we can always extend functions on subspaces

Proof (of Lemma)

First, note that if $t^{'} \in M^{'} (= M + C x)$ then there exists unique $t \in M$ and $a \in C$ such that $t^{'} = t + a x$ .

(uniqueness)

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

Thus, upon choosing $λ \in C$ , the map

u^{'} (t + a x) = u (t) + a λ t \in M, a \in C

is well-defined on $M^{'}$ and $u^{'} : M^{'} \to C$ is linear.

WLOG, suppose $C = 1$ . We want to choose $λ \in C$ such that

\forall t \in M, a \in C | u^{'} (t) + a λ | \leq | | t + a λ | |

Once $λ$ is fixed, $u^{'}$ is our continuous extension. When $a = 0$ , this inequality holds regardless of $λ$ (because it holds on $M$ ). Thus we only need to choose $λ$ for $a \neq 0$ . When $a \neq 0$ , we have

\begin{aligned} | u^{'} (t) + a λ | \leq | | t + a x | | & ⟺ | | u (\frac{t}{- a}) - λ | | \leq | | \frac{t}{- a} - x | | \forall t \in M \\ ⟺ | u (t) - λ | \leq | | t - x | | \forall t \in M \end{aligned}

Since $\frac{t}{- a} \in M$ . We first show there exists some $α \in R$ such that

| w (t) - α | \leq | | t - x | | \forall t \in M

where $w (t) = \frac{u (t) - \overset{―}{u (t)}}{2} = Re (u (t))$ . Now, note that

\begin{aligned} | w (t) | & = | Re (u (t)) | \\ \leq | u (t) | \\ \leq | | t | | \\ ⟹ w (t_{1}) - w (t_{2}) & = w (t_{1} - t_{2}) \forall t_{1}, t_{2} \in M \\ \leq | w (t_{1} - t_{2}) | \\ \leq | | t_{1} - t_{2} | | \\ \leq | | t_{1} - x | | + | | t_{2} - x | | \\ ⟹ w (t_{1}) - | | t_{1} - x | | & \leq w (t_{2}) + | | t_{2} - x | | \forall t_{1}, t_{2} \in M \\ ⟹ sup_{t \in M} w (t) - | | t - x | | & \leq w (t_{2}) + | | t_{2} - x | | \forall t_{2} \in M \\ ⟹ sup_{t \in M} w (t) - | | t - x | | & \leq inf_{t \in M} w (t) + | | t - x | | \end{aligned}

So we can choose an $α$ between these two quantities such that for all $t \in M$ we have

\begin{aligned} w (t) - | | t - x | | & \leq α & \leq w (t) + | | t - x | | \\ ⟹ - | | t - x | | & \leq α - w (t) & \leq | | t - x | | \\ ⟹ | w (t) - α | & \leq | | t - x | | \end{aligned}

Using the same process for the imaginary part of $u (t)$ and repeating the argument with $i x$ , we can find a $λ$ 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 + a x) = u (t) + a λ

on all of $M + C x$ . Thus we are done.

\begin{matrix} ◼ \end{matrix}

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we can always extend functions on subspaces

References

See Also

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