20211006 integral, differential and projection belong to linear transformation

Two criteria for judging linear transformation ： Number multiplication and addition .
therefore , Integral and differential belong to linear transformation .

example $1.11$ In linear space $P_n$ in , Differentiation is a linear transformation , here use $D$ Express , namely
$$Df(t)=f^{\prime }(t)\left(\forall f(t)\in P_n\right)$$
in fact , For any $f(t),g(t)\in P_n$ And $k,l\in \mathbf{R}$, Yes
$$\begin{array}{rl}D(kf(t)+\lg (t))=& (kf(t)+\lg (t){)}^{\prime }=\\ & k{f}^{\prime }(t)+{\lg }^{\prime }(t)=\\ & k(Df(t))+l(Dg(t))\end{array}$$
example $1.12$ Defined in a closed interval $[a,b]$ The set of all real continuous functions on $C(a,b)$ constitute $\mathbf{R}$ A linear space on . stay $C(a,b)$ Define transform on $J$, namely
$$J(f(t))={\int }_a^{t}f(u)\mathrm{d}u(\forall f(t)\in C(a,b))$$
be $J$ yes $C(a,b)$ A linear transformation of .
in fact , Because there is
$$\begin{array}{rl}J(kf(t)+\lg (t))=& {\int }_a^t(kf(u)+\lg (u))\mathrm{d}u=\\ & k{\int }_a^{t}f(u)\mathrm{d}u+l{\int }_a^{t}g(u)\mathrm{d}u=\\ & k(Jf(t))+l(Jg(t))\end{array}$$

Projection ：$y$ Is the unit vector ,
$y{y}^T{x}_1+y{y}^T{x}_2=y{y}^T\left(x_1+x_2\right)$
$y{y}^T\left(k{x}_1\right)=y{y}^T\left(k{x}_1\right)$