Let $M: P_{1}\rightarrow P_{1}$ be defined by $M(f)=f^{'}+f$, i.e. $M(a_{0}+a_{1}x)=a_{1}+a_{0}+a_{1}x$. Find the adjoint $M^{*}$ of $M$, i.e find $M^{*}(a_{0}+a_{1}x_{1})$, assuming that $L^{2}(0,1)$ inner product is imposed on $P_{1}$. $\langle f(x),g(x)\rangle =\int^{1}_{0}f(x)g(x)dx$.

• This looks like a homework question; please see how to ask such questions. In any case, people here don't tend to like textbook-style problems with no work shown and phrased as commands. – Nate Eldredge Jun 10 '13 at 0:37

Luckily, $\|1\|=\sqrt{\langle 1,1\rangle}=\sqrt{\int_0^11}=1$. But $x$ is not orthogonal to $1$, so we have to modify it: $$\langle x,1\rangle =\int_0^11\cdot x\,dx=\frac12\,,$$ so that $(x-\frac12)\perp 1$. So is its double, $p:=\,2x-1$. Its norm can be obtained: $\|p\|^2=\|(2x-1)\|^2=\int_0^1(2x-1)^2\,dx=\frac43-2+1=\frac13$.
Then, use the general fact that a vector $v$ can be decomposed in the orthogonal basis $e_1,e_2,...$ as $$v=\frac{\langle v,e_1\rangle}{\|e_1\|^2} e_1+\frac{\langle v,e_2\rangle}{\|e_2\|^2} e_2+\dots$$
So that, now we can apply this to arrive at $M^*$: $$M^*(1)= \langle M^*(1),\,1\rangle\, +3\,\langle M^*(1),\,2x-1\rangle\cdot (2x-1)= \\ = \langle 1,M(1)\rangle+3\,\langle1,M(2x-1)\rangle\cdot(2x-1)\,.$$ Similarly one can obtain $M^*(x)$.
• can you make use of $<M(f),g>=<f,M^{*}(g)>$? – 81235 Jun 9 '13 at 23:43
• One can also introduce an orthonormal basis, and work with matrices. Now we can fix $(1,\sqrt3\cdot(2x-1))$, now both these has norm $1$, and are orthogonal to each other (w.r.t. the $L^2$ inner product). Then, coordinating $M$ in this basis, we get $$[M]=\pmatrix{1&2\sqrt3\\0&1}\,$$ so its adjoint (in the same basis!) is $$[M^*]=\pmatrix{1&0\\2\sqrt3&1}\,.$$ Looks odd, but it gives the same as my solution. – Berci Jun 10 '13 at 0:00
• My approach: $<M(f),g>=<f,M^{*}(g)>$, $f=a_{0}+a_{1}x$,$g=b_{0}+b_{1}x$, $<M(f),g>=\int^{1}_{0}(a_{1}+a_{0}+a_{1}x)(b_0+b_{1}x)dx$, but I don't how to continue. Can you prove it following my idea? – 81235 Jun 10 '13 at 0:02
• I have read your answer carefully, according your answer,$M^{*}(x)=7x-3$,am I right? – 81235 Jun 10 '13 at 0:15