Prove in $L_2[0,1]$ that $\int_0^1 e^{2x}dx - |\int_0^1 e^x \overline{\phi(x)}|^2= \int_0^1|e^x - \int_0^1 e^t \overline{\phi(t)}dt\phi(x)|^2dx$ where $\int_0^1 |\phi(x)|^2=1$
denote $f=e^x$ so this can be rewritten as $$\langle f , \overline{f} \rangle - |\langle f,\phi \rangle |^2 = \parallel f - \langle f, \phi \rangle \phi \parallel_2^2$$
How can I finish the proof from here? tried opening RHS though didn't amount to significant progress.
Thanks!
I think you are missing a square on the left-hand side of your identity. In fact, let's start from the right-hand side. First of all, we denote by $$ a:=\int_0^1 e^x\overline{\phi}(x)dx. $$ Thus, the right-hand side can be rewritten as: $$ \quad \quad \quad \mathrm{RHS}=\int_0^1e^{2x}dx-\overline{a}\int_0^1 e^x\overline{\phi}(x)dx-a\int e^x\phi(x)dx+\vert a\vert^2\int \vert\phi\vert^2(x)dx. \quad \quad \quad (*) $$ Now notice that, by using the definition of $a$ we have: $$ -\overline{a}\int_0^1 e^x\overline{\phi}(x)dx-a\int e^x\phi(x)dx=-2\vert a\vert^2. $$ Hence, replacing into $(*)$ and recalling that $\Vert \phi\Vert_{L^2}=1,$ we obtain $$ \mathrm{RHS}=\int_0^1e^{2x}dx-\vert a\vert^2=\int_0^1e^{2x}dx-\left\vert \int_0^1e^x\overline{\phi}(x)dx\right\vert^2, $$ which is exactly the latter identity you wrote on your post (exactly the left-hand side of it, which is not exactly the same as the one on the title).