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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?

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

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