## Elementary Linear Algebra 7th Edition

(a) $|\langle f,g \rangle|=1\leq\| f \|\| g \|=\sqrt{\frac{1}{3}}\sqrt{\frac{1}{2}(e^2-1)}$ (b) $\| f+g \| =\sqrt{\frac{11}{6}+\frac{e^2}{2}} \leq\| f \|+\| g \|=\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{2}(e^2-1)}$.
Let $f(x)=x, \quad g(x)=e^x$, $\langle f, g\rangle=\int_{0}^1 f(x)g(x) d x$, we have $\langle f, g\rangle=\int_{0}^1 xe^x d x=\left[xe^x-e^x\right]_{0}^1=1,$ $\langle f, f\rangle=\int_{0}^{1} x^2 d x= \frac{1}{3}\left[x^3\right]_{0}^{1}=\frac{1}{3},$ $\langle g, g\rangle=\int_{0}^{1}e^{2x}d x= \frac{1}{2}\left[e^{2x}\right]_{0}^{1}=\frac{1}{2}(e^2-1),$ by making use of the calculations above, we get \begin{aligned}\langle f+g, f+g\rangle &=\int_{0}^{1}( x +e^{x})^2d x\\ &=\int_{0}^{1}\left(x^2+e^{2x}+2x e^{x} \right) d x \\ &=\left[\frac{1}{3}x^3+\frac{1}{2}e^{2x}+2 (xe^x-e^x)\right]_{0}^{1} \\ &=\frac{11}{6}+\frac{e^2}{2}\end{aligned}. Since for any $f$ we have $\| f \| =\sqrt{\langle f, f\rangle}$, then (a) $|\langle f,g \rangle|=1\leq\| f \|\| g \|=\sqrt{\frac{1}{3}}\sqrt{\frac{1}{2}(e^2-1)}$ (b) $\| f+g \| =\sqrt{\frac{11}{6}+\frac{e^2}{2}} \leq\| f \|+\| g \|=\sqrt{\frac{1}{3}}+\sqrt{\frac{1}{2}(e^2-1)}$.