# Linear Functional: Continuous? [duplicate]

Given a Banach space: $E$
and chosen a Hamel basis: $\mathcal{B}$

Any vector induces a (noncanonical) algebraic linear functional by: $$\delta:E\to E^*:\delta_b(b'):=\delta_{b,b'}\text{ defined linearly and extended linearly}$$ How to show that the induced linear functionals are continuous iff the Banach spaces is finite dimensional?

## marked as duplicate by Nate Eldredge, Davide Giraudo, Asaf Karagila♦ general-topology StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 6 '14 at 22:39

• Well one direction is easy, every linear function defined on a finite dimensional Banach space is continuous. The other direction isn't immediately clear to me yet. – Alex Schiff Jun 6 '14 at 15:25
• Why is this so? – C-Star-W-Star Jun 6 '14 at 15:27
• Because the weak topology and the strong topology are always equivalent on a finite dimensional Banach space. – Alex Schiff Jun 6 '14 at 15:31
• Ok ^^ but isn't that precisely the question? – C-Star-W-Star Jun 6 '14 at 15:35
• See here for a proof that every linear functional on a finite dimensional banach space is continuous. – Alex Schiff Jun 6 '14 at 15:49

Suppose that $\mathcal B$ contains a sequence $(b_k)_{k\geqslant 1}$ such that $\lVert b_k\rVert=1$ for each $k$. Define $$L_n(x)=\sum_{k=1}^nk\cdot \delta_{b_k}(x).$$ Then for each $x$, $\sup_{n\geqslant 1}|L_n(x)|$ is finite. Since $\lVert L_n\rVert\geqslant n$, the principle of uniform boundedness implies a contradiction.
• Sorry, one more caveat: I think all we can say is $\Vert L_n\Vert\ge n$, unless I'm missing something. – David Mitra Jun 6 '14 at 19:18
• You are right. A priori nothing seems to prevent to have a norm greater than $n$ (it will depend on the norm of linear combinations of $b_k$'s). – Davide Giraudo Jun 6 '14 at 19:20