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Let $H = L^2(0,T;V)$ where $V$ is separable in $H$. We have $y_n(t) = \sum_{i=1}^n c_{i,n}(t)b_i$ where $b_i$ are basis vectors in $V$. Suppose we have the estimate $$\lVert y_n \rVert_H^2 = \int_0^T \lVert y_n(t) \rVert^2_{V} \leq C\left(\lVert y_0 \rVert^2_H + \int_0^T \lVert f(t) \rVert^2_{V^*}\right)$$ for all $n$. Suppose also $y_n \rightharpoonup y$ in $H$.

How do I show that $$\lVert y \rVert_{H}^2 = \int_0^T \lVert y(t) \rVert^2_{V} \leq C\left(\lVert y_0 \rVert^2_H + \int_0^T \lVert f(t) \rVert^2_{V^*}\right)$$ also holds?

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weak lower semicontinuity of norm will not give you the result? – Saugata Nov 18 '12 at 16:24
up vote 1 down vote accepted

The first equality is the definition of the norm in $H$. For the inequality, we have the following result:

Let $H$ a Hilbert space, $\{x_n\}$ a sequence which converges weakly to $x$. Then $$\lVert x\rVert\leqslant \liminf_{n\to +\infty}\lVert x_n\rVert.$$

Proof: Since $|\langle x,x_n\rangle|\leqslant \lVert x_n\rVert\cdot\lVert x\rVert$, we get $$\lVert x\rVert^2=\liminf_{n\to +\infty}|\langle x,x_n\rangle|\leqslant \liminf_{n\to +\infty}\lVert x_n\rVert\cdot\lVert x\rVert.$$ This gives the result (if $x=0$, it's obvious, otherwise divide by $\lVert x\rVert$).

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Note that it also holds when $H$ is a normed space. – Davide Giraudo Nov 18 '12 at 17:52

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