# Bound on the norm of a bounded linear functional $f:C[0,1] \rightarrow \mathbb{R}$ defined by $f(\varphi)=\int_0^1\varphi(x)dx$.

How can one show that the linear functional defined by $$f(\varphi)=\int_0^1\varphi(x)dx$$ has norm $$\|f\| \leq 1$$? Since it is a bounded linear functional, the norm is given by $$\|f\|=\sup\limits_{\|x\|=1}|f(x)|$$. Supposedly since

$$|f(\varphi)|=\left\lvert\int_0^1\varphi(x)dx\right\rvert \leq \int_0^1|\varphi(x)|dx \leq \sup\limits_{x \in [0,1]}|\varphi(x)| = \|\varphi\|_\infty,$$

we know that $$\|f\| \leq 1$$. I fail to see this implication. Surely $$\|\varphi\|_\infty$$ is not bounded above for $$\varphi \in C[0,1]$$ in general, and so the above inequality does not imply that $$|f(\varphi)|$$ is bounded. Why is the norm bounded by 1?

• Please again refer to the definition of a bounded operator and it's norm. – Akash Yadav Apr 5 at 19:08

Hint: It is an easy exercise to prove that $$\|f\|=\sup_{\|\phi\|\leq1}|f(\varphi)|=\inf\{c>0: |f(\varphi)|\leq c\|\varphi\| \text{ for all }\varphi \}.$$
Since you have shown that $$|f(\varphi)|\leq\|\varphi\|$$ for all $$\varphi$$, this shows that $$\|f\|$$, which is the least constant with this ability, is less than $$1$$, which (as you showed) has this property.