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Let $X = Y = C([0,1], \mathbb{K})$ with $\|f \|_X = \| f \|_Y = \|f \|_1 = \int_{[0,1]} | f(t) |dt $ and let $(Tf)(x) = \int_0^x f(t) dt, \ x \in [0,1], \ f \in X$. Compute the norm of T.

I have managed to show that $\|T \| \leq 1$. So I wish to show that $\| T \| \geq 1$. But I dont know how to do that, so I took a look at the solutions, where they introduce $f_n (x) = (2n - 2n^2x) \mathbb{1}_{[0,1/n]} (x) $ with this function they manage to show that $1 - 1/n \leq \| Tf_n \|_1 \leq 1$.

How should I be able to "guess" such a function $f_n(x)$, to me it's seems that it's just coming from above or something, how could I come up with this function, like in a critical situation like an exam?

The question itself is from a previous exam, so I guess I should be able to "see" this function $f_n$ without much effort.

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  • $\begingroup$ This is very good question. $\endgroup$
    – Ran Wang
    Mar 2, 2017 at 13:16
  • $\begingroup$ By the way, what is $\mathbb{K}$? $\endgroup$
    – Ran Wang
    Mar 2, 2017 at 13:25
  • $\begingroup$ Does not say, but I think it would be ok to assume $\mathbb{R}, \mathbb{C}$. @RanWang $\endgroup$
    – Olba12
    Mar 2, 2017 at 13:34
  • $\begingroup$ Intuitively, you're looking for functions $g$ with $g(0) = 0$ for which the area under $g$ is large compared to the area under $g'$. $\endgroup$ Mar 2, 2017 at 13:52
  • $\begingroup$ Could you please elaborate? @Omnomnomnom $\endgroup$
    – Olba12
    Mar 2, 2017 at 14:03

2 Answers 2

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It is instructive to first consider in what manner the operator $T$ impacts the norm of an operator, to this end we compute, for arbitrary $f \in X$: \begin{align*} \|Tf\| &= \int_{0}^{1} \bigg|\int_{0}^{x} f(t) \text{d} t\bigg| \text{d} x \\ &= \int_{0}^{1} \bigg| \int_{0}^{1} f(t)\theta(t-x) \text{d}t \bigg| \text{d}x \\ &\leqslant \int_{0}^{1} \int_{0}^{1} |f(t)| \theta(t-x) \text{d}t \text{d}x \\ &= \|(1-t)f\|, \end{align*} where $\theta$ is the Heaviside theta function. Some steps manipulating the integrals above are left to the reader.

Note that the inequality is an equality, precisely when the function $f$ is either non-positive or non-negative.

Hence we see that, as far as the norm is concerned the operator $T$ just multiplies the function $f$ by the function $(1-t)$. From this, one may argue that the more a function is concentrated around the point $0$, the less the operator $T$ affects the norm of this function.

The sequence $f_{n}$ given in the answer may be seen as an attempt to find an element that is totally concentrated at $x = 0$. This is where the indicator $1_{[0,1/n]}$ comes in. The remainder of the functions $f_{n}$ are presumably a matter of experimentation/experience of the author.

Let me elaborate a bit on how one might come up with these functions. We are looking for a sequence of functions that are non-negative, and where each function has norm $1$, that is the area under the curve is equal to $1$. The support of the functions should approach the point $0$. One way one might think of doing this is by drawing triangles of constant area as depicted in the following figure.

Triangles of constant area with support approaching zero

One may verify that the functions $f_{n}$ precisely give the triangles as drawn in this picture.

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  • $\begingroup$ Great! It's worth mentioning that the trick you've "left to the reader" consists primarily of switching the order of integration. $\endgroup$ Mar 2, 2017 at 13:56
  • $\begingroup$ Shouldnt your first equality be a $\leq$? Since $\| Tf \| = \int_{[0,1]} |Tf| dy = \int_{[0,1]} \left| \int_0^x f(t) dt \right| dy \leq \int_{[0,1]} \int_0^x |f(t)| dt dy ... $ or this is included in the "manipulation" by reader? $\endgroup$
    – Olba12
    Mar 2, 2017 at 22:01
  • $\begingroup$ You are right, good catch! I have corrected my mistake and added some explanation. $\endgroup$
    – Peter
    Mar 3, 2017 at 8:51
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Build on the excellent answer by the previous answerer, let me complete the part why your teacher has chosen this function.

As has been shown, any sequence of continuous functions that converges to a point mass at $0$ will work. In this case, we want to approximate the point mass by function with the type $1_{[0, 1/n]}g_n(x)$. This functions needs to satisfy

  1. $g_n(0)$ are constants
  2. $g_n(1/n)=0$

An obvious choice is the linear function $g_n(x) = 1 - nx$. Now to simplify the computation, your teacher might want that $||f_n||_1 = 1$, however $\int_{0}^{1/n}g_n(x) = \frac{1}{2n}$, and we can scale $g_n$ by a factor of $2n$ so that the norm of $f_n$ is indeed $1$. Now it remains to show that this function indeed gives the identity.

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