Convergence in Hilbert space question

quasar987
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Homework Statement


Is it true/possible to show that in a Hilbert space, if z_n is a sequence (not known to converge a priori) such that (z_n,y)-->0 for all y, then z_n-->0 ?


The Attempt at a Solution



I've shown that if z_n converges, then it must be to 0. But does it converge?
 
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Consider an orthonormal basis of the Hilbert space, {e_n} and let the sequence be the basis vectors. Now (e_n,y) is the nth component of y, which must tend to zero as n goes to infinity for any y. Yet e_n clearly does not converge. So no, you can't show that.
 
Good example!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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