{k =1 \\k \text{ even}}^n \frac{2a_k}{k+3} + x^2 \sum_{k = 1 \\ k \text{ even}}^n \frac{2a_k}{k+1} +x \sum_{k = 1 \\ k \text{ odd}}^n \frac{-4a_k}{k+2} \\
&= \Big( \sum_{k = 1 \\ k \text{ even}}^n \frac{2a_k}{k+1} \Big) x^2 + \Big( \sum_{k = 1 \\ k \text{ odd}}^n \frac{-4a_k}{k+2}\Big) x + \sum_{k =1 \\k \text{ even}}^n \frac{2a_k}{k+3}
\end{align*}
Notice that this is actually a polynomial in $x$. So $T(f)(x) = ax^2 +bx +c$ is actually a linear system of equations in $a_1,a_2,...a_n$
Also notice the equations are linearly independent for $n$ greater than $4$.
If $n = 4$ and the variables $a_1 = x, a_2 = y, a_3 = z, a_4 =t$ we have the following system
Let $a,b,c \in \mathbb{R}$
\begin{align*}
\frac{2}{3} y + \frac{2}{5} t &= a \\
\frac{-4}{3} x +  \frac{-4}{5} z&= b \\
\frac{2}{5} y + \frac{2}{7} t  &= c
\end{align*}
By solving the two equations in $y$ and $t$  first we get a unique solution for $y$ and $t$. Then we can always chose $x$ and $z$ such that the last equation is satisfied.
Therefore 
 $$ Im(T) = \{f(x) \in \mathbb{R}[X] : deg(f)\leq 2\}.$$
