Consider the linear operator $T : C[0,1] \to \mathbb{R}$ defined by \begin{align*} T(x) := x(0) - \int_{0}^{1} x(t)\phantom{.}dt \end{align*} Show that $T$ is bounded and find its norm $\|T\|$.

Boundedness: $$\|Tx\| = \bigg| x(0) - \int_{0}^{1} x(t)\phantom{.}dt \bigg| \leqslant |x(0)| + \int_{0}^{1}|x(t)|\phantom{.}dt \leqslant 2\|x\|_{\infty} $$
which shows that $T$ is bounded and $\|T\| \leqslant 2$. I want to show that $\|T\| \geqslant 2$ which will imply $\|T\| = 2$. I believe this entails constructing a sequence of $C[0,1]$ functions but I am not immediately seeing how to construct these sequences. Thank you.

  • $\begingroup$ @Chilango, the function you proposed has norm $4>1$, so it doesn't help us find the norm of $T.$ $\endgroup$ – Nikolaos Skout Mar 24 '16 at 0:14
  • $\begingroup$ the space of (real of complex) continuous functions on $[0,1]$ is a Banach space $C([0,1])$ for the $\|.\|_\infty$ norm. this space has no name ? I can only find it mentioned there en.wikipedia.org/wiki/Banach_space#Classical_spaces (because of the Fourier series, after symetrizing it to get a $2$ periodic continuous function, we see that it is related to the Banach space of $l^\infty(\mathbb{N})$ sequences converging to $0$ when $n \to \infty$) $\endgroup$ – reuns Mar 24 '16 at 0:16

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You can use the sequence $(x_n)$, whose graph is shown in the picture (the exact formula, if you insist, is left as an exercise). Since $\int_{0}^{1}x_n(t)dt \rightarrow -1$ and $x_n(0)=1$, you get $Tx_n\rightarrow 2$ and since $\|x_n\|_\infty=1$ for all $n,$ you get $|Tx_n|\leq \|T\|$ for all $n$, so $2\leq \|T\|.$

  • $\begingroup$ $x_\epsilon(t) = -1$ for $t > \epsilon$, $x_\epsilon(t) = 1-2\frac{t}{\epsilon}$ for $t \in [0,\epsilon[$. we get $\|x_\epsilon\|= 1$, $\int_0^1 x(t) dt = -1 + \epsilon$ and $T x_\epsilon = 2- \epsilon$. hence $\lim_{\epsilon \to 0} |Tx_\epsilon| = 2$ $\endgroup$ – reuns Mar 24 '16 at 0:13

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