Proof. Let \(f,g\in C[0,1]\) and \(\alpha , \beta \in \mathbb {R}\). Then \begin {align*} \Lambda _x(\alpha f+\beta g) &=(\alpha f+\beta g)(x)\\ &=\alpha f(x)+\beta g(x)\hspace {0.5cm} \text {f,g are continuous and linear}\\ &=\alpha \Lambda _x(f)+\beta \Lambda _x(g) \end {align*}

\(\implies \Lambda _x\) is linear.

Now \(|\Lambda _x(f)|=|f(x)|\leq ||f||\) so that \[||\Lambda _x||=\sup _{f\in C[0,1]}\Bigg \{\frac {|\Lambda _x(f)|}{||f||}\Bigg \}\leq 1\] proving that \(\Lambda _x\) is bounded.
Therefore \(\Lambda _x\in X^*\). □

Proof. Let \(x,z\in l^p\) and \(\alpha , \beta \in \mathbb {C}\). Then \begin {align*} \Lambda _y(\alpha x+\beta z) &=\sum ^{\infty }_{n=1}(\alpha x_n+\beta z_n)\overline {y_n}\\ &=\sum _{n=1}^{\infty }\alpha x_n\overline {y_n}+\sum _{n=1}^{\infty }\beta z_n\overline {y_n}\\ &=\alpha \sum ^{\infty }_{n=1}x_n\overline {y_n}+\beta \sum ^{\infty }_{n=1}z_n\overline {y_n}\\ &=\alpha \Lambda _y(x)+\beta \Lambda _y(z) \end {align*}

Thus \(\Lambda _y\) is linear.

To show boundedness, we first note that Holder’s inequality
\(\implies |\Lambda _y(x)|\leq ||x||_p||y||_q\hspace {0.3cm}\implies \displaystyle {\frac {|\Lambda _y(x)|}{||x||_p}\leq ||y||_q}\) \[\implies ||\Lambda _y||=\sup _{x\in l^p}\Bigg \{\frac {|\Lambda _y(x)|}{||x||_p}\Bigg \}\leq ||y||_q\qquad (1)\]
Now let \(x_n=0\) when \(y_n=0\), otherwise let \(\displaystyle {x_n=\frac {|y_n|^{1+\frac {q}{p}}}{\overline {y_n}}}.\) Then \begin {align*} ||x||_p &=\Bigg (\sum ^{\infty }_{n=1}|x_n|^p\Bigg )^{\frac {1}{p}} =\Bigg (\sum ^{\infty }_{n=1}\Bigg |\frac {|y_n|^{1+\frac {q}{p}}}{\overline {y_n}}\Bigg |^p\Bigg )^{\frac {1}{p}}\\ &=\Bigg (\sum _{n=1}^{\infty }|y_n|^q\Bigg )^{\frac {1}{p}}\\ &=\Bigg \{\Bigg (\sum ^{\infty }_{n=1}|y_n|^q\Bigg )^{\frac {1}{q}}\Bigg \}\\ &=||y||^{\frac {q}{p}}_q \end {align*}

So that \begin {align*} \Lambda _y(x) &=\sum ^{\infty }_{n=1}x_n\overline {y_n} =\sum ^{\infty }_{n=1}|y_n|^{1+\frac {q}{p}}\\ &=\sum ^{\infty }_{n=1}|y_n|^q\\ &=||y||^q_q =||y||^{1+\frac {q}{p}}_q\\ &=||y||_q||y||_q^{\frac {q}{p}}\\ &=||y||_q||x||_p \end {align*}

This implies that \(\displaystyle {||\Lambda _y||=\sup _{x\in l^p}\Bigg \{\frac {|\Lambda _y(x)|}{||x||_p}\Bigg \}\geq ||y||_q\qquad (2)}.\hspace {0.3cm}\) combining \((1)\) and \((2)\), \[\implies ||\Lambda _y||=||y||_q.\] □

