Vector norm of integral inequality

Squaring the first inequality gives $$ \left\Vert\int_0^1 v(t)\,dt\right\Vert^2\leq\left(\int_0^1\Vert v(t)\Vert \, dt \right)^2 \, . $$ Now apply Inequality releating squared absolute value of an integral to the integral of the squared absolute values of the integrand to the scalar-valued function $f(t) = \Vert v(t)\Vert$ to conclude that the right-hand side is $$ \le \int_0^1\Vert v(t)\Vert^2 \, dt \, . $$

Tags:

Inequality

Integration

Norm

Vector Analysis

Integral Inequality

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