2

If $T:C(X,\mathbb{R})\rightarrow C(Y,\mathbb{R})$ is an positive linear application, then $T$ is continuous and $||T||=||T(1)||_{\infty}$, where $1\in C(X,\mathbb{R})$ and $||f||_{\infty}=\max_{x\in X}|f|$.

Now, let $\varphi (x,t):[0,1]\times[0,1]\rightarrow[0,+\infty[$ continuous function such that $\dfrac{\partial \varphi}{\partial x}:[0,1]\times[0,1]\rightarrow[0,+\infty[$ exists and is continuous.

Show that the application $T:C([0,1],\mathbb{R})\rightarrow C([0,1],\mathbb{R})$ defined by $T(f)(x)=\int_0^1 \varphi (x,t)f(t)dt$ has norm $||T||=\int_0^1\varphi (1,t)dt$.

My partial solution:

$||T||=||T(1)||_{\infty}=\max_{x\in [0,1]}|T(1)(x)|=\max_{x\in [0,1]}|\int_0^1 \varphi (x,t)1(t)dt|=\max_{x\in [0,1]}|\int_0^1 \varphi (x,t)dt|$

Since $\varphi$ is postive $\Rightarrow ||T||=\max_{x\in [0,1]}\int_0^1 \varphi (x,t)dt$ and let $g(x)=\int_0^1 \varphi (x,t)dt$ with which you have $||T||=\max_{x\in [0,1]} g(x)$.

How to search for the maximum of the function then calculate the derivative of the function. Here I tried to use the derivation theorem under the sign of integration, but I cannot "find" the expression $\int_0^1\varphi (1,t)dt$.

Can you help me?

Alex Pozo
  • 1,290

1 Answers1

1

Let $F(x)=\displaystyle\int_{0}^{1}\varphi(x,t)dt$. For each fixed $t\in[0,1]$, the function $x\rightarrow\varphi(x,t)$, $x\in[0,1]$ is increasing, this can be seen by the Mean Value Theorem that $\varphi(x,t)-\varphi(y,t)=\dfrac{\partial\varphi}{\partial x}(\eta_{x,y,t},t)(x-y)\geq 0\cdot(x-y)=0$ for $x\geq y$ and the assumption that $\dfrac{\partial\varphi}{\partial x}\geq 0$. So $F(x)\leq\displaystyle\int_{0}^{1}\varphi(1,t)dt=F(1)$. So $\max_{x\in[0,1]}F(x)=F(1)=\displaystyle\int_{0}^{1}\varphi(1,t)dt$.

user284331
  • 55,591