Proof. Suppose \(\Lambda \in (l^p)^*\). Let \(e^n\) be the sequence in \(l^p\) which has all its terms equal to zero except for a one in the \(n^{th}\) place. i.e \[ e^n= \begin {cases} 1 &{on\hspace {0.3cm} the\hspace {0.3cm} n^{th}\hspace {0.3cm} place}\\\\ 0 &\text {otherwise} \end {cases} \] \begin {align*} e^1 &=\{1,0,0,\dots ,0\}\\ e^3 &=\{0,0,1,0,0,\dots ,0\}\\ \vdots \\ e^5 &=\{0,0,0,0,1,0,\dots ,0\} \end {align*}

Then \(\displaystyle { ||e^n||_p =\Bigg (\sum _{k=1}^{\infty }|e^n_k|^p\Bigg )^{\frac {1}{p}} =|e^n_1|+|e^n_2|+\dots =1\hspace {3cm} \forall p\in [1,\infty )}\)

Let \(\overline {y_n}=\Lambda (e^n)\) and \(y=\{y_n\}^{\infty }_{n=1}\). We show that \(y\in l^q\) for all \(p\in [1,\infty )\).

If \(p=1\), then \(q=\infty \) and \(|y_n|=|\Lambda (e^n)|\leq ||\Lambda ||||e^n||_1=||\Lambda ||<\infty .\) This implies that \[||y||_{\infty }=\sup _{n}|y_n|<\infty \] \(\implies y\in l^q\).

If \(p>1\), then \begin {align*} \sum ^{k}_{n=1}|y_n|^q &=\Bigg |\sum ^k_{n=1}\frac {|y_n|^q}{\overline {y_n}}.\overline {y_n}\Bigg | =\Bigg |\sum ^k_{n=1}\frac {|y_n|^q}{\overline {y_n}}.\Lambda (e^n)\Bigg |\\ &=\Bigg |\Lambda \Bigg (\sum ^k_{n=1}\frac {|y_n|^q}{\overline {y_n}}.e^n\Bigg )\Bigg |\\ &\leq ||\Lambda ||\Bigg |\Bigg |\sum ^k_{n=1}\frac {||y_n||^q}{\overline {y_n}}.e^n\Bigg |\Bigg |_p =||\Lambda ||\Bigg (\sum ^k_{n=1}|y_n|^{(q-1)p}\Bigg )^{\frac {1}{p}}\\ &=||\Lambda ||\Bigg (\sum ^k_{n=1}|y_n|^q\Bigg )^{\frac {1}{p}} \end {align*}

This implies that \begin {align*} \Bigg (\sum ^k_{n=1}|y_n|^q\Bigg )^{\frac {1}{q}} &=\Bigg (\sum ^k_{n=1}|y_n|^q\Bigg )^{1-\frac {1}{p}}\\ &=\Bigg (\sum ^k_{n=1}|y_n|^q\Bigg )\Bigg (\sum ^k_{n=1}|y_n|^q\Bigg )^{1-\frac {1}{p}}\\ &\leq ||\Lambda || <\infty \end {align*}

\(\implies y\in l^q\).

It remains to show that \(\Lambda =\Lambda _y\). Let \(x=\{x_n\}^{\infty }_{n=1} \in l^p\) for \(1\leq p<\infty \). Then \begin {align*} \Big |\Big |x-\sum ^k_{n=1}x_ne^n\Big |\Big |_p &=\Bigg (\sum ^{\infty }_{n=k+1}|x_n|^p\Bigg )^{\frac {1}{p}}\\ &\rightarrow 0\hspace {0.3cm} \text {as}\hspace {0.3cm} k\rightarrow \infty . \end {align*}

and so \(\displaystyle { x=\sum ^{\infty }_{n=1}x_ne^n}\hspace {0.1cm}.\hspace {0.2cm}\) Now \(\forall k\), we have \begin {align*} \Lambda \Bigg (\sum ^k_{n=1}x_ne^n\Bigg ) &=\sum ^k_{n=1}x_n\Lambda (e^n)\\ &=\sum ^k_{n=1}x_n\overline {y_n}\\ &=\Lambda _y \Bigg (\sum ^k_{n=1}x_ne^n\Bigg ) \end {align*}

