If the operator $T$ is defined by $Tf(x)=\int_0^xf(t)\,dt$, show that $Tf \in C[0,1]$

Question

Consider the operator $T$ on $L^2[0,1]$ defined by $Tf(x)=\displaystyle \int_0^xf(t)\,dt.$ Show that $Tf \in C[0,1].$

I have one question before this:

What are the implications between $L^p$ spaces, i.e if $f \in L^p$ does this imply that $f \in L^{p+1}$?

My attempt:

Assume $Tf$ is not in $C[0,1]$, so there exists $y \in [0,1]$ and $\epsilon>0$ such that for all $\delta >0$, we can find $x_0$, with $|x-x_0|<\delta$ but

$$\bigg\rVert\int_0^x f-\int_0^{x_0} f\,\bigg\rVert>\epsilon$$

WLOG assume $x>x_0$, so $$\bigg\lVert \int_{x_0}^x f\,\bigg\rVert \geq \epsilon.$$ So for $\delta_n=\frac{1}{n},$ we can find $x_n \in [0,1]$ such that $|x-x_0|<\frac{1}{n}$ and $$\bigg\lVert\int_{x_n}^xf\,\bigg\rVert \geq \epsilon$$ I don't know if that will lead me to a contradiction.

I would appreciate any help or hints with that.

Answer 1

Your first question is actually relevant here: the relevant inclusion is that a $L^p$ function on a finite measure space is in $L^r$ for all $r<p$. Thus a $L^2([0,1])$ function is also in $L^1([0,1])$. This is the main ingredient you need, along with the theorem that if $f \in L^1$ then for all $\varepsilon > 0$ there exists $\delta > 0$ such that if $\mu(A)<\delta$ then $\int_A |f| d \mu < \varepsilon$.

One way to prove this theorem more or less "by hand" is to use the bounded convergence theorem: given $g$ such that $\| f - g \|_{L^1} < \varepsilon/2$ and $|g| \leq M=M(\varepsilon)$, you have $\int_A |f| d \mu \leq \int_A |g| d \mu + \int_A |f-g| d \mu < M \mu(A) + \varepsilon/2$, so you can select $\delta=\varepsilon/(2M)$.

Another way to proceed is to use the dominated convergence theorem, by noting that $f(x) 1_{[0,x_n]}(x) \to f(x) 1_{[0,x_0]}(x)$ pointwise if $x_n \to x_0$.

Answer 2

Let $x_0 \in [0,1]$ and $x_n \to x_0$

Then $f_n:=f(x)1_{[0,x_n]}(x) \to f(x)1_{[0,x_0]}(x)$

Also $|f_n| \leq |f| \in L^2$

So by dominated convergence we have that $Tf(x_n) \to Tf(x)$

Answer 3

Since a bounded linear functional is continuous, it suffices to show that $Tf$ is bounded. Set $k(x, t) = \chi_{[0, x]}(t) $. We have:

$||Tf||_{L^{2}}^{2} = \langle Tf, Tf \rangle = \int_{0}^{1} \left[ \int_{0}^{1} k(x,t) f(t)dt\right]^{2}dx = \int_{0}^{1} \left[ \int_{0}^{1} |k(x,t)| |f(t)|dt\right]^{2}dx $

$\leq \int_{0}^{1} \left[ \left( \int_{0}^{1} |k(x,t)|^{2} dt\right)^{\frac{1}{2}} \cdot \left( \int_{0}^{1} |f(t)|^{2} dt \right)^{\frac{1}{2}}\right]^{2}dx = \int_{0}^{1} \int_{0}^{1} |k(x, t)|^{2} dt dx \cdot ||f||_{L^{2}}^{2} = \frac{1}{2} ||f||_{L^{2}}^{2}$.

We get $||Tf||_{L^{2}} \leq \sqrt{\frac{1}{2}} ||f||_{L^{2}}$ and we have $Tf$ is bounded and so it must be continuous.