we can always extend functions on subspaces

[[concept]]

Topics

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Lemma

If $V$ is a normed vector space and $M\subset V$ is a subspace and $u:M\to \mathbb{C}$ is linear such that $|u(t)|\le C||t||$ for all $t\in M$, and

• $x\notin M$ Then there exists a function $u^{\prime }:M^{\prime }\to \mathbb{C}$ which is linear
• (Where $M^{\prime }=M+\mathbb{C}x=\{t+ax:t\in M,a\in \mathbb{C}\}$ ) Such that $u^{\prime }{|}_M=u$ and for all $t^{\prime }\in M^{\prime }$ we have $|u^{\prime }(t^{\prime })|\le C||t^{\prime }||$

Proof (of Lemma)

First, note that if $t^{\prime }\in M^{\prime }(=M+\mathbb{C}x)$ then there exists unique $t\in M$ and $a\in \mathbb{C}$ such that $t^{\prime }=t+ax$.

(uniqueness) $t+ax=\tilde{t}+\tilde{a}x⟹(a-\tilde{a})x=\tilde{t}-t\in M$. And if $a-\tilde{a}\ne 0$ then $x\in M$ so we must have $a=\tilde{a}$. But then we have $\tilde{t}=t$, so $t,a$ must be unique.

If

Thus, upon choosing $\lambda \in \mathbb{C}$, the map $u^{\prime }(t+ax)=u(t)+a\lambda t\in M,a\in \mathbb{C}$ is well-defined on $M^{\prime }$ and $u^{\prime }:M^{\prime }\to \mathbb{C}$ is linear.

WLOG, suppose $C=1$. We want to choose $\lambda \in \mathbb{C}$ such that $\forall t\in M,a\in \mathbb{C}|u^{\prime }(t)+a\lambda |\le ||t+a\lambda ||$ Once $\lambda$ is fixed, $u^{\prime }$ is our continuous extension. When $a=0$, this inequality holds regardless of $\lambda$ (because it holds on $M$). Thus we only need to choose $\lambda$ for $a\ne 0$. When $a\ne 0$, we have

\lvert u'(t) + a\lambda \rvert \leq \lvert \lvert t + ax \rvert \rvert&\iff \left\lvert \left\lvert u\left( \frac{t}{-a} \right) -\lambda \right\rvert \right\rvert \leq \left\lvert \left\lvert \frac{t}{-a} -x \right\rvert \right\rvert \quad\forall t \in M \\ &\iff \lvert u(t) - \lambda \rvert \leq \lvert \lvert t - x \rvert \rvert \quad\forall t \in M \end{align}$$ Since $\frac{t}{-a} \in M$. We first show there exists some $\alpha \in \mathbb{R}$ such that $$\lvert w(t) - \alpha \rvert \leq \lvert \lvert t-x \rvert \rvert \quad \forall t \in M$$ where $w(t) = \frac{u(t) - \overline{u(t)}}{2} = \mathrm{Re}(u(t))$. Now, note that $$\begin{align} \lvert w(t) \rvert & = \lvert \mathrm{Re}(u(t)) \rvert \\ & \leq \lvert u(t) \rvert \\ & \leq \lvert \lvert t \rvert \rvert \\ \implies w(t_{1}) - w(t_{2}) & =w(t_{1}-t_{2}) \quad \forall t_{1},t_{2} \in M\\ & \leq \lvert w(t_{1}-t_{2}) \rvert \\ & \leq \lvert \lvert t_{1}-t_{2} \rvert \rvert \\ & \leq \lvert \lvert t_{1}-x \rvert \rvert + \lvert \lvert t_{2}-x \rvert \rvert\\ \implies w(t_{1}) - \lvert \lvert t_{1}-x \rvert \rvert & \leq w(t_{2}) + \lvert \lvert t_{2}-x \rvert \rvert \quad\forall t_{1},t_{2}\in M \\ \implies \sup_{t \in M} w(t) -\lvert \lvert t-x \rvert \rvert & \leq w(t_{2}) + \lvert \lvert t_{2} - x \rvert \rvert \quad \forall t_{2} \in M\\ \implies \sup_{t \in M} w(t) -\lvert \lvert t-x \rvert \rvert & \leq\inf_{t \in M} w(t) + \lvert \lvert t-x \rvert \rvert \end{align}$$ So we can choose an $\alpha$ between these two quantities such that for all $t \in M$ we have $$\begin{align} w(t) - \lvert \lvert t-x \rvert \rvert &\leq \alpha &\leq w(t) + \lvert \lvert t-x \rvert \rvert \\ \implies - \lvert \lvert t-x \rvert \rvert &\leq \alpha - w(t) &\leq \lvert \lvert t-x \rvert \rvert \\ \implies \lvert w(t) - \alpha \rvert &\leq \lvert \lvert t-x \rvert \rvert \end{align}$$ Using the same process for the imaginary part of $u(t)$ and repeating the argument with $ix$, we can find a $\lambda$ such that the desired bound holds. > [!NOTE] > To see this, we can verify that since the bound holds on both the real and imaginary components of $x$, it holds for all complex multiples of $x$ This defines our function $$u'(t+ax) = u(t) + a\lambda$$ on all of $M +\mathbb{C}x$. Thus we are done. $$\tag*{$\blacksquare$}$$

We use this lemma in our proof for the Hahn-Banach theorem

Strategy for using this lemma to prove Hahn-Banach

1. Place a partial order on all continuous extensions of $u$
2. Apply Zorn’s lemma to this set, giving us a maximal element $U$
3. Use this lemma to show that this maximal extension is defined on all of our desired space $V$

• the proof of this will be similar to our proof that we indeed had a basis in the proof that every vector space has a hamel basis

References

References

See Also

Mentions

Mentions

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