[proofplan] We prove first that evaluation at $x_0$ respects addition and scalar multiplication, hence is linear. We then bound $|\operatorname{ev}_{x_0}(f)|$ by the [uniform norm](/page/Uniform%20Norm) of $f$, which gives boundedness and the upper estimate for the operator norm. Finally, we test the functional on the constant function $1$ to show that this upper estimate is sharp. [/proofplan] [step:Verify that point evaluation is linear] Let $f,g \in B(E)$ and let $\lambda \in \mathbb{R}$. Since addition and scalar multiplication in $B(E)$ are defined pointwise, \begin{align*} \operatorname{ev}_{x_0}(f+g) = (f+g)(x_0) = f(x_0)+g(x_0) = \operatorname{ev}_{x_0}(f)+\operatorname{ev}_{x_0}(g). \end{align*} Also, \begin{align*} \operatorname{ev}_{x_0}(\lambda f) = (\lambda f)(x_0) = \lambda f(x_0) = \lambda \operatorname{ev}_{x_0}(f). \end{align*} Thus $\operatorname{ev}_{x_0}: B(E) \to \mathbb{R}$ is linear. [/step] [step:Bound point evaluation by the uniform norm] Let $f \in B(E)$. By the definition of the supremum over $E$ and since $x_0 \in E$, \begin{align*} |\operatorname{ev}_{x_0}(f)| = |f(x_0)| \le \sup_{x \in E} |f(x)| = \|f\|_\infty. \end{align*} Therefore $\operatorname{ev}_{x_0}$ is bounded, and the definition of the operator norm gives \begin{align*} \|\operatorname{ev}_{x_0}\|_{\mathcal{L}(B(E),\mathbb{R})} \le 1. \end{align*} [guided] The purpose of this step is to show that evaluating a bounded function at one point cannot be larger than taking its largest absolute value over the whole set. Let $f \in B(E)$. Since $x_0 \in E$, the number $|f(x_0)|$ is one of the values included in the set \begin{align*} \{|f(x)| : x \in E\}. \end{align*} By the defining property of the supremum, \begin{align*} |f(x_0)| \le \sup_{x \in E} |f(x)|. \end{align*} Using the definition of $\operatorname{ev}_{x_0}$ and the definition of the uniform norm, this becomes \begin{align*} |\operatorname{ev}_{x_0}(f)| = |f(x_0)| \le \|f\|_\infty. \end{align*} This inequality holds for every $f \in B(E)$. Since $\operatorname{ev}_{x_0}$ is already known to be linear, it proves boundedness of the linear functional. Moreover, by the definition of the operator norm, \begin{align*} \|\operatorname{ev}_{x_0}\|_{\mathcal{L}(B(E),\mathbb{R})} = \sup_{\|f\|_\infty \le 1} |\operatorname{ev}_{x_0}(f)|. \end{align*} The estimate just proved gives $|\operatorname{ev}_{x_0}(f)| \le 1$ whenever $\|f\|_\infty \le 1$, hence \begin{align*} \|\operatorname{ev}_{x_0}\|_{\mathcal{L}(B(E),\mathbb{R})} \le 1. \end{align*} [/guided] [/step] [step:Test on the constant function to obtain the sharp lower bound] Define the constant function \begin{align*} \mathbf{1}_E: E &\to \mathbb{R} \end{align*} \begin{align*} x &\mapsto 1. \end{align*} This function belongs to $B(E)$, and since $E$ is nonempty, \begin{align*} \|\mathbf{1}_E\|_\infty = \sup_{x \in E} |\mathbf{1}_E(x)| = 1. \end{align*} Also, \begin{align*} |\operatorname{ev}_{x_0}(\mathbf{1}_E)| = |\mathbf{1}_E(x_0)| = 1. \end{align*} Therefore the operator norm satisfies \begin{align*} \|\operatorname{ev}_{x_0}\|_{\mathcal{L}(B(E),\mathbb{R})} \ge 1. \end{align*} Combining this lower bound with the upper bound from the previous step gives \begin{align*} \|\operatorname{ev}_{x_0}\|_{\mathcal{L}(B(E),\mathbb{R})} = 1. \end{align*} Thus $\operatorname{ev}_{x_0}$ is a bounded linear functional of norm $1$. [/step]