Since \(\Lambda \) and \(\Lambda _y\) are continuous, this implies that \begin {align*} \Lambda (x) &=\lim _{k\rightarrow \infty }\Lambda \Bigg (\sum ^k_{n=1}x_ne^n\Bigg )\\ &=\lim _{k\rightarrow \infty }\Lambda _y\Bigg (\sum ^k_{n=1}x_ne^n\Bigg )\\ &=\Lambda _y(x).\\\\ \end {align*} □

Proof. Let \(x\in X\), \(u(x)=\textbf {Re}\Lambda (x)\) and \(v(x)=\textbf {Im}\Lambda (x)\). Then \(\Lambda (x)=u(x)+iv(x)\hspace {0.2cm}\) so that \begin {align*} iu(x)-v(x) &=i\Lambda (x)\\ &=\Lambda (ix)\\ &=u(ix)+iv(x) \end {align*}

It follows that \(u(ix)=-v(x)\), and \(\Lambda (x)=u(x)-iu(ix)\) \[\Lambda (x)=u(x)-iu(ix).\] Now since for all \(x\in X\), \(|\Lambda (x)|\geq |u(x)|\). It implies that \(||\Lambda ||\geq ||u||\qquad (1)\)

Conversely, let \(\Lambda (x)=re^{i\theta }\). Then \begin {align*} |\Lambda (x)| &= r=e^{-i\theta }\Lambda (x)\\ &=\Lambda \Big (e^{-i\theta }x\Big ) \end {align*}

from which it follows that \(\Lambda \Big (e^{-i\theta }x\Big )\) is real and positive. Thus \(\displaystyle {\Lambda \Big (e^{-i\theta }x\Big )=u\Big (xe^{-i\theta }\Big )}\hspace {0.2cm}\) and \begin {align*} |\Lambda (x)| &=u\Big (xe^{-i\theta }\Big )\\ &\leq ||u||||xe^{-i\theta }||\\ &=||u||||x||. \end {align*}

\[\implies ||\Lambda ||=\sup _{x\in X}\Bigg \{\frac {|\Lambda (x)|}{||x||}\Bigg \}\leq ||u||\qquad (2)\] Combining \((1)\) and \((2)\) \(\implies ||\Lambda ||=||u||\). □

Proof. We treat the real case; the complex case follows by applying it to the real part and recovering the imaginary part from \(u(x)=\operatorname {Re}u(x)-i\operatorname {Re}u(ix)\).

Assume without loss that \(||u||=1\), rescaling otherwise. Every element of \(M_0\) is uniquely \(x+\alpha x_0\) with \(x\in M\) and \(\alpha \) scalar, since \(x_0\notin M\). So an extension is determined by the single number \(c=u_0(x_0)\), and we must choose \(c\) so that \[|u(x)+\alpha c|\leq ||x+\alpha x_0||\qquad \text {for all }x\in M,\ \alpha .\] Dividing by \(|\alpha |\) and renaming, it suffices to arrange \[|u(x)+c|\leq ||x+x_0||\qquad \text {for all }x\in M .\]

Such a \(c\) exists. For any \(x,y\in M\), \[u(x)-u(y)=u(x-y)\leq ||x-y||\leq ||x+x_0||+||y+x_0|| ,\] using \(||u||=1\) and the triangle inequality. Rearranging, \[-u(y)-||y+x_0||\ \leq \ -u(x)+||x+x_0||\qquad \text {for all }x,y\in M .\] Hence \[a=\sup _{y\in M}\big [-u(y)-||y+x_0||\big ]\ \leq \ \inf _{x\in M}\big [-u(x)+||x+x_0||\big ]=b ,\] and any \(c\in [a,b]\) satisfies the required inequality. Setting \(u_0(x_0)=c\) gives an extension with \(||u_0||\leq 1=||u||\); and \(||u_0||\geq ||u||\) because \(u_0\) agrees with \(u\) on \(M\). So \(||u_0||=||u||\). □

Proof. First assume \(\mathbb {F}=\mathbb {R}\). For \(x\in M\) and \(x_0\in X/M\), we define \(u_0(x+\alpha x_0)=u(x)+\alpha t\hspace {0.2cm}\) where \(t\) is such that \(||u_0||=||u||\).

