I'm learning about bounded operators and need to compute the norm of the following:
The operator $V:C[0,1]\to C[0,1]$ defined by $$(Vf)(t)=\int_0^t f(s)\ ds \qquad t\in [0,1];$$ here $C[0,1]$ is equipped with $\lVert\cdot\rVert_p$ for $p\in [1,\infty)$.
My attempt: Using Hölder's inequality, \begin{align} \lVert Vf\rVert_p^p & = \int_0^1 |(Vf)(t)|^p\ dt \\ & =\int_0^1 \left|\int_0^t f(s)\ ds\right|^p\ dt\\ &\leq \int_0^1 \left[\int_0^t |f(s)|\ ds\right]^p\ dt\\ &\leq \int_0^1\lVert f\rVert_p^p\,\lVert\mathbf{1}\rVert_q^p \ dt\\ &=\lVert f\rVert^p, \end{align} here $q$ is the conjugate exponent of $p$ and $\mathbf{1}$ is the constant one function. Thus $V$ is bounded and $\,\lVert V\rVert\leqslant1$.
On the other hand, \begin{align} \lVert V\mathbf{1}\rVert_p^p & = \int_0^1 |(V\mathbf{1})(t)|^p\ dt \\ & =\int_0^1 \left|\int_0^t \mathbf{1}(s)\ ds\right|^p\ dt\\ & =\int_0^1 t^p\ dt\\ &=\frac{1}{p+1}. \end{align} So $\,\lVert V\mathbf{1}\rVert_p = \left(\frac{1}{p+1}\right)^{1/p}\ge 1$. Thus, $\lVert V\rVert\geqslant1$.
It follows that $\lVert V\rVert=1$.