We show that this \(t\) exists. Now \begin {align*} |u(x)+\alpha t| &=|u_0(x+\alpha x_0)|\\ &\leq ||u_0||||x+\alpha x_0||\\ &=||u||||x+\alpha x_0|| \end {align*}

It follows from this that (let \(\alpha =1\)) \(|u(x)+t|\leq ||u||||x+x_0||\hspace {0.2cm}\) which implies that
\(-||u||||x+x_0||\leq u(x)+t\leq ||u||||x+x_0||\hspace {0.2cm}\) or \[-(u(x)+||u||||x+x_0||)\leq t\leq -u(x)+||u||||x+x_0||\] That is, \(u(x)-||u||||x+x_0||\leq t\leq u(x)+||u||||x+x_0||\). This shows that for this \(t\) and \(\forall x\in M\) \[u(x)-||u||||x+x_0||\leq u(x)+||u||||x+x_0||\] This is valid since if \(x,y\in M\), then \begin {align*} u(x)-u(y) &=u(x-y)\\ &=u(x-y)\leq |u(x-y)|\\ &\leq ||u||||x-y||\\ &\leq ||u||(||x+x_0||+||y+x_0||) \end {align*}

So that \(u(x)-||u||||x+x_0||\leq u(y)+||u||||y+x_0||\)

To show that \(u_0\) is linear. Let \(x,y\in M\) and \(\alpha _1, \alpha _2, \beta _1, \beta _2 \in \mathbb {R}\). Then \begin {align*} u_0(\beta _1(x+\alpha _1x_0)+\beta _2(y+\alpha _2 x_0)) &=u_0(\beta _1 x+\beta _2 y+\beta _1 \alpha _1 x_0+\beta _2 \alpha _2 x_0)\\ &=u_0(\beta _1 x+\beta _2 y+(\beta _1 \alpha _1 +\beta _2 \alpha _2)x_0)\\ &=u(\beta _1 x+\beta _2 y)+(\beta _1 \alpha _1 +\beta _2 \alpha _2)t\\ &=\beta _1 (u(x)+\alpha _1 t)+\beta _2 (u(y)+\alpha _2 t)\\ &=\beta _1 u_0(x+\alpha _1 x_0)+\beta _2 u_0(y+\alpha _2 x_0) \end {align*}

Thus \(u_0\) is linear.

Boundedness of \(u_0\) follows from its definition and the fact that \(u\) is bounded.

Now assume \(\mathbb {F}=\mathbb {C}\). Let \(a(x)=\textbf {Re}u(x)\) and \(b(x)=\textbf {Im}u(x)\), \[a(x)=ib(x)=u(x)=a(x)-ia(ix)\] since \(a\in M^*\), there is \(a_0\in M^*\) such that \(a_0(x)=a(x)\) and \(||a_0||=||a||\).
Therefore, if we define \(u_0(x)=a_0(x)-ia_0(ix)\) then \(u_0\in M^*_0\), \(u_0(x)=u(x)\) and \(||u_0||=||u||\). □

Proof. The lemma extends a functional by one dimension without increasing its norm. Zorn’s lemma turns that into an extension to the whole space.

Consider the collection of all pairs \((N,v)\) where \(N\) is a subspace containing \(M\) and \(v\in N^*\) extends \(u\) with \(||v||=||u||\). It is not empty, containing \((M,u)\). Order it by \[(N_1,v_1)\preceq (N_2,v_2)\iff N_1\subseteq N_2 \text { and } v_2|_{N_1}=v_1 .\]

Every chain has an upper bound: take the union of the subspaces in the chain, which is again a subspace because the chain is totally ordered, and define the functional on it by using whichever member of the chain contains the given point. This is well defined precisely because the functionals in a chain agree wherever both are defined, and its norm is still \(||u||\).

By Zorn’s lemma there is a maximal element \((N_0,u_0)\). If \(N_0\neq X\), choose \(x_0\in X\setminus N_0\) and apply the lemma to extend \(u_0\) to \(N_0+\mathbb {F}x_0\) with the same norm — contradicting maximality. Hence \(N_0=X\), and \(u_0\in X^*\) is the required extension